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椭圆积分的勒让德正规化

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青青子衿 Posted at 2024-4-18 04:33:57 |Read mode
Last edited by 青青子衿 at 2024-9-23 17:39:00
\begin{gather*}

\int_{\beta}^{x}\frac{t}{\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau^{2})}}{\mathrm{d}}t\\
\\
\begin{split}
&=\frac{1}{2\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\alpha+\beta-\frac{(A-B)^{2}}{\alpha+\beta-2\sigma}}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\quad+\frac{1}{2\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{(A-B)^{2}}{\alpha+\beta-2\sigma}}{\left(1-\kappa_1^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad\quad+\frac{1}{2\sqrt{AB}}\int_{0}^{f}\frac{\beta-\alpha}{\left(1-\kappa_1^2t^{2}\right)\sqrt{1-k^2t^{2}}}{\mathrm{d}}t\\

&=\frac{1}{2\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\alpha+\beta-\frac{(A-B)^{2}}{\alpha+\beta-2\sigma}}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\quad+\frac{1}{2\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{(A-B)^{2}}{\alpha+\beta-2\sigma}}{\left(1-\kappa_1^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\qquad\quad+\operatorname{arctanh}\left(\frac{\scriptsize\frac{\beta-\alpha}{2\sqrt{AB}}f}{\sqrt{1-k^2f^{2}}}\right)\\

&=\frac{1}{2\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\alpha+\beta-\frac{(A-B)^{2}}{\alpha+\beta-2\sigma}}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\quad+\frac{1}{2\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{(A-B)^{2}}{\alpha+\beta-2\sigma}}{\left(1-\kappa_1^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\qquad\quad+\operatorname{arctanh}\Big(\sqrt{\scriptsize\frac{(x-\alpha)(x-\beta)}{(x-\sigma)^{2}+\tau^{2}}}\,\Big)\\
\end{split}\\
\\

\qquad\left\{\begin{split}
f&={\raise1.5px\scriptsize\tfrac{2\sqrt{AB(x-\alpha)(x-\beta)}}{(x-\beta)A+(x-\alpha)B}}\\

k^2&={\raise1.5px\scriptsize\tfrac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}}\\
&={\raise1.5px\scriptsize\frac{1}{4}\big(2+\tfrac{A^{2}+B^{2}-(\beta-\alpha)^{2}}{AB}\big)}\\

&={\raise1.5px\scriptsize\frac{1}{2}\Big(1+\tfrac{(\alpha-\sigma)(\beta-\sigma)+\tau^{2}}{\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}\Big)}\\

\kappa_1^2&={\raise1.5px\scriptsize\tfrac{(A+B)^{2}}{4AB}}\\

&={\raise1.5px\scriptsize\frac{1}{4}\big(2+\tfrac{A^{2}+B^{2}}{AB}\big)}\\

&={\raise1.5px\scriptsize\frac{1}{2}\Big(1+\tfrac{(\alpha -\sigma )^2+(\beta -\sigma )^2+2 \tau ^2}{\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}\Big)}\\

A&=\scriptsize{\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}\\
B&=\scriptsize{\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}
\end{split}\right.
\\

\end{gather*}


  1. \int_{\beta}^{x}\frac{t}{\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau_{0}^{2})}}dt
  2. \frac{1}{2\sqrt{AB}}\int_{0}^{f}\frac{\alpha+\beta-\frac{(A-B)^{2}}{\alpha+\beta-2\sigma}+\frac{(A-B)^{2}}{(\alpha+\beta-2\sigma)(1-\frac{(A+B)^{2}}{4AB}t^{2})}+\frac{(\beta-\alpha)\sqrt{1-t^{2}}}{1-\frac{(A+B)^{2}}{4AB}t^{2}}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt
  3. \frac{1}{2\sqrt{AB}}\int_{0}^{f}\frac{\alpha+\beta-\frac{(A-B)^{2}}{\alpha+\beta-2\sigma}+\frac{(A-B)^{2}}{(\alpha+\beta-2\sigma)(1-\frac{(A+B)^{2}}{4AB}t^{2})}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt+\operatorname{arctanh}\left(\frac{\frac{\beta-\alpha}{2\sqrt{AB}}f}{\sqrt{1-\frac{1}{4}\left(2+\frac{A^{2}+B^{2}-(\beta-\alpha)^{2}}{AB}\right)f^{2}}}\right)
  4. \frac{1}{2\sqrt{AB}}\int_{0}^{f}\frac{\alpha+\beta-\frac{(A-B)^{2}}{\alpha+\beta-2\sigma}+\frac{(A-B)^{2}}{(\alpha+\beta-2\sigma)(1-\frac{(A+B)^{2}}{4AB}t^{2})}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt+\operatorname{arctanh}\left(\sqrt{\frac{(x-\alpha)(x-\beta)}{(x-\sigma)^{2}+\tau_{0}^{2}}}\right)
  5. f=\frac{2\sqrt{AB(x-\alpha)(x-\beta)}}{(x-\beta)A+(x-\alpha)B}
  6. A=\sqrt{(\alpha-\sigma)^{2}+\tau_{0}^{2}}
  7. B=\sqrt{(\beta-\sigma)^{2}+\tau_{0}^{2}}
  8. \alpha=0.88
  9. \beta=1.33
  10. \sigma=4.36
  11. \tau_{0}=5.79
Copy the Code


(两实两虚)


\begin{gather*}
\int_{\beta}^{+\infty}\frac{1}{\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau^{2})}}{\mathrm{d}}t\\
\\
=\int_{0}^{\frac{2(\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}-\sqrt{(\beta-\sigma)^{2}+\tau^{2}})\left((\alpha-\sigma)^{2}+\tau^{2}\right)^{1/4}\left((\beta-\sigma)^{2}+\tau^{2}\right)^{1/4}}{(\beta-\alpha)(2\sigma-\alpha-\beta)}}\frac{\frac{1}{\left((\alpha-\sigma)^{2}+\tau^{2}\right)^{1/4}\left((\beta-\sigma)^{2}+\tau^{2}\right)^{1/4}}}{\sqrt{(1-t^{2})(1-\frac{1}{2}(\raise1.5px{\scriptsize1+\frac{(\alpha-\sigma)(\beta-\sigma)+\tau^{2}}{\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}})t^{2})}}{\mathrm{d}}t\\
\\
(\alpha<\beta)

\end{gather*}

\begin{align*}
&\quad\int_{\beta}^{x}\frac{1}{\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau^{2})}}{\mathrm{d}}t\\
\\
&=\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{1}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&=\frac{1}{\sqrt{AB}}\int_{0}^{
\sqrt{\scriptsize1-\bar{f}^{2}}
}\frac{1}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
\\
&\qquad\quad
\left\{\begin{split}
f&={\raise1.5px\scriptsize\tfrac{2\sqrt{AB(x-\alpha)(x-\beta)}}{(x-\beta)A+(x-\alpha)B}}\\
&={\raise1.5px\scriptsize\tfrac{2\sqrt{(x-\alpha)(x-\beta)\sqrt{((\alpha-\sigma)^{2}+\tau^{2})((\beta-\sigma)^{2}+\tau^{2})}}}
{(x-\beta)\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}+(x-\alpha)\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}}\\

\bar{f}&={\raise1.5px\scriptsize\tfrac{(x-\alpha)B-(x-\beta)A}{(x-\alpha)B+(x-\beta)A}}\\
&={\raise1.5px\scriptsize\tfrac{(x-\alpha)\sqrt{(\beta-\sigma)^{2}+\tau^{2}}-(x-\beta)\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}{(x-\alpha)\sqrt{(\beta-\sigma)^{2}+\tau^{2}}+(x-\beta)\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}}\\

k^2&={\raise1.5px\scriptsize\tfrac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}}\\

&={\raise1.5px\scriptsize\frac{1}{4}\big(2+\tfrac{A^{2}+B^{2}-(\beta-\alpha)^{2}}{AB}\big)}\\

&={\raise1.5px\scriptsize\frac{1}{2}\Big(1+\tfrac{(\alpha-\sigma)(\beta-\sigma)+\tau^{2}}{\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}\Big)}\\

A&=\scriptsize{\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}\\
B&=\scriptsize{\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}
\end{split}\right.\\
\\
&\qquad\qquad\qquad(\alpha<\beta<x)
\end{align*}




\begin{align*}
&\qquad\int_{x}^{+\infty}\frac{1}{\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau^{2})}}{\mathrm{d}}t\\
\\
&=\frac{1}{\sqrt{AB}}\int_{0}^{\mathscr{g}}\frac{1}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
\\
\\
&\quad
\left\{\begin{split}
{\mathscr{g}}&={\Tiny\tfrac{2((\alpha+\beta-2\sigma)x+\sigma^{2}+\tau^{2}-\alpha\beta) \sqrt{AB}}{((\alpha+\beta-2\sigma)x+\sigma^{2}+\tau^{2}-\alpha\beta-AB)\sqrt{(x-\alpha)(x-\beta)}
+((\alpha+\beta-2\sigma)x+\sigma^{2}+\tau^{2}-\alpha\beta+AB)\sqrt{(x-\sigma)^{2}+\tau^{2}}}}\\


k^2&={\raise1.5px\scriptsize\tfrac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}}\\

&={\raise1.5px\scriptsize\frac{1}{4}\big(2+\tfrac{A^{2}+B^{2}-(\beta-\alpha)^{2}}{AB}\big)}\\

&={\raise1.5px\scriptsize\frac{1}{2}\Big(1+\tfrac{(\alpha-\sigma)(\beta-\sigma)+\tau^{2}}{\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}\Big)}\\

A&=\scriptsize{\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}\\
B&=\scriptsize{\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}
\end{split}\right.\\
\\
&\qquad\qquad\qquad\quad(\alpha<\beta<x)
\end{align*}






  1. (2 Sqrt[Sqrt[((\[Alpha] - \[Sigma])^2 + \[Tau]^2) ((\[Beta] - \[Sigma])^2 +
  2.       \[Tau]^2)] (x - \[Alpha]) (x - \[Beta])])/(
  3. Sqrt[(\[Alpha] - \[Sigma])^2 + \[Tau]^2] (x - \[Beta]) +
  4.   Sqrt[(\[Beta] - \[Sigma])^2 + \[Tau]^2] (x - \[Alpha])) /. {\[Alpha] ->
  5.    1, \[Beta] -> 4, \[Sigma] -> 1, \[Tau] -> 3}
  6. (D[%, x]/Sqrt[Sqrt[((\[Alpha] - \[Sigma])^2 + \[Tau]^2) ((\[Beta] - \[Sigma])^2 + \[Tau]^2)]]/
  7.    Sqrt[(1 - (%)^2) (1 - (
  8.        1 + ((\[Sigma] - \[Alpha]) (\[Sigma] - \[Beta]) + \[Tau]^2)/
  9.         Sqrt[((\[Alpha] - \[Sigma])^2 + \[Tau]^2) ((\[Beta] - \[Sigma])^2 + \[Tau]^2)])/
  10.        2  (%)^2)])^2 /. {\[Alpha] -> 1, \[Beta] -> 4, \[Sigma] -> 1,
  11.    \[Tau] -> 3} // Factor
  12. 1/( (x^2 - (\[Alpha] + \[Beta]) x + \[Alpha]*\[Beta]) (\[Tau]^2+\[Sigma]^2 + 2 \[Sigma]*x +
  13.      x^2)) /. {\[Alpha] -> 1, \[Beta] -> 4, \[Sigma] -> 1, \[Tau] -> 3} // Factor
Copy the Code

  1. \int_{\beta}^{x}\frac{1}{\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau^{2})}}dt
  2. \int_{0}^{\frac{2\sqrt{(x-\alpha)(x-\beta)\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}}{(x-\beta)\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}+(x-\alpha)\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}}\frac{\frac{1}{\left((\sigma-\alpha)^{2}+\tau^{2}\right)^{1/4}\left((\sigma-\beta)^{2}+\tau^{2}\right)^{1/4}}}{\sqrt{(1-t^{2})(1-\frac{1}{2}\left(1+\frac{(\sigma-\alpha)(\sigma-\beta)+\tau^{2}}{\sqrt{\left((\sigma-\alpha)^{2}+\tau^{2}\right)\left((\sigma-\beta)^{2}+\tau^{2}\right)}}\right)t^{2})}}dt
  3. \int_{0}^{\frac{2(\sqrt{(\sigma-\alpha)^{2}+\tau^{2}}-\sqrt{(\sigma-\beta)^{2}+\tau^{2}})\left((\sigma-\alpha)^{2}+\tau^{2}\right)^{1/4}\left((\sigma-\beta)^{2}+\tau^{2}\right)^{1/4}}{(\beta-\alpha)(2\sigma-\alpha-\beta)}}\frac{\frac{1}{\left((\sigma-\alpha)^{2}+\tau^{2}\right)^{1/4}\left((\sigma-\beta)^{2}+\tau^{2}\right)^{1/4}}}{\sqrt{(1-t^{2})(1-\frac{1}{2}\left(1+\frac{(\sigma-\alpha)(\sigma-\beta)+\tau^{2}}{\sqrt{\left((\sigma-\alpha)^{2}+\tau^{2}\right)\left((\sigma-\beta)^{2}+\tau^{2}\right)}}\right)t^{2})}}dt-\int_{0}^{\frac{2\sqrt{(x-\alpha)(x-\beta)\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}}{(x-\beta)\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}+(x-\alpha)\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}}\frac{\frac{1}{\left((\sigma-\alpha)^{2}+\tau^{2}\right)^{1/4}\left((\sigma-\beta)^{2}+\tau^{2}\right)^{1/4}}}{\sqrt{(1-t^{2})(1-\frac{1}{2}\left(1+\frac{(\sigma-\alpha)(\sigma-\beta)+\tau^{2}}{\sqrt{\left((\sigma-\alpha)^{2}+\tau^{2}\right)\left((\sigma-\beta)^{2}+\tau^{2}\right)}}\right)t^{2})}}dt
  4. \int_{0}^{\frac{2(\sigma^{2}+\tau^{2}-\alpha\beta-x(2\sigma-\alpha-\beta))((\sigma-\alpha)^{2}+\tau^{2})^{1/4}((\sigma-\beta)^{2}+\tau^{2})^{1/4}}{\sqrt{(x-\alpha)(x-\beta)}\left(\sigma^{2}+\tau^{2}-\alpha\beta-\sqrt{((\sigma-\alpha)^{2}+\tau^{2})((\sigma-\beta)^{2}+\tau^{2})}-x(-\alpha+2\sigma-\beta)\right)+\sqrt{(\sigma-x)^{2}+\tau^{2}}\left(\sigma^{2}+\tau^{2}-\alpha\beta+\sqrt{((\sigma-\alpha)^{2}+\tau^{2})((\sigma-\beta)^{2}+\tau^{2})}-x(2\sigma-\alpha-\beta)\right)}}\frac{\frac{1}{\left((\sigma-\alpha)^{2}+\tau^{2}\right)^{1/4}\left((\sigma-\beta)^{2}+\tau^{2}\right)^{1/4}}}{\sqrt{(1-t^{2})(1-\frac{1}{2}\left(1+\frac{(\sigma-\alpha)(\sigma-\beta)+\tau^{2}}{\sqrt{\left((\sigma-\alpha)^{2}+\tau^{2}\right)\left((\sigma-\beta)^{2}+\tau^{2}\right)}}\right)t^{2})}}dt
  5. \frac{1}{\tau}\int_{\frac{1}{\sqrt{\frac{(\sigma-\alpha)(\sigma-\beta)+\tau^{2}+\sqrt{\left((\sigma-\alpha)^{2}+\tau^{2}\right)\left((\sigma-\beta)^{2}+\tau^{2}\right)}}{2\tau^{2}}}}}^{\frac{\sqrt{\frac{(x-\alpha)(x-\beta)}{(x-\sigma)^{2}+\tau^{2}}}}{\sqrt{\frac{(\sigma-\alpha)(\sigma-\beta)+\tau^{2}+\sqrt{\left((\sigma-\alpha)^{2}+\tau^{2}\right)\left((\sigma-\beta)^{2}+\tau^{2}\right)}}{2\tau^{2}}}}}\frac{\frac{1}{\sqrt{-\frac{(\sigma-\alpha)(\sigma-\beta)+\tau^{2}-\sqrt{\left((\sigma-\alpha)^{2}+\tau^{2}\right)\left((\sigma-\beta)^{2}+\tau^{2}\right)}}{2\tau^{2}}}}}{\sqrt{(1-t^{2})(1-\frac{\frac{(\sigma-\alpha)(\sigma-\beta)+\tau^{2}+\sqrt{\left((\sigma-\alpha)^{2}+\tau^{2}\right)\left((\sigma-\beta)^{2}+\tau^{2}\right)}}{2\tau^{2}}}{\frac{(\sigma-\alpha)(\sigma-\beta)+\tau^{2}-\sqrt{\left((\sigma-\alpha)^{2}+\tau^{2}\right)\left((\sigma-\beta)^{2}+\tau^{2}\right)}}{2\tau^{2}}}t^{2})}}dt
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 Author| 青青子衿 Posted at 2024-4-18 14:27:31
Last edited by 青青子衿 at 2024-11-25 20:10:00
\begin{gather*}
\int_{-\left(1+\sqrt{3}-\frac{3^{1/4}+3^{3/4}}{\sqrt{2}}\right)}^{x}\frac{t}{\sqrt{1+4t-6t^{2}+4t^{3}+t^{4}}}{\mathrm{d}}t\\
\\
=\int_{0}^{\frac{(\sqrt{3}-1)\sqrt{1+2(1+\sqrt{3})x+x^{2}}}{1+x}}\left(\tfrac{\frac{\sqrt{2}+\sqrt{6}}{4}}{\left(1-\frac{2+\sqrt{3}}{2}t^{2}\right)\sqrt{1-\frac{2+\sqrt{3}}{4}t^{2}}}
-\tfrac{1+\sqrt{3}}{2\cdot3^{1/4}}\left(\tfrac{1-\frac{1+\sqrt{3}}{4}t^{2}}{\left(1-\frac{2+\sqrt{3}}{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-\frac{2+\sqrt{3}}{4}t^{2}\right)}}\right)\right){\mathrm{d}}t
\end{gather*}


  1. \int_{-\left(1+\sqrt{3}-\frac{3^{1/4}+3^{3/4}}{\sqrt{2}}\right)}^{x}\frac{t}{\sqrt{1+4t-6t^{2}+4t^{3}+t^{4}}}dt
  2. \int_{0}^{\frac{(\sqrt{3}-1)\sqrt{1+2(1+\sqrt{3})x+x^{2}}}{1+x}}\left(\frac{\frac{\sqrt{2}+\sqrt{6}}{4}}{\left(1-\frac{2+\sqrt{3}}{2}t^{2}\right)\sqrt{1-\frac{2+\sqrt{3}}{4}t^{2}}}-\frac{1+\sqrt{3}}{2\cdot3^{1/4}}\left(\frac{1-\frac{1+\sqrt{3}}{4}t^{2}}{\left(1-\frac{2+\sqrt{3}}{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-\frac{2+\sqrt{3}}{4}t^{2}\right)}}\right)\right)dt
Copy the Code



\begin{gather*}
{\Large\int}_0^{x}\frac{1-{\raise1px\small\frac{1+\sqrt{3}}{4}}t^{2}}{\left(1-\!{\raise1.5px\scriptsize\color{blue}{\frac{2+\sqrt{3}}{2}}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-\!{\raise2px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}{\mathrm{d}}t\\
=\tfrac{\sqrt{3}-1}{3^{3/4}}\operatorname{arctanh}\left(\tfrac{\frac{3^{3/4}(1+\sqrt{3})}{2}x}{1+\frac{5+3\sqrt{3}}{4}x^{2}}\sqrt{\tfrac{1-\frac{2+\sqrt{3}}{4}x^{2}}{1-x^{2}}}\,\right)
\end{gather*}

\begin{align*}
\frac{\displaystyle\int_{0}^{\scriptsize\frac{1}{\sqrt{\color{blue}{\frac{2+\sqrt{3}}{2}}}}}\frac{{\mathrm{d}}t}{\scriptsize\sqrt{\left(1-t^{2}\right)(1-\frac{2+\sqrt{3}}{4}t^{2})}}}{\displaystyle{\int}_{0}^{1}\frac{{\mathrm{d}}t}{\scriptsize\sqrt{\left(1-t^{2}\right)(1-{\raise1px\Tiny\frac{2+\sqrt{3}}{4}}t^{2})}}}=\dfrac{1}{3}
\end{align*}

\begin{align*}
\tfrac{1}{6}\ln({\scriptsize\,x^6+12 x^5+45 x^4+44 x^3-33 x^2+43+\left(x^4+10 x^3+30 x^2+22 x-11\right) \sqrt{x^4+4 x^3-6 x^2+4 x+1}}\,)
\end{align*}


\begin{gather*}
\int_{0}^{x}\tfrac{{\mathrm{d}}t}{\left(1-{\raise2px\scriptsize\frac{2+\sqrt{3}}{2}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-{\raise2px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}
\\
=\tfrac{2}{\,\,3^{1/4}}\operatorname{artanh}\left(\tfrac{{\raise1px\scriptsize\frac{3^{1/4}}{2}}\>\!x}{\sqrt{\left(1-x^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}x^{2}\right)}}\right)-\int_{0}^{x}\tfrac{{\raise1px\scriptsize\frac{1}{2}}t^{2}}{\left(1-{\raise1px\scriptsize\frac{1}{2}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}{\raise2px\scriptsize{\mathrm{d}}t}

\end{gather*}





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 Author| 青青子衿 Posted at 2024-4-22 03:47:21

\begin{align*}
&\quad\>\>\>\int_0^{x}\tfrac{1-{\raise1px\scriptsize\frac{1+\sqrt{3}}{4}}t^{2}}{\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{2}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}{\raise2px\scriptsize{\mathrm{d}}t}\\

&=\int_0^{x}\tfrac{1-{\raise1px\scriptsize\frac{\sqrt{3}-1}{2}}+{\raise1px\scriptsize\frac{\sqrt{3}-1}{2}}-{\raise1px\scriptsize\frac{\sqrt{3}-1}{2}}\cdot{\raise1px\scriptsize\frac{2+\sqrt{3}}{2}}t^{2}}{\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{2}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}{\raise2px\scriptsize{\mathrm{d}}t}\\

&=\int_0^{x}\tfrac{{\raise1px\scriptsize\frac{3-\sqrt{3}}{2}}+{\raise1px\scriptsize\frac{\sqrt{3}-1}{2}}(1-{\raise1px\small\frac{2+\sqrt{3}}{2}}t^{2})}{\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{2}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}{\raise2px\scriptsize{\mathrm{d}}t}\\

&=\int_0^{x}\tfrac{{\raise1px\small\frac{3-\sqrt{3}}{2}}}{\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{2}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}{\raise2px\scriptsize{\mathrm{d}}t}\\
&\qquad
+\int_0^{x}\tfrac{{\raise1px\small\frac{\sqrt{3}-1}{2}}}{\sqrt{\left(1-t^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}{\raise2px\scriptsize{\mathrm{d}}t}\\


&=\tfrac{3-\sqrt{3}}{\,\,3^{1/4}}\operatorname{artanh}\left(\tfrac{{\raise1px\scriptsize\frac{3^{1/4}}{2}}\>\!x}{\sqrt{\left(1-x^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}x^{2}\right)}}\right)\\
&\qquad
-\int_{0}^{x}\tfrac{{\raise1px\small\frac{3-\sqrt{3}}{4}}\,t^{2}-{\raise1px\small\frac{\sqrt{3}-1}{2}}\left(1-{\raise1px\scriptsize\frac{1}{2}}t^{2}\right)}{\left(1-{\raise1px\scriptsize\frac{1}{2}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}{\raise2px\scriptsize{\mathrm{d}}t}\\
&=\tfrac{3-\sqrt{3}}{\,\,3^{1/4}}\operatorname{artanh}\left(\tfrac{{\raise1px\scriptsize\frac{3^{1/4}}{2}}\>\!x}{\sqrt{\left(1-x^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}x^{2}\right)}}\right)\\
&\qquad
+\int_{0}^{x}\tfrac{{\raise1px\small\frac{\sqrt{3}-1}{2}}-{\raise1px\small\frac{1}{2}}t^{2}}{\left(1-{\raise1px\scriptsize\frac{1}{2}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}{\raise2px\scriptsize{\mathrm{d}}t}\\
&=\tfrac{3-\sqrt{3}}{\,\,3^{1/4}}\operatorname{artanh}\left(\tfrac{{\raise1px\scriptsize\frac{3^{1/4}}{2}}\>\!x}{\sqrt{\left(1-x^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}x^{2}\right)}}\right)\\
&\qquad
+{\tfrac{\sqrt{3}-1}{2}}\int_{0}^{x}\tfrac{1-{\raise1px\small\frac{1+\sqrt{3}}{2}}t^{2}}{\left(1-{\raise1px\scriptsize\frac{1}{2}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}t^{2}\right)}}{\raise2px\scriptsize{\mathrm{d}}t}\\
&=\tfrac{3(\sqrt{3}-1)}{\,\,3^{3/4}}\operatorname{artanh}\left(\tfrac{{\raise1px\scriptsize\frac{3^{1/4}}{2}}\>\!x}{\sqrt{\left(1-x^{2}\right)\left(1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}x^{2}\right)}}\right)\\
&\qquad
+\tfrac{\sqrt{3}-1}{3^{3/4}}\operatorname{artanh}\left(\tfrac{
{\raise0.8px\small\frac{3^{3/4}}{2}}x\sqrt{(1-x^{2})(1-{\raise0.8px\small\frac{2+\sqrt{3}}{4}}x^{2})}}
{1-{\raise0.8px\small\frac{4-\sqrt{3}}{4}}x^{2}-{\raise0.8px\small\frac{1+\sqrt{3}}{8}}x^{4}}\right)\\
&=\tfrac{\sqrt{3}-1}{3^{3/4}}\operatorname{artanh}\left(\tfrac{{\raise1px\scriptsize\frac{3^{3/4}(1+\sqrt{3})}{2}}x}{1+{\raise1px\scriptsize\frac{5+3\sqrt{3}}{4}}x^{2}}\sqrt{\tfrac{1-{\raise1px\scriptsize\frac{2+\sqrt{3}}{4}}x^{2}}{1-x^{2}}}\,\right)
\end{align*}



  1. \int_{0}^{x}\frac{1-\frac{1+\sqrt{3}}{4}t^{2}}{\left(1-\frac{2+\sqrt{3}}{2}t^{2}\right)\sqrt{(1-t^{2})(1-\frac{2+\sqrt{3}}{4}t^{2})}}dt
  2. \int_{0}^{x}\frac{\frac{3-\sqrt{3}}{2}}{\left(1-\frac{2+\sqrt{3}}{2}t^{2}\right)\sqrt{(1-t^{2})(1-\frac{2+\sqrt{3}}{4}t^{2})}}dt+\int_{0}^{x}\frac{\frac{\sqrt{3}-1}{2}}{\sqrt{(1-t^{2})(1-\frac{2+\sqrt{3}}{4}t^{2})}}dt
  3. \frac{3-\sqrt{3}}{3^{1/4}}\operatorname{artanh}\left(\frac{\frac{3^{1/4}}{2}x}{\sqrt{(1-x^{2})(1-\frac{2+\sqrt{3}}{4}x^{2})}}\right)-\int_{0}^{x}\frac{\frac{3-\sqrt{3}}{4}t^{2}-\frac{\sqrt{3}-1}{2}\left(1-\frac{1}{2}t^{2}\right)}{\left(1-\frac{1}{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-\frac{2+\sqrt{3}}{4}t^{2}\right)}}dt
  4. \frac{3-\sqrt{3}}{3^{1/4}}\operatorname{artanh}\left(\frac{\frac{3^{1/4}}{2}x}{\sqrt{(1-x^{2})(1-\frac{2+\sqrt{3}}{4}x^{2})}}\right)+\frac{\sqrt{3}-1}{2}\int_{0}^{x}\frac{1-\frac{1+\sqrt{3}}{2}t^{2}}{\left(1-\frac{1}{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-\frac{2+\sqrt{3}}{4}t^{2}\right)}}dt
  5. 3\left(\frac{\sqrt{3}-1}{3^{3/4}}\right)\operatorname{artanh}\left(\frac{\frac{3^{1/4}}{2}x}{\sqrt{(1-x^{2})(1-\frac{2+\sqrt{3}}{4}x^{2})}}\right)+\frac{\sqrt{3}-1}{3^{3/4}}\operatorname{artanh}\left(\frac{\frac{3^{3/4}}{2}x\sqrt{\left(1-x^{2}\right)\left(1-\frac{2+\sqrt{3}}{4}x^{2}\right)}}{1-\frac{4-\sqrt{3}}{4}x^{2}-\frac{1+\sqrt{3}}{8}x^{4}}\right)
  6. \frac{\sqrt{3}-1}{3^{3/4}}\operatorname{artanh}\left(\frac{\frac{3^{3/4}(1+\sqrt{3})}{2}x}{1+\frac{5+3\sqrt{3}}{4}x^{2}}\sqrt{\frac{1-\frac{2+\sqrt{3}}{4}x^{2}}{1-x^{2}}}\right)
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 Author| 青青子衿 Posted at 2024-5-10 05:27:21
Last edited by 青青子衿 at 2024-11-20 10:01:00
\begin{gather*}
\int_{-\frac{a^{2}+b^{2}-c^{2}-d^{2}+AB}{2(c-a)}}^{x}\frac{t}{\sqrt{((t-a)^{2}+b^{2})((t-c)^{2}+d^{2})}}{\mathrm{d}}t\\

=
\frac{A-B}{2bd}\int_{0}^{\frac{\sqrt{\frac{1}{2}(1+\frac{(a-c)^{2}-b^{2}+d^{2}}{AB})}\left(x+\frac{a^{2}+b^{2}-c^{2}-d^{2}+AB}{2(c-a)}\right)}{\sqrt{(x-a)^{2}+b^{2}}}}\frac{a-\frac{\frac{AB-(a-c)^{2}+b^{2}-d^{2}}{2(c-a)}}{1-\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b^{2}(a-c)^{2}}t^{2}}+\frac{\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b(a-c)^{2}}t\sqrt{1-t^{2}}}{1-\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b^{2}(a-c)^{2}}t^{2}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4b^{2}d^{2}}t^{2})}}{\mathrm{d}}t\\
\\
\left\{
\begin{split}
A&=\sqrt{(a-c)^2+(b+d)^2}\\
B&=\sqrt{(a-c)^2+(b-d)^2}
\end{split}\right.
\end{gather*}


  1. NIntegrate[
  2. 1/Sqrt[(1 - t^2) (1 - k^2 t^2)] /. k -> 1/Sqrt[3], {t, 0,
  3.   a + b*I /. {k -> 1/Sqrt[3], a -> 1/7, b -> 1/3}}]
  4. NIntegrate[
  5. 1/Sqrt[(1 - t^2) (1 - k^2 t^2)] /. k -> 1/Sqrt[3], {t, 0,
  6.   Sqrt[((1 + (a^2 + b^2)) -
  7.       Sqrt[((1 - a)^2 + b^2) ((1 + a)^2 + b^2) ]) ( (1 +
  8.         k^2 (a^2 + b^2) ) -
  9.       Sqrt[(b^2 k^2 + (1 - a k)^2) (b^2 k^2 + (1 + a k)^2)])]/(
  10.    2 a*k) /. {k -> 1/Sqrt[3], a -> 1/7, b -> 1/3}}]
  11. 1/2 NIntegrate[
  12.   1/Sqrt[(1 - t^2) (1 - k^2 t^2)] /. k -> 1/Sqrt[3], {t,
  13.    0, (a*Sqrt[2 Sqrt[P ] + 2 Q] - b*Sqrt[2 Sqrt[P ] - 2 Q])/(
  14.      1 - (a^2 +
  15.          b^2)^2 k^2) /. {P -> ((1 - a)^2 + b^2) ((1 + a)^2 +
  16.           b^2) ((1 - k*a)^2 + b^2 k^2) ((1 + k*a)^2 + b^2 k^2),
  17.       Q -> (1 - a^2 + b^2) (1 - a^2 k^2 + b^2 k^2) -
  18.         4 a^2 b^2 k^2} /. {k -> 1/Sqrt[3], a -> 1/7, b -> 1/3}}]
Copy the Code



\begin{gather*}
\int_{-\frac{a^{2}+b^{2}-c^{2}-d^{2}+AB}{2(c-a)}}^{x}\frac{t}{\sqrt{((t-a)^{2}+b^{2})((t-c)^{2}+d^{2})}}{\mathrm{d}}t\\

=
\frac{A-B}{2bd}\int_{0}^{f}\frac{a}{\sqrt{(1-t^{2})(1-{\raise1px\scriptsize\frac{AB(A-B)^{2}}{4b^{2}d^{2}}}t^{2})}}{\mathrm{d}}t\\

-\frac{A-B}{2bd}\int_{0}^{f}\frac{\frac{AB-(a-c)^{2}+b^{2}-d^{2}}{2(c-a)}}{\left(1-{\raise1px\scriptsize\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b^{2}(a-c)^{2}}}t^{2}\right)\sqrt{(1-t^{2})(1-{\raise1px\scriptsize\frac{AB(A-B)^{2}}{4b^{2}d^{2}}}t^{2})}}{\mathrm{d}}t\\
+\frac{A-B}{2bd}\int_{0}^{f}\frac{\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b(a-c)^{2}}t}{\left(1-{\raise1px\scriptsize\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b^{2}(a-c)^{2}}}t^{2}\right)\sqrt{1-{\raise1px\scriptsize\frac{AB(A-B)^{2}}{4b^{2}d^{2}}}t^{2}}}{\mathrm{d}}t\\
\\
\\
f=\tfrac{\sqrt{\frac{1}{2}(1+\frac{(a-c)^{2}-b^{2}+d^{2}}{AB})}\left(x+\frac{a^{2}+b^{2}-c^{2}-d^{2}+AB}{2(c-a)}\right)}{\sqrt{(x-a)^{2}+b^{2}}}\\
\left\{
\begin{split}
A&=\scriptsize{\sqrt{(\lambda-\sigma)^2+(\mu+\tau)^2}}\\
B&=\scriptsize{\sqrt{(\lambda-\sigma)^2+(\mu-\tau)^2}}
\end{split}\right.
\end{gather*}

\begin{gather*}
\int_{-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}+AB}{2(\sigma-\lambda)}}^{x}\frac{t}{\sqrt{((t-\lambda)^{2}+\mu^{2})((t-\sigma)^{2}+\tau^{2})}}{\mathrm{d}t}\\
\\
\begin{split}
&=\frac{A-B}{2\mu\tau}\int_{0}^{f}\frac{\lambda}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\quad-\frac{A-B}{2\mu\tau}\int_{0}^{f}\frac{\scriptsize\frac{AB-(\lambda-\sigma)^{2}+\mu^{2}-\tau^{2}}{2(\sigma-\lambda)}}{\left(1-\kappa_{3}^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad\quad+\frac{A-B}{2\mu\tau}\int_{0}^{f}\frac{\scriptsize\frac{AB(AB-(\lambda-\sigma)^{2}+\mu^{2}-\tau^{2})}{2\mu(\lambda-\sigma)^{2}}t}{\left(1-\kappa_3^2t^{2}\right)\sqrt{1-k^2t^{2}}}{\mathrm{d}}t\\

&=\frac{A-B}{2\mu\tau}\int_{0}^{f}\frac{\lambda}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\quad-\frac{A-B}{2\mu\tau}\int_{0}^{f}\frac{\scriptsize\frac{AB-(\lambda-\sigma)^{2}+\mu^{2}-\tau^{2}}{2(\sigma-\lambda)}}{\left(1-\kappa_{3}^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad\quad+\operatorname{arctanh}\left({\raise{1.5px}\scriptsize\frac{A-B}{2\tau\sqrt{1-k^2f^{2}}}}\right)-\operatorname{arctanh}\left({\raise{1.5px}\scriptsize\frac{A-B}{2\tau}}\right)\\

&=\frac{A-B}{2\mu\tau}\int_{0}^{f}\frac{\lambda}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\quad-\frac{A-B}{2\mu\tau}\int_{0}^{f}\frac{\scriptsize\frac{AB-(\lambda-\sigma)^{2}+\mu^{2}-\tau^{2}}{2(\sigma-\lambda)}}{\left(1-\kappa_{3}^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad\quad+\operatorname{arctanh}\bigg(\sqrt{\scriptsize\frac{(x-\lambda)^{2}+\mu^{2}}{(x-\sigma)^{2}+\tau_{0}^{2}}}\,\bigg)-\operatorname{arctanh}\bigg({\scriptsize\frac{A-B}{2\tau_{0}}}\bigg)

\end{split}\\
\\

\qquad\left\{\begin{split}
f&={\raise1.5px\scriptsize\frac{\sqrt{\frac{(A+B)^{2}-4\mu^{2}}{AB}}\left(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}\right)}{2\sqrt{(x-\lambda)^{2}+\mu^{2}}}}\\

k^2&={\raise1.5px\scriptsize\tfrac{AB(A-B)^{2}}{4\mu^{2}\tau^{2}}}\\


\kappa_3^2&={\raise1px\scriptsize\tfrac{AB(AB-(\lambda-\sigma)^{2}+\mu^{2}-\tau^{2})}{2\mu^{2}(\lambda-\sigma)^{2}}}\\


A&=\scriptsize{\sqrt{(\lambda-\sigma)^{2}+(\mu+\tau)^{2}}}\\
B&=\scriptsize{\sqrt{(\lambda-\sigma)^{2}+(\mu-\tau)^{2}}}
\end{split}\right.
\\

\end{gather*}



  1. \int_{-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}}^{x}\frac{t}{\sqrt{((t-\lambda)^{2}+\mu^{2})((t-\sigma)^{2}+\tau_{0}^{2})}}dt
  2. \frac{A-B}{2\mu\tau_{0}}\int_{0}^{f}\frac{\lambda}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt-\frac{A-B}{2\mu\tau_{0}}\int_{0}^{f}\frac{\frac{AB-(\lambda-\sigma)^{2}+\mu^{2}-\tau_{0}^{2}}{2(\sigma-\lambda)}}{\left(1-\frac{AB(AB-(\lambda-\sigma)^{2}+\mu^{2}-\tau_{0}^{2})}{2\mu^{2}(\lambda-\sigma)^{2}}t^{2}\right)\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt+\frac{A-B}{2\mu\tau_{0}}\int_{0}^{f}\frac{\frac{AB(AB-(\lambda-\sigma)^{2}+\mu^{2}-\tau_{0}^{2})}{2\mu(\lambda-\sigma)^{2}}t}{\left(1-\frac{AB(AB-(\lambda-\sigma)^{2}+\mu^{2}-\tau_{0}^{2})}{2\mu^{2}(\lambda-\sigma)^{2}}t^{2}\right)\sqrt{1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2}}}dt
  3. f=\frac{\sqrt{\frac{1}{2}(1+\frac{(\lambda-\sigma)^{2}-\mu^{2}+\tau_{0}^{2}}{AB})}\left(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}\right)}{\sqrt{(x-\lambda)^{2}+\mu^{2}}}
  4. A=\sqrt{(\lambda-\sigma)^{2}+(\mu+\tau_{0})^{2}}
  5. B=\sqrt{(\lambda-\sigma)^{2}+(\mu-\tau_{0})^{2}}
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  1. \int_{-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}}^{x}\frac{t}{\sqrt{((t-\lambda)^{2}+\mu^{2})((t-\sigma)^{2}+\tau_{0}^{2})}}dt
  2. \frac{A-B}{2\mu\tau_{0}}\left(-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}\right)\int_{0}^{f}\frac{\frac{1+\frac{(\lambda^{2}+\mu^{2})^{2}-2\lambda\sigma(\lambda^{2}+\mu^{2})+(\lambda^{2}-\mu^{2})(\sigma^{2}+\tau_{0}^{2})-(\lambda^{2}+\mu^{2})AB}{2\mu(\lambda^{2}\sigma+\mu^{2}\sigma-\lambda\sigma^{2}-\lambda\tau_{0}^{2})}\frac{t}{\sqrt{1-t^{2}}}}{1+\frac{(\lambda-\sigma)^{2}-\mu^{2}+\tau_{0}^{2}-AB}{2\mu(\lambda-\sigma)}\frac{t}{\sqrt{1-t^{2}}}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt
  3. 10\int_{-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}}^{x}\frac{1}{t\sqrt{((t-\lambda)^{2}+\mu^{2})((t-\sigma)^{2}+\tau_{0}^{2})}}dt
  4. 10\frac{\frac{A-B}{2\mu\tau_{0}}}{-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}}\int_{0}^{f}\frac{\frac{1+\frac{(\lambda-\sigma)^{2}-\mu^{2}+\tau_{0}^{2}-AB}{2\mu(\lambda-\sigma)}\frac{t}{\sqrt{1-t^{2}}}}{1+\frac{(\lambda^{2}+\mu^{2})^{2}-2\lambda\sigma(\lambda^{2}+\mu^{2})+(\lambda^{2}-\mu^{2})(\sigma^{2}+\tau_{0}^{2})-(\lambda^{2}+\mu^{2})AB}{2\mu(\lambda^{2}\sigma+\mu^{2}\sigma-\lambda\sigma^{2}-\lambda\tau_{0}^{2})}\frac{t}{\sqrt{1-t^{2}}}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt
  5. A=\sqrt{(\lambda-\sigma)^{2}+(\mu+\tau_{0})^{2}}
  6. B=\sqrt{(\lambda-\sigma)^{2}+(\mu-\tau_{0})^{2}}
  7. f=\frac{\sqrt{\frac{1}{2}(1+\frac{(\lambda-\sigma)^{2}-\mu^{2}+\tau_{0}^{2}}{AB})}\left(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}\right)}{\sqrt{(x-\lambda)^{2}+\mu^{2}}}
  8. 15\frac{\frac{A-B}{2\mu\tau_{0}}}{-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}}\int_{0}^{f}\frac{\frac{1+PQ}{1+Q^{2}}-\frac{\frac{Q(P-Q)}{1+Q^{2}}}{1-(1+Q^{2})t^{2}}+\frac{(P-Q)t\sqrt{1-t^{2}}}{1-(1+Q^{2})t^{2}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt
  9. 15\frac{A-B}{2\mu\tau_{0}(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\lambda+\frac{\mu Q}{1-(1+Q^{2})t^{2}}-\frac{\mu(1+Q^{2})t\sqrt{1-t^{2}}}{1-(1+Q^{2})t^{2}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt
  10. P=\frac{(\lambda-\sigma)^{2}-\mu^{2}+\tau_{0}^{2}-AB}{2\mu(\lambda-\sigma)}
  11. Q=\frac{(\lambda^{2}+\mu^{2})^{2}-2\lambda\sigma(\lambda^{2}+\mu^{2})+(\lambda^{2}-\mu^{2})(\sigma^{2}+\tau_{0}^{2})-(\lambda^{2}+\mu^{2})AB}{2\mu(\lambda^{2}\sigma+\mu^{2}\sigma-\lambda\sigma^{2}-\lambda\tau_{0}^{2})}
Copy the Code





\begin{align*}
{\Large\int }
\frac{x-\frac{1}{44}(7+2\sqrt{2})}{\sqrt{x^{4}-\frac{1}{2}(3+\sqrt{2})x^{3}+\frac{1}{16}(11+4\sqrt{2})x^{2}+\frac{1}{32}(2-5\sqrt{2})x+\frac{1}{128}}}
{\large\mathrm{d}x}
\end{align*}


\begin{align*}
{\large\int}\frac{t(5-t^3)}{\left(3+t^6\right) \sqrt{1+4 t^3}}\mathrm{d}t
\end{align*}





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 Author| 青青子衿 Posted at 2024-7-15 03:58:30
Last edited by 青青子衿 at 2024-7-15 15:10:00
青青子衿 发表于 2024-5-10 05:27
\begin{gather*}
\int_{-\frac{a^{2}+b^{2}-c^{2}-d^{2}+AB}{2(c-a)}}^{x}\frac{t}{\sqrt{((t-a)^{2}+b^{2})((t-c)^{2}+d^{2})}}{\mathrm{d}}t\\

=
\frac{A-B}{2bd}\int_{0}^{f}\frac{a}{\sqrt{(1-t^{2})(1-{\raise1px\scriptsize\frac{AB(A-B)^{2}}{4b^{2}d^{2}}}t^{2})}}{\mathrm{d}}t\\

-\frac{A-B}{2bd}\int_{0}^{f}\frac{\frac{AB-(a-c)^{2}+b^{2}-d^{2}}{2(c-a)}}{\left(1-{\raise1px\scriptsize\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b^{2}(a-c)^{2}}}t^{2}\right)\sqrt{(1-t^{2})(1-{\raise1px\scriptsize\frac{AB(A-B)^{2}}{4b^{2}d^{2}}}t^{2})}}{\mathrm{d}}t\\
+\frac{A-B}{2bd}\int_{0}^{f}\frac{\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b(a-c)^{2}}t}{\left(1-{\raise1px\scriptsize\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b^{2}(a-c)^{2}}}t^{2}\right)\sqrt{1-{\raise1px\scriptsize\frac{AB(A-B)^{2}}{4b^{2}d^{2}}}t^{2}}}{\mathrm{d}}t\\
\\
\\
f=\frac{\sqrt{\frac{1}{2}(1+\frac{(a-c)^{2}-b^{2}+d^{2}}{AB})}\left(x+\frac{a^{2}+b^{2}-c^{2}-d^{2}+AB}{2(c-a)}\right)}{\sqrt{(x-a)^{2}+b^{2}}}\\
\left\{
\begin{split}
A&=\sqrt{(a-c)^2+(b+d)^2}\\
B&=\sqrt{(a-c)^2+(b-d)^2}
\end{split}\right.
\end{gather*}

\begin{align*}
\int_{0}^{\frac{5+2i}{4}}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-(\frac{\sqrt{10}-\sqrt{2}+2\sqrt{\sqrt{5}-1}}{4})^{2}t^{2})}}-\int_{0}^{\frac{5-2i}{4}}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-(\frac{\sqrt{10}-\sqrt{2}+2\sqrt{\sqrt{5}-1}}{4})^{2}t^{2})}}

\end{align*}


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 Author| 青青子衿 Posted at 2024-9-22 02:00:45
Last edited by 青青子衿 at 2024-9-26 16:01:00
青青子衿 发表于 2024-5-10 05:27
\begin{gather*}
\int_{-\frac{a^{2}+b^{2}-c^{2}-d^{2}+AB}{2(c-a)}}^{x}\frac{t}{\sqrt{((t-a)^{2}+b^{2})((t-c)^{2}+d^{2})}}{\mathrm{d}}t\\

=
\frac{A-B}{2bd}\int_{0}^{\frac{\sqrt{\frac{1}{2}(1+\frac{(a-c)^{2}-b^{2}+d^{2}}{AB})}\left(x+\frac{a^{2}+b^{2}-c^{2}-d^{2}+AB}{2(c-a)}\right)}{\sqrt{(x-a)^{2}+b^{2}}}}\frac{a-\frac{\frac{AB-(a-c)^{2}+b^{2}-d^{2}}{2(c-a)}}{1-\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b^{2}(a-c)^{2}}t^{2}}+\frac{\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b(a-c)^{2}}t\sqrt{1-t^{2}}}{1-\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b^{2}(a-c)^{2}}t^{2}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4b^{2}d^{2}}t^{2})}}{\mathrm{d}}t\\
\\
\left\{
\begin{split}
A&=\sqrt{(a-c)^2+(b+d)^2}\\
B&=\sqrt{(a-c)^2+(b-d)^2}
\end{split}\right.
\end{gather*}

\begin{align*}
&\qquad\qquad\qquad\qquad\qquad\int_{x}^{+\infty}\frac{{\mathrm{d}}t}{t\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}\\

&=-\int_{0}^{\mathscr{g}}\frac{\frac{2(\delta-\gamma)}{\gamma\delta\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{(1-\frac{\gamma(\delta-\alpha)}{\delta(\gamma-\alpha)}t^{2})\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}{\mathrm{d}}t
+\int_{0}^{\mathscr{g}}\frac{\frac{2}{\gamma\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad-\frac{2}{\sqrt{\alpha\beta\gamma\delta}}\operatorname{arctanh}\left(\frac{\sqrt{\alpha\beta\gamma\delta}(x-\delta)}{\delta(x^{2}-\delta\>\!x+\sqrt{(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)})}\right)\\
\\
\\
&\qquad\qquad\qquad\qquad\qquad\int_{x}^{+\infty}\frac{{\mathrm{d}}t}{\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}\\

&\qquad\qquad\qquad\qquad\qquad=\int_{0}^{\mathscr{g}}\frac{\frac{2}{\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}{\mathrm{d}}t\\
\\
\\
&\qquad\qquad\qquad\qquad\quad\>{\mathscr{g}}=\tfrac{\sqrt{(\delta-\beta)(\gamma-\alpha)(x-\alpha)(x-\beta)}-\sqrt{(\delta-\beta)(\gamma-\alpha)(x-\gamma)(x-\delta)}}{(\delta+\gamma-\beta-\alpha)x+\alpha\beta-\gamma\delta}\\


\end{align*}


  1. \int_{\delta}^{x}\frac{1}{t\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}dt
  2. \int_{0}^{\sqrt{\frac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}}\frac{\frac{2}{\gamma\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}dt-\int_{0}^{\sqrt{\frac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}}\frac{\frac{2(\delta-\gamma)}{\gamma\delta\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{(1-\frac{\gamma(\delta-\alpha)}{\delta(\gamma-\alpha)}t^{2})\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}dt
  3. \int_{x}^{100000}\frac{1}{t\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}dt+\int_{0}^{\frac{\sqrt{(\delta-\beta)(\gamma-\alpha)(x-\alpha)(x-\beta)}-\sqrt{(\delta-\beta)(\gamma-\alpha)(x-\gamma)(x-\delta)}}{(\delta+\gamma-\beta-\alpha)x+\alpha\beta-\gamma\delta}}\frac{\frac{2(\delta-\gamma)}{\gamma\delta\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{(1-\frac{\gamma(\delta-\alpha)}{\delta(\gamma-\alpha)}t^{2})\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}dt
  4. \int_{0}^{\frac{\sqrt{(\delta-\beta)(\gamma-\alpha)(x-\alpha)(x-\beta)}-\sqrt{(\delta-\beta)(\gamma-\alpha)(x-\gamma)(x-\delta)}}{(\delta+\gamma-\beta-\alpha)x+\alpha\beta-\gamma\delta}}\frac{\frac{2}{\gamma\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}dt-\frac{2}{\sqrt{\alpha\beta\gamma\delta}}\operatorname{arctanh}\left(\frac{\sqrt{\alpha\beta\gamma\delta}(x-\delta)}{\delta(x^{2}-\delta\ x+\sqrt{(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)})}\right)
  5. \int_{x}^{100000}\frac{1}{\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}dt
  6. \int_{0}^{\frac{\sqrt{(\delta-\beta)(\gamma-\alpha)(x-\alpha)(x-\beta)}-\sqrt{(\delta-\beta)(\gamma-\alpha)(x-\gamma)(x-\delta)}}{(\delta+\gamma-\beta-\alpha)x+\alpha\beta-\gamma\delta}}\frac{\frac{2}{\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}dt
  7. \int_{\delta}^{x}\frac{t}{\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}dt
  8. \int_{0}^{\sqrt{\frac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}}\frac{\frac{2(\delta-\gamma)}{\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{(1-\frac{\delta-\alpha}{\gamma-\alpha}t^{2})\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}dt+\int_{0}^{\sqrt{\frac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}}\frac{\frac{2\gamma}{\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}dt
  9. \int_{\delta}^{x}\frac{1}{\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}dt
  10. \int_{0}^{\sqrt{\frac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}}\frac{\frac{2}{\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}dt
  11. \alpha=1.12
  12. \beta=1.44
  13. \gamma=1.85
  14. \delta=3.06
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\begin{align*}
&\qquad\qquad\quad\int_{\delta}^{x}\frac{{\mathrm{d}}t}{t\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}\\

&=\int_{0}^{\sqrt{\frac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}}\frac{-\frac{2(\delta-\gamma)}{\gamma\delta\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{(1-\frac{\gamma(\delta-\alpha)}{\delta(\gamma-\alpha)}t^{2})\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}{\mathrm{d}}t\\

&\qquad\quad+\int_{0}^{\sqrt{\frac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}}\frac{\frac{2}{\gamma\sqrt{(\delta-\beta)(\gamma-\alpha)}}  }{\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}{\mathrm{d}}t\\
\\
\\
&\qquad\qquad\quad\int_{\delta}^{x}\frac{t}{\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}{\mathrm{d}}t\\

&=\int_{0}^{\sqrt{\frac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}}\frac{\frac{2(\delta-\gamma)}{\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{(1-\frac{\delta-\alpha}{\gamma-\alpha}t^{2})\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}{\mathrm{d}}t\\

&\qquad\quad+\int_{0}^{\sqrt{\frac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}}\frac{\frac{2\gamma}{\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}{\mathrm{d}}t\\
\\
\\
&\qquad\qquad\quad\int_{\delta}^{x}\frac{{\mathrm{d}}t}{\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}\\
&\qquad\quad=\int_{0}^{\sqrt{\frac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}}\frac{\frac{2}{\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{\sqrt{(1-t^{2})(1-\frac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}t^{2})}}{\mathrm{d}}t

\end{align*}





\begin{align*}
&\qquad\qquad\qquad\int_{\beta}^{x}\frac{1}{t\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau_{0}^{2})}}dt\\
&=
\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\frac{\frac{\alpha+\beta}{2\alpha\beta}-\frac{\alpha\beta-\sigma^{2}-\tau_{0}^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})}}{1-\frac{(\alpha B+\beta A)^{2}}{4\alpha\beta AB}t^{2}}+\frac{\alpha\beta-\sigma^{2}-\tau_{0}^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})}-\frac{\frac{\beta-\alpha}{2\alpha\beta}\sqrt{1-t^{2}}}{1-\frac{(\alpha B+\beta A)^{2}}{4\alpha\beta AB}t^{2}}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt\\
&=\frac{\frac{\alpha+\beta}{2\alpha\beta}}{\sqrt{AB}}\int_{0}^{f}\frac{\frac{1-\frac{(A+B)(\alpha B+A\beta)}{2AB(\alpha+\beta)}t^{2}+\frac{\alpha-\beta}{\alpha+\beta}\sqrt{1-t^{2}}}{1-\frac{(\alpha B+\beta A)^{2}}{4\alpha\beta AB}t^{2}}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt
\end{align*}

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 Author| 青青子衿 Posted at 2024-9-24 15:56:59
Last edited by 青青子衿 at 2024-9-28 10:53:00
\begin{gather*}

\int_{\beta}^{x}\frac{{\mathrm{d}}t}{t\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau^{2})}}\\
\\
\begin{split}
&=\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{\alpha\beta-\sigma^{2}-\tau^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau^{2})}}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\quad+\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{\alpha+\beta}{2\alpha\beta}-\frac{\alpha\beta-\sigma^{2}-\tau^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau^{2})}}{\left(1-\kappa_2^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad\quad-\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{\beta-\alpha}{2\alpha\beta}}{\left(1-\kappa_2^2t^{2}\right)\sqrt{1-k^2t^{2}}}{\mathrm{d}}t\\

&=\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{\alpha\beta-\sigma^{2}-\tau^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau^{2})}}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\quad+\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{\alpha+\beta}{2\alpha\beta}-\frac{\alpha\beta-\sigma^{2}-\tau^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau^{2})}}{\left(1-\kappa_2^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad\quad-\frac{1}{\sqrt{\scriptsize\alpha\beta(\sigma^{2}+\tau^{2})}}\operatorname{arctanh}\left(\tfrac{\frac{\beta-\alpha}{2}\sqrt{\frac{\sigma^{2}+\tau^{2}}{\alpha\beta AB}}f}{\sqrt{1-k^2f^{2}}}\right)\\

&=\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{\alpha\beta-\sigma^{2}-\tau^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau^{2})}}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\quad+\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{\alpha+\beta}{2\alpha\beta}-\frac{\alpha\beta-\sigma^{2}-\tau^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau^{2})}}{\left(1-\kappa_2^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\qquad\quad-\frac{1}{\sqrt{\scriptsize\alpha\beta(\sigma^{2}+\tau^{2})}}\operatorname{arctanh}\bigg({\sqrt{\scriptsize\frac{(\sigma^{2}+\tau^{2})(x-\alpha)(x-\beta)}{\alpha\beta((x-\sigma)^{2}+\tau^{2})}}}\,\bigg)

\end{split}\\
\\

\qquad\left\{\begin{split}
f&={\raise1.5px\scriptsize\tfrac{2\sqrt{AB(x-\alpha)(x-\beta)}}{(x-\beta)A+(x-\alpha)B}}\\

k^2&={\raise1.5px\scriptsize\tfrac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}}\\
&={\raise1.5px\scriptsize\frac{1}{4}\big(2+\tfrac{A^{2}+B^{2}-(\beta-\alpha)^{2}}{AB}\big)}\\

&={\raise1.5px\scriptsize\frac{1}{2}\Big(1+\tfrac{(\alpha-\sigma)(\beta-\sigma)+\tau^{2}}{\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}\Big)}\\

\kappa_2^2&={\raise1px\scriptsize\tfrac{(\alpha\,\!B+\beta\,\!A)^2}{4\alpha\beta\,\!AB}}\\

&={\raise1.5px\scriptsize\frac{1}{4} \big(2+\tfrac{\alpha ^2B^2+\beta ^2A^2}{\alpha\beta A B}\big)}\\

&={\raise1.5px\scriptsize\frac{1}{4}\Big(2+\tfrac{\alpha^{2}(\beta-\sigma)^{2}+\beta^{2}(\alpha-\sigma)^{2}+(\alpha^{2}+\beta^{2})\tau^{2}}{\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}\Big)}\\

A&=\scriptsize{\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}\\
B&=\scriptsize{\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}
\end{split}\right.
\\

\end{gather*}


  1. \int_{\beta}^{x}\frac{1}{t\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau_{0}^{2})}}dt
  2. \frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\frac{\frac{\alpha+\beta}{2\alpha\beta}-\frac{\alpha\beta-\sigma^{2}-\tau_{0}^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})}}{1-\frac{(\alpha B+\beta A)^{2}}{4\alpha\beta AB}t^{2}}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt+\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\frac{\alpha\beta-\sigma^{2}-\tau_{0}^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt-\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\frac{\frac{\beta-\alpha}{2\alpha\beta}}{1-\frac{(\alpha B+\beta A)^{2}}{4\alpha\beta AB}t^{2}}}{\sqrt{1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2}}}dt
  3. \frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\frac{\frac{\alpha+\beta}{2\alpha\beta}-\frac{\alpha\beta-\sigma^{2}-\tau_{0}^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})}}{1-\frac{(\alpha B+\beta A)^{2}}{4\alpha\beta AB}t^{2}}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt+\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\frac{\alpha\beta-\sigma^{2}-\tau_{0}^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt-\frac{1}{\sqrt{\alpha\beta(\sigma^{2}+\tau_{0}^{2})}}\operatorname{arctanh}\left(\frac{\frac{\beta-\alpha}{2}\sqrt{\frac{\sigma^{2}+\tau_{0}^{2}}{\alpha\beta AB}}f}{\sqrt{1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}f^{2}}}\right)
  4. \frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\frac{\frac{\alpha+\beta}{2\alpha\beta}-\frac{\alpha\beta-\sigma^{2}-\tau_{0}^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})}}{1-\frac{(\alpha B+\beta A)^{2}}{4\alpha\beta AB}t^{2}}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt+\frac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\frac{\alpha\beta-\sigma^{2}-\tau_{0}^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt-\frac{1}{\sqrt{\alpha\beta(\sigma^{2}+\tau_{0}^{2})}}\operatorname{arctanh}\left(\sqrt{\frac{(\sigma^{2}+\tau_{0}^{2})(x-\alpha)(x-\beta)}{\alpha\beta((x-\sigma)^{2}+\tau_{0}^{2})}}\right)
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 Author| 青青子衿 Posted at 2024-9-28 11:22:21
青青子衿 发表于 2024-9-24 15:56
\begin{align*}
\int_{\beta}^{x}\frac{{\mathrm{d}}t}{t\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau^{2})}}\\
\\
\end{align*}

\begin{gather*}

\int_{x}^{+\infty}\frac{{\mathrm{d}}t}{t\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau^{2})}}\\
\\
\begin{split}
&=\frac{1}{\sqrt{AB}}\int_{0}^{\mathscr{g}}\frac{\scriptsize\frac{\alpha\beta-\sigma^{2}-\tau^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau^{2})}}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\quad+\frac{1}{\sqrt{AB}}\int_{0}^{{\mathscr{g}}}\frac{\scriptsize\frac{\alpha+\beta}{2\alpha\beta}-\frac{\alpha\beta-\sigma^{2}-\tau^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau^{2})}}{\left(1-\kappa_2^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\


&\qquad\qquad\quad-\frac{1}{\sqrt{\scriptsize\alpha\beta(\sigma^{2}+\tau^{2})}}\operatorname{arctanh}\bigg({\sqrt{\tfrac{\sigma^{2}+\tau^{2}}{\alpha\beta}}
\tfrac{\phi_1(x)+2\alpha\beta\sqrt{(x-\alpha)(x-\beta)((x-\sigma)^{2}+\tau^{2})}}{\phi_2(x)+\psi_2(x)\sqrt{(x-\alpha)(x-\beta)((x-\sigma)^{2}+\tau^{2})}}
}\,\bigg)\\


\end{split}\\
\\

\qquad\left\{\begin{split}
\mathscr{g}&={\Tiny\tfrac{2((\alpha+\beta-2\sigma)x+\sigma^{2}+\tau^{2}-\alpha\beta) \sqrt{AB}}{((\alpha+\beta-2\sigma)x+\sigma^{2}+\tau^{2}-\alpha\beta-AB)\sqrt{(x-\alpha)(x-\beta)}
+((\alpha+\beta-2\sigma)x+\sigma^{2}+\tau^{2}-\alpha\beta+AB)\sqrt{(x-\sigma)^{2}+\tau^{2}}}}\\


k^2&={\raise1.5px\scriptsize\tfrac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}}\\
&={\raise1.5px\scriptsize\frac{1}{4}\big(2+\tfrac{A^{2}+B^{2}-(\beta-\alpha)^{2}}{AB}\big)}\\

&={\raise1.5px\scriptsize\frac{1}{2}\Big(1+\tfrac{(\alpha-\sigma)(\beta-\sigma)+\tau^{2}}{\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}\Big)}\\

\kappa_2^2&={\raise1px\scriptsize\tfrac{(\alpha\,\!B+\beta\,\!A)^2}{4\alpha\beta\,\!AB}}\\

&={\raise1.5px\scriptsize\frac{1}{4} \big(2+\tfrac{\alpha ^2B^2+\beta ^2A^2}{\alpha\beta A B}\big)}\\

&={\raise1.5px\scriptsize\frac{1}{4}\Big(2+\tfrac{\alpha^{2}(\beta-\sigma)^{2}+\beta^{2}(\alpha-\sigma)^{2}+(\alpha^{2}+\beta^{2})\tau^{2}}{\sqrt{\left((\alpha-\sigma)^{2}+\tau^{2}\right)\left((\beta-\sigma)^{2}+\tau^{2}\right)}}\Big)}\\

\phi_1(x)&=\scriptsize{(\alpha^{2}+\beta^{2}-2\sigma(\alpha+\beta))x^{2}}\\
&\qquad\scriptsize{+(4\alpha\beta\sigma-(\alpha+\beta)(\alpha\beta-\sigma^{2}-\tau^{2}))x-2\alpha\beta(\sigma^{2}+\tau^{2})}\\

\phi_2(x)&=\scriptsize{(\alpha+\beta-2\sigma)x^{3}-2\alpha\beta x^{2}}\\
&\qquad\scriptsize{+((\alpha+\beta)(\sigma^{2}+\tau^{2})+2\alpha\beta\sigma)x-2\alpha\beta(\sigma^{2}+\tau^{2})}\\

\psi_2(x)&=\scriptsize{\left((\alpha+\beta-2\sigma)x+2(\sigma^{2}+\tau^{2})\right)}\\

A&=\scriptsize{\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}\\
B&=\scriptsize{\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}
\end{split}\right.
\\

\end{gather*}



  1. \int_{x}^{+10000000}\frac{1}{t\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau_{0}^{2})}}dt
  2. \alpha=1.35
  3. \beta=2.13
  4. \sigma=0.68
  5. \tau_{0}=1.44
  6. A=\sqrt{(\alpha-\sigma)^{2}+\tau_{0}^{2}}
  7. B=\sqrt{(\beta-\sigma)^{2}+\tau_{0}^{2}}
  8. g\left(x\right)=\frac{2\sqrt{AB}\left((\alpha+\beta-2\sigma)x-\alpha\beta+\sigma^{2}+\tau_{0}^{2}\right)}{\left((\alpha+\beta-2\sigma)x-\alpha\beta-AB+\sigma^{2}+\tau_{0}^{2}\right)\sqrt{(x-\alpha)(x-\beta)}+\left((\alpha+\beta-2\sigma)x-\alpha\beta+AB+\sigma^{2}+\tau_{0}^{2}\right)\sqrt{(x-\sigma)^{2}+\tau_{0}^{2}}}
  9. \frac{\frac{\alpha+\beta}{2\alpha\beta}-\frac{\alpha\beta-\sigma^{2}-\tau_{0}^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})}}{\sqrt{AB}}\int_{0}^{g\left(x\right)}\frac{\frac{1}{1-\frac{(\alpha B+\beta A)^{2}}{4\alpha\beta AB}t^{2}}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt+\frac{1}{\sqrt{AB}}\int_{0}^{g\left(x\right)}\frac{\frac{\alpha\beta-\sigma^{2}-\tau_{0}^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})}}{\sqrt{(1-t^{2})(1-\frac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}t^{2})}}dt-\frac{1}{\sqrt{\alpha\beta(\sigma^{2}+\tau_{0}^{2})}}\operatorname{arctanh}\left(\sqrt{\frac{\sigma^{2}+\tau_{0}^{2}}{\alpha\beta}}\frac{(\alpha^{2}+\beta^{2}-2\sigma(\alpha+\beta))x^{2}+(4\alpha\beta\sigma-(\alpha+\beta)(\alpha\beta-\sigma^{2}-\tau_{0}^{2}))x-2\alpha\beta(\sigma^{2}+\tau_{0}^{2})+2\alpha\beta\sqrt{(x-\alpha)(x-\beta)((x-\sigma)^{2}+\tau_{0}^{2})}}{(\alpha+\beta-2\sigma)x^{3}-2\alpha\beta x^{2}+((\alpha+\beta)(\sigma^{2}+\tau_{0}^{2})+2\alpha\beta\sigma)x-2\alpha\beta(\sigma^{2}+\tau_{0}^{2})+\left((\alpha+\beta-2\sigma)x+2(\sigma^{2}+\tau_{0}^{2})\right)\sqrt{(x-\alpha)(x-\beta)((x-\sigma)^{2}+\tau_{0}^{2})}}\right)
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 Author| 青青子衿 Posted at 2024-10-6 11:37:23
Last edited by 青青子衿 at 2024-10-14 15:48:00
青青子衿 发表于 2024-5-10 05:27
\begin{gather*}
\int_{-\frac{a^{2}+b^{2}-c^{2}-d^{2}+AB}{2(c-a)}}^{x}\frac{t}{\sqrt{((t-a)^{2}+b^{2})((t-c)^{2}+d^{2})}}{\mathrm{d}}t\\

=
\frac{A-B}{2bd}\int_{0}^{\frac{\sqrt{\frac{1}{2}(1+\frac{(a-c)^{2}-b^{2}+d^{2}}{AB})}\left(x+\frac{a^{2}+b^{2}-c^{2}-d^{2}+AB}{2(c-a)}\right)}{\sqrt{(x-a)^{2}+b^{2}}}}\frac{a-\frac{\frac{AB-(a-c)^{2}+b^{2}-d^{2}}{2(c-a)}}{1-\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b^{2}(a-c)^{2}}t^{2}}+\frac{\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b(a-c)^{2}}t\sqrt{1-t^{2}}}{1-\frac{AB(AB-(a-c)^{2}+b^{2}-d^{2})}{2b^{2}(a-c)^{2}}t^{2}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4b^{2}d^{2}}t^{2})}}{\mathrm{d}}t\\
\\
\left\{
\begin{split}
A&=\sqrt{(a-c)^2+(b+d)^2}\\
B&=\sqrt{(a-c)^2+(b-d)^2}
\end{split}\right.
\end{gather*}


\begin{gather*}
\int_{0}^{x}\frac{{\mathrm{d}t}}{\sqrt{((t-\lambda)^{2}+\mu^{2})((t-\sigma)^{2}+\tau^{2})}}\\
\\
\begin{split}

&=\frac{A-B}{2\mu\tau}\int_{0}^{\mathscr{g}}\frac{{\mathrm{d}}t}{\sqrt{(1-t^{2})(1-k^2t^{2})}}\\


\end{split}\\
\\

\qquad\left\{\begin{split}
\mathscr{g}&=\scriptsize\tfrac{\frac{(A+B)\ x\big((\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2})x+2(\lambda\sigma^{2}+\lambda\tau^{2}-\lambda^{2}\sigma-\mu^{2}\sigma)\big)}{4(\sigma-\lambda)(\lambda\sigma^{2}+\lambda\tau^{2}-\lambda^{2}\sigma-\mu^{2}\sigma)\sqrt{(\lambda^{2}+\mu^{2})(\sigma^{2}+\tau^{2})}}}{\frac{\Big(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}+AB}{2(\sigma-\lambda)}\Big)\sqrt{(x-\lambda)^{2}+\mu^{2}}}{(\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}+AB)\sqrt{\lambda^{2}+\mu^{2}}}+\frac{\Big(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}-AB}{2(\sigma-\lambda)}\Big)\sqrt{(x-\sigma)^{2}+\tau^{2}}}{(\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}-AB)\sqrt{\sigma^{2}+\tau^{2}}}}\\


k^2&={\raise1.5px\scriptsize\tfrac{AB(A-B)^{2}}{4\mu^{2}\tau^{2}}}\\


A&=\scriptsize{\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}\\
B&=\scriptsize{\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}
\end{split}\right.
\\

\end{gather*}








\begin{gather*}
\int_{x}^{+\infty}\frac{{\mathrm{d}t}}{\sqrt{((t-\lambda)^{2}+\mu^{2})((t-\sigma)^{2}+\tau^{2})}}\\
\\
\begin{split}

&=\frac{A-B}{2\mu\tau}\int_{0}^{\mathscr{g}}\frac{{\mathrm{d}}t}{\sqrt{(1-t^{2})(1-k^2t^{2})}}\\


\end{split}\\
\\

\qquad\left\{\begin{split}
\mathscr{g}&=\scriptsize\tfrac{(A+B)\Big(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}}{2(\sigma-\lambda)}\Big)}{\Big(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}+AB}{2(\sigma-\lambda)}\Big)\sqrt{(x-\lambda)^{2}+\mu^{2}}+\Big(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}-AB}{2(\sigma-\lambda)}\Big)\sqrt{(x-\sigma)^{2}+\tau^{2}}}\\


k^2&={\raise1.5px\scriptsize\tfrac{AB(A-B)^{2}}{4\mu^{2}\tau^{2}}}\\


A&=\scriptsize{\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}\\
B&=\scriptsize{\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}
\end{split}\right.
\\

\end{gather*}



\begin{gather*}
\int_{G}^{x}\frac{1}{t\sqrt{((t-\lambda)^{2}+\mu^{2})((t-\sigma)^{2}+\tau^{2})}}{\mathrm{d}t}\\
\\
\begin{split}
&=\frac{A-B}{2\mu\tau(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\lambda}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\quad+\frac{A-B}{2\mu\tau(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\scriptsize\mu\,\!Q}{\left(1-(1+Q^{2})t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad\quad-\frac{A-B}{2\mu\tau(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\scriptsize\mu(1+Q^{2})t}{\left(1-(1+Q^{2})t^{2}\right)\sqrt{1-k^2t^{2}}}{\mathrm{d}}t\\

&=\frac{A-B}{2\mu\tau(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\lambda}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\quad+\frac{A-B}{2\mu\tau(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\scriptsize\mu\,\!Q}{\left(1-(1+Q^{2})t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad\quad-\frac{A+B}{2}\sqrt{\frac{\lambda^{2}+\mu^{2}}{\sigma^{2}+\tau^{2}}}\operatorname{arctanh}\left(\raise{1.5px}\scriptsize\frac{\frac{A-B}{2\tau}\sqrt{\frac{\sigma^{2}+\tau^{2}}{\lambda^{2}+\mu^{2}}}}{\sqrt{1-k^2f^{2}}}\right)\\
&\qquad\qquad\qquad+\frac{A+B}{2}\sqrt{\frac{\lambda^{2}+\mu^{2}}{\sigma^{2}+\tau^{2}}}\operatorname{arctanh}\left(\scriptsize\frac{A-B}{2\tau}\sqrt{\frac{\sigma^{2}+\tau^{2}}{\lambda^{2}+\mu^{2}}}\,\right)\\

&=\frac{A-B}{2\mu\tau(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\lambda}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\quad+\frac{A-B}{2\mu\tau(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\scriptsize\mu\,\!Q}{\left(1-(1+Q^{2})t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad\quad-\frac{A+B}{2}\sqrt{\frac{\lambda^{2}+\mu^{2}}{\sigma^{2}+\tau^{2}}}\operatorname{arctanh}\left(\raise{1.5px}\scriptsize\sqrt{\frac{(\sigma^{2}+\tau^{2})((x-\lambda)^{2}+\mu^{2})}{(\lambda^{2}+\mu^{2})((x-\sigma)^{2}+\tau^{2})}}\,\right)\\
&\qquad\qquad\qquad+\frac{A+B}{2}\sqrt{\frac{\lambda^{2}+\mu^{2}}{\sigma^{2}+\tau^{2}}}\operatorname{arctanh}\left(\scriptsize\frac{A-B}{2\tau}\sqrt{\frac{\sigma^{2}+\tau^{2}}{\lambda^{2}+\mu^{2}}}\,\right)\\


\end{split}\\
\\

\qquad\left\{\begin{split}
f&={\raise1.5px\scriptsize\frac{\sqrt{\frac{(A+B)^{2}-4\mu^{2}}{AB}}\left(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}+AB}{2(\sigma-\lambda)}\right)}{2\sqrt{(x-\lambda)^{2}+\mu^{2}}}}\\

k^2&={\raise1.5px\scriptsize\tfrac{AB(A-B)^{2}}{4\mu^{2}\tau^{2}}}\\

Q&={\raise1px\scriptsize\tfrac{(\lambda^{2}+\mu^{2})^{2}-2\lambda\sigma(\lambda^{2}+\mu^{2})+(\lambda^{2}-\mu^{2})(\sigma^{2}+\tau^{2})-(\lambda^{2}+\mu^{2})AB}{2\mu(\lambda^{2}\sigma+\mu^{2}\sigma-\lambda\sigma^{2}-\lambda\tau^{2})}}\\


A&=\scriptsize{\sqrt{(\lambda-\sigma)^{2}+(\mu+\tau)^{2}}}\\
B&=\scriptsize{\sqrt{(\lambda-\sigma)^{2}+(\mu-\tau)^{2}}}
\end{split}\right.
\\

\end{gather*}







  1. 12\int_{-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}}^{x}\frac{1}{t\sqrt{((t-\lambda)^{2}+\mu^{2})((t-\sigma)^{2}+\tau_{0}^{2})}}dt
  2. 12\frac{\frac{A-B}{2\mu\tau_{0}}}{-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}}\int_{0}^{f}\frac{\frac{1+\frac{(\lambda-\sigma)^{2}-\mu^{2}+\tau_{0}^{2}-AB}{2\mu(\lambda-\sigma)}\frac{t}{\sqrt{1-t^{2}}}}{1+\frac{(\lambda^{2}+\mu^{2})^{2}-2\lambda\sigma(\lambda^{2}+\mu^{2})+(\lambda^{2}-\mu^{2})(\sigma^{2}+\tau_{0}^{2})-(\lambda^{2}+\mu^{2})AB}{2\mu(\lambda^{2}\sigma+\mu^{2}\sigma-\lambda\sigma^{2}-\lambda\tau_{0}^{2})}\frac{t}{\sqrt{1-t^{2}}}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt
  3. 15\frac{\frac{A-B}{2\mu\tau_{0}}}{-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}}\int_{0}^{f}\frac{\frac{1+PQ}{1+Q^{2}}-\frac{\frac{Q(P-Q)}{1+Q^{2}}}{1-(1+Q^{2})t^{2}}+\frac{(P-Q)t\sqrt{1-t^{2}}}{1-(1+Q^{2})t^{2}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt
  4. 15\frac{A-B}{2\mu\tau_{0}(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\lambda+\frac{\mu Q}{1-(1+Q^{2})t^{2}}-\frac{\mu(1+Q^{2})t\sqrt{1-t^{2}}}{1-(1+Q^{2})t^{2}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt
  5. 15\frac{A-B}{2\mu\tau_{0}(\lambda^{2}+\mu^{2})}\left(\int_{0}^{f}\frac{\lambda}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt+\int_{0}^{f}\frac{\frac{\mu Q}{1-(1+Q^{2})t^{2}}}{\sqrt{(1-t^{2})(1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}t^{2})}}dt-\frac{A+B}{2}\sqrt{\frac{\lambda^{2}+\mu^{2}}{\sigma^{2}+\tau_{0}^{2}}}\operatorname{arctanh}\left(\frac{\frac{A-B}{2\tau_{0}}\sqrt{\frac{\sigma^{2}+\tau_{0}^{2}}{\lambda^{2}+\mu^{2}}}}{\sqrt{1-\frac{AB(A-B)^{2}}{4\mu^{2}\tau_{0}^{2}}f^{2}}}\right)+\frac{A+B}{2}\sqrt{\frac{\lambda^{2}+\mu^{2}}{\sigma^{2}+\tau_{0}^{2}}}\operatorname{arctanh}\left(\frac{A-B}{2\tau_{0}}\sqrt{\frac{\sigma^{2}+\tau_{0}^{2}}{\lambda^{2}+\mu^{2}}}\right)\right)
  6. A=\sqrt{(\lambda-\sigma)^{2}+(\mu+\tau_{0})^{2}}
  7. B=\sqrt{(\lambda-\sigma)^{2}+(\mu-\tau_{0})^{2}}
  8. P=\frac{(\lambda-\sigma)^{2}-\mu^{2}+\tau_{0}^{2}-AB}{2\mu(\lambda-\sigma)}
  9. Q=\frac{(\lambda^{2}+\mu^{2})^{2}-2\lambda\sigma(\lambda^{2}+\mu^{2})+(\lambda^{2}-\mu^{2})(\sigma^{2}+\tau_{0}^{2})-(\lambda^{2}+\mu^{2})AB}{2\mu(\lambda^{2}\sigma+\mu^{2}\sigma-\lambda\sigma^{2}-\lambda\tau_{0}^{2})}
  10. f=\frac{\sqrt{\frac{1}{2}(1+\frac{(\lambda-\sigma)^{2}-\mu^{2}+\tau_{0}^{2}}{AB})}\left(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau_{0}^{2}+AB}{2(\sigma-\lambda)}\right)}{\sqrt{(x-\lambda)^{2}+\mu^{2}}}
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\begin{gather*}
\quad\int_{\delta}^{x}\frac{{\mathrm{d}}t}{t\sqrt{(t-\alpha)(t-\beta)(t-\gamma)(t-\delta)}}\\
\\

\begin{split}
&=\int_{0}^{f}\frac{\scriptsize\frac{2}{\gamma\sqrt{(\delta-\beta)(\gamma-\alpha)}}  }{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad
-\int_{0}^{f}\frac{\scriptsize\frac{2(\delta-\gamma)}{\gamma\delta\sqrt{(\delta-\beta)(\gamma-\alpha)}}}{(1-\kappa^2t^{2})\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t
\end{split}\\
\\

\left\{\begin{split}
f^2&={\raise1.5px\scriptsize\tfrac{(\gamma-\alpha)(x-\delta)}{(\delta-\alpha)(x-\gamma)}}\\

k^2&={\raise1.5px\scriptsize\tfrac{(\delta-\alpha)(\gamma-\beta)}{(\delta-\beta)(\gamma-\alpha)}}\\

\kappa^2&={\raise1px\scriptsize\tfrac{\gamma(\delta-\alpha)}{\delta(\gamma-\alpha)}}\\
\end{split}\right.
\\
\end{gather*}


\begin{gather*}

\int_{\beta}^{x}\frac{{\mathrm{d}}t}{t\sqrt{(t-\alpha)(t-\beta)((t-\sigma)^{2}+\tau^{2})}}\\
\\
\begin{split}
&=\tfrac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{\alpha\beta-\sigma^{2}-\tau^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau^{2})}}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\quad+\tfrac{1}{\sqrt{AB}}\int_{0}^{f}\frac{\scriptsize\frac{\alpha+\beta}{2\alpha\beta}-\frac{\alpha\beta-\sigma^{2}-\tau^{2}-AB}{2\alpha\beta\sigma-(\alpha+\beta)(\sigma^{2}+\tau^{2})}}{\left(1-\kappa^2t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\qquad\quad-\tfrac{1}{\sqrt{\alpha\beta(\sigma^{2}+\tau^{2})}}\operatorname{arctanh}\bigg({\sqrt{\scriptsize\frac{(\sigma^{2}+\tau^{2})(x-\alpha)(x-\beta)}{\alpha\beta((x-\sigma)^{2}+\tau^{2})}}}\,\bigg)

\end{split}\\
\\

\qquad\left\{\begin{split}
f&={\raise1.5px\scriptsize\tfrac{2\sqrt{AB(x-\alpha)(x-\beta)}}{(x-\beta)A+(x-\alpha)B}}\\

k^2&={\raise1.5px\scriptsize\tfrac{(A+B)^{2}-(\beta-\alpha)^{2}}{4AB}}\\
\kappa^2&={\raise1px\scriptsize\tfrac{(\alpha\,\!B+\beta\,\!A)^2}{4\alpha\beta\,\!AB}}\\

A&=\scriptsize{\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}\\
B&=\scriptsize{\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}
\end{split}\right.
\\

\end{gather*}


\begin{gather*}
\int_{-\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}+AB}{2(\sigma-\lambda)}}^{x}\frac{\mathrm{d}t}{t\sqrt{((t-\lambda)^{2}+\mu^{2})((t-\sigma)^{2}+\tau^{2})}}\\
\\
\begin{split}

&=\tfrac{A-B}{2\mu\tau(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\lambda}{\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\
&\qquad\quad+\tfrac{A-B}{2\mu\tau(\lambda^{2}+\mu^{2})}\int_{0}^{f}\frac{\scriptsize\mu\,\!Q}{\left(1-(1+Q^{2})t^{2}\right)\sqrt{(1-t^{2})(1-k^2t^{2})}}{\mathrm{d}}t\\

&\qquad\qquad\quad-\tfrac{A+B}{2}\sqrt{\tfrac{\lambda^{2}+\mu^{2}}{\sigma^{2}+\tau^{2}}}\operatorname{arctanh}\left(\raise{1.5px}\scriptsize\sqrt{\frac{(\sigma^{2}+\tau^{2})((x-\lambda)^{2}+\mu^{2})}{(\lambda^{2}+\mu^{2})((x-\sigma)^{2}+\tau^{2})}}\,\right)\\
&\qquad\qquad\qquad\quad+\tfrac{A+B}{2}\sqrt{\tfrac{\lambda^{2}+\mu^{2}}{\sigma^{2}+\tau^{2}}}\operatorname{arctanh}\left(\scriptsize\frac{A-B}{2\tau}\sqrt{\frac{\sigma^{2}+\tau^{2}}{\lambda^{2}+\mu^{2}}}\,\right)
\end{split}\\
\\

\quad\>\left\{\begin{split}
f&={\raise1.5px\scriptsize\frac{\sqrt{\frac{(A+B)^{2}-4\mu^{2}}{AB}}\left(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}+AB}{2(\sigma-\lambda)}\right)}{2\sqrt{(x-\lambda)^{2}+\mu^{2}}}}\\

k^2&={\raise1.5px\scriptsize\tfrac{AB(A-B)^{2}}{4\mu^{2}\tau^{2}}}\\
G&={\raise1px\scriptsize-\tfrac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}+AB}{2(\sigma-\lambda)}}\\
Q&={\raise1px\scriptsize\tfrac{(\lambda^{2}+\mu^{2})^{2}-2\lambda\sigma(\lambda^{2}+\mu^{2})+(\lambda^{2}-\mu^{2})(\sigma^{2}+\tau^{2})-(\lambda^{2}+\mu^{2})AB}{2\mu(\lambda^{2}\sigma+\mu^{2}\sigma-\lambda\sigma^{2}-\lambda\tau^{2})}}\\

A&=\scriptsize{\sqrt{(\lambda-\sigma)^{2}+(\mu+\tau)^{2}}}\\
B&=\scriptsize{\sqrt{(\lambda-\sigma)^{2}+(\mu-\tau)^{2}}}
\end{split}\right.
\\

\end{gather*}






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 Author| 青青子衿 Posted at 2024-10-21 22:46:58
Last edited by 青青子衿 at 2024-11-5 20:15:00
青青子衿 发表于 2024-10-6 11:37
\begin{gather*}
\int_{0}^{x}\frac{{\mathrm{d}t}}{\sqrt{((t-\lambda)^{2}+\mu^{2})((t-\sigma)^{2}+\tau^{2})}}\\
\\
\begin{split}

&=\frac{A-B}{2\mu\tau}\int_{0}^{\mathscr{g}}\frac{{\mathrm{d}}t}{\sqrt{(1-t^{2})(1-k^2t^{2})}}\\


\end{split}\\
\\

\qquad\left\{\begin{split}
\mathscr{g}&=\scriptsize\tfrac{\frac{(A+B)\ x\big((\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2})x+2(\lambda\sigma^{2}+\lambda\tau^{2}-\lambda^{2}\sigma-\mu^{2}\sigma)\big)}{4(\sigma-\lambda)(\lambda\sigma^{2}+\lambda\tau^{2}-\lambda^{2}\sigma-\mu^{2}\sigma)\sqrt{(\lambda^{2}+\mu^{2})(\sigma^{2}+\tau^{2})}}}{\frac{\Big(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}+AB}{2(\sigma-\lambda)}\Big)\sqrt{(x-\lambda)^{2}+\mu^{2}}}{(\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}+AB)\sqrt{\lambda^{2}+\mu^{2}}}+\frac{\Big(x+\frac{\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}-AB}{2(\sigma-\lambda)}\Big)\sqrt{(x-\sigma)^{2}+\tau^{2}}}{(\lambda^{2}+\mu^{2}-\sigma^{2}-\tau^{2}-AB)\sqrt{\sigma^{2}+\tau^{2}}}}\\


k^2&={\raise1.5px\scriptsize\tfrac{AB(A-B)^{2}}{4\mu^{2}\tau^{2}}}\\


A&=\scriptsize{\sqrt{(\alpha-\sigma)^{2}+\tau^{2}}}\\
B&=\scriptsize{\sqrt{(\beta-\sigma)^{2}+\tau^{2}}}
\end{split}\right.
\\

\end{gather*}

\begin{align*}
&\quad\frac{\mathrm{d}}{\mathrm{d}x}\left(-2\operatorname{arctanh}\left(\frac{\left(x+\frac{\alpha}{4}\right)^{2}-\frac{3\alpha^{2}-8\beta}{16}}{\sqrt{x^{4}+\alpha x^{3}+\beta x^{2}+\gamma x+\delta}}\right)\right)\\
&=
\frac{\frac{\left(\frac{\alpha}{4}+\frac{(\alpha^{2}-4\beta)^{2}-64\delta}{8(\alpha^{3}-4\alpha\beta+8\gamma)}\right)^{2}-\frac{3\alpha^{2}-8\beta}{16}}{x-\frac{(\alpha^{2}-4\beta)^{2}-64\delta}{8(\alpha^{3}-4\alpha\beta+8\gamma)}}-\frac{\alpha}{2}+\frac{(\alpha^{2}-4\beta)^{2}-64\delta}{8(\alpha^{3}-4\alpha\beta+8\gamma)}-3x}{\sqrt{x^{4}+\alpha x^{3}+\beta x^{2}+\gamma x+\delta}}\\
\\
&\quad\frac{\mathrm{d}}{\mathrm{d}x}\left(-\operatorname{arctanh}\left(\frac{x^{2}+\frac{\alpha}{2}x+\xi}{\sqrt{x^{4}+\alpha x^{3}+\beta x^{2}+\gamma x+\delta}}\right)\right)\\
&=
\frac{\frac{\left(\frac{\alpha}{4}-\frac{2(\alpha\xi-\gamma)}{\alpha^{2}-4\beta+8\xi}\right)^{2}-\frac{\alpha^{2}-16\xi}{16}}{x+\frac{2(\alpha\xi-\gamma)}{\alpha^{2}-4\beta+8\xi}}-\frac{2(\alpha\xi-\gamma)}{\alpha^{2}-4\beta+8\xi}-x}{\sqrt{x^{4}+\alpha x^{3}+\beta x^{2}+\gamma x+\delta}}\\
\\
&\qquad(\alpha\xi-\gamma)^{2}+(\delta-\xi^{2})(\alpha^{2}-4\beta+8\xi)=0\\
\\
&\quad\frac{\mathrm{d}}{\mathrm{d}x}\left(-2\operatorname{arctanh}\left(\frac{x^{2}+\rho x+\eta}{\sqrt{x^{4}+\alpha x^{3}+\beta x^{2}+\gamma x+\delta}}\right)\right)\\
&=
\frac{\frac{\frac{3(\alpha\eta-\gamma)}{\alpha-2\rho}-\frac{(\beta-\rho^{2}-2\eta)(3\alpha\rho-4\beta-2\rho^{2}+8\eta)}{3(\alpha-2\rho)^{2}}}{x+\frac{\beta-\rho^{2}-2\eta}{3(\alpha-2\rho)}}+\rho-\frac{\beta-\rho^{2}-2\eta}{\alpha-2\rho}-x}{\sqrt{x^{4}+\alpha x^{3}+\beta x^{2}+\gamma x+\delta}}\\
\\
&\quad
\left\{\begin{split}
(\beta-\rho^{2}-2\eta)(\gamma-2\rho\eta)&=9(\alpha-2\rho)(\delta-\eta^{2})\\
(\beta-\rho^{2}-2\eta)^{2}&=3(\alpha-2\rho)(\gamma-2\rho\eta)
\end{split}\right.
\\
\end{align*}



\begin{align*}
{\large\int}\frac{1}{x\sqrt{x^{4}+{\raise{0.5pt}\scriptsize\frac{4(7+\sqrt[3]{10}+\sqrt[3]{100})}{9}}x^{3}-{\raise{0.5pt}\scriptsize\frac{2(41-10\sqrt[3]{10}+5\sqrt[3]{100})}{27}}x^{2}+{\raise{0.5pt}\scriptsize\frac{4(2\sqrt[3]{100}-1-\sqrt[3]{10})}{27}}x+{\raise{0.5pt}\scriptsize\frac{2\sqrt[3]{10}+2\sqrt[3]{100}-13}{27}}}}{\mathrm{d}x}
\end{align*}


\begin{align*}
\int_{0}^{1}\frac{{\mathrm{d}}t}{(t+1)\sqrt{t^{4}-2t^{3}-7t^{2}+8}}=\frac{2}{5}\ln\left(1+\sqrt{2}\right)
\end{align*}


\begin{gather*}
\int\frac{3x^{2}-\left(x_{1}+x_{2}+x_{3}\right)x-\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)-\frac{\beta}{2}-\Big(\frac{x_{1}^{3}+\frac{\beta x_{1}}{2}+\psi}{x-x_{1}}+\frac{x_{2}^{3}+\frac{\beta x_{2}}{2}+\psi}{x-x_{2}}+\frac{x_{3}^{3}+\frac{\beta x_{3}}{2}+\psi}{x-x_{3}}\Big)}
{\sqrt{x^{6}+\beta x^{4}+\gamma x^{3}+\delta x^{2}+\varepsilon x+\phi}}{\mathrm{d}}x\\
\\
=2\operatorname{arctanh}\left(\frac{x^3+\frac{\beta  x}{2}+\psi }{\sqrt{x^6+\beta  x^4+\gamma  x^3+\delta  x^2+\varepsilon x +\varphi }}\right)\\
\\
\Big(x_i^3+\frac{\beta  x_i}{2}+\psi\Big)^2=x_i^6+\beta  x_i^4+\gamma  x_i^3+\delta  x_i^2+\varepsilon x_i +\varphi
\end{gather*}



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 Author| 青青子衿 Posted at 2024-12-4 12:37:52
Last edited by 青青子衿 at 2024-12-4 14:22:00
\begin{gather*}
\int_{0}^{x}\frac{{\mathrm{d}}t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}\\
=
\frac{2}{3}\int_{0}^{\frac{(1+2x-2k^{2}x^{3}-k^{2}x^{4})\sqrt{1-x}-(1-2x+2k^{2}x^{3}-k^{2}x^{4})\sqrt{1+x}}{(1+2kx-2kx^{3}-k^{2}x^{4})\sqrt{1-kx}+(1-2kx+2kx^{3}-k^{2}x^{4})\sqrt{1+kx}}}\frac{{\mathrm{d}}t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}
\end{gather*}


  1. \int_{0}^{x}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}dt
  2. \frac{2}{3}\int_{0}^{\frac{\left(1+2x-2k^{2}x^{3}-k^{2}x^{4}\right)\sqrt{1-x}-\left(1-2x+2k^{2}x^{3}-k^{2}x^{4}\right)\sqrt{1+x}}{\left(1+2kx-2kx^{3}-k^{2}x^{4}\right)\sqrt{1-kx}+\left(1-2kx+2kx^{3}-k^{2}x^{4}\right)\sqrt{1+kx}}}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}dt
  3. \frac{2}{3}\int_{0}^{\frac{(1+2x-2k^{2}x^{3}-k^{2}x^{4})\sqrt{1-x}-(1-2x+2k^{2}x^{3}-k^{2}x^{4})\sqrt{1+x}}{(1+2kx-2kx^{3}-k^{2}x^{4})\sqrt{1-kx}+(1-2kx+2kx^{3}-k^{2}x^{4})\sqrt{1+kx}}}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}dt
  4. \frac{1}{3}\int_{0}^{\frac{x(3-4x^{2}-4k^{2}x^{2}+6k^{2}x^{4}-k^{4}x^{8})}{1-6k^{2}x^{4}+4k^{2}x^{6}+4k^{4}x^{6}-3k^{4}x^{8}}}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}dt
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 Author| 青青子衿 Posted at 2025-2-22 00:20:51
Last edited by 青青子衿 at 2025-3-12 17:16:23
\begin{align*}
&\qquad\quad{\large{\int}}\frac{77x^{3}-177x^{2}+259x+801}{(x-3)(x-1)(49x+111)\sqrt{x^{3}-7x+10}}{\mathrm{d}}x\\
\\
&=\operatorname{artanh}\left(\frac{x^{2}-4x+7}{2\sqrt{x^{3}-7x+10}}\right)-\operatorname{artanh}\left(\frac{5x-7}{2\sqrt{x^{3}-7x+10}}\right)\\
&\qquad\qquad-\operatorname{artanh}\left(\frac{37x+31}{14\sqrt{x^{3}-7x+10}}\right)+C
\end{align*}

\begin{gather*}
{\large\int}\frac{143 x^2+9614 x+47231}{(x+1) (121 x+593) \sqrt{x^3+69 x+470}}{\mathrm{d}}x\\
\\
=4\operatorname{artanh}\left(\frac{11 x^2+4 x-207}{10 \sqrt{x^3+69 x+470}}\right)-2 \operatorname{artanh}\left(\frac{3 x+15}{\sqrt{x^3+69 x+470}}\right)+C
\end{gather*}



  1. \int_{x}^{1000}\frac{1}{\left(t-p\right)\sqrt{(t-a)((t-u)^{2}+v^{2})}}dt
  2. \frac{1}{(A-a+p)\sqrt{A}}\int_{0}^{\frac{2\sqrt{A(x-a)}}{x-a+A}}\left(\frac{1-\frac{A-a+p}{2(p-a)}}{1-\frac{(A-a+p)^{2}}{4A(p-a)}t^{2}}-1+\frac{\frac{(A-a+p)}{2(p-a)}\sqrt{1-t^{2}}}{1-\frac{(A-a+p)^{2}}{4A(p-a)}t^{2}}\right)\frac{1}{\sqrt{(1-t^{2})(1-\frac{A+u-a}{2A}t^{2})}}dt
  3. a=-\frac{(135-6\sqrt{249})^{1/3}+(135+6\sqrt{249})^{1/3}}{3}
  4. u=\frac{(135-6\sqrt{249})^{1/3}+(135+6\sqrt{249})^{1/3}}{6}
  5. v=\frac{(135+6\sqrt{249})^{1/3}-(135-6\sqrt{249})^{1/3}}{6}\sqrt{3}
  6. A=\sqrt{((a-u)^{2}+v^{2})}
  7. p=3
  8. \int_{x}^{1000}\frac{1}{\left(t-q\right)\sqrt{(t-a)((t-u)^{2}+v^{2})}}dt
  9. \frac{1}{(A-a+q)\sqrt{A}}\int_{0}^{\frac{2\sqrt{A(x-a)}}{x-a+A}}\left(\frac{1-\frac{A-a+q}{2(q-a)}}{1-\frac{(A-a+q)^{2}}{4A(q-a)}t^{2}}-1+\frac{\frac{(A-a+q)}{2(q-a)}\sqrt{1-t^{2}}}{1-\frac{(A-a+q)^{2}}{4A(q-a)}t^{2}}\right)\frac{1}{\sqrt{(1-t^{2})(1-\frac{A+u-a}{2A}t^{2})}}dt
  10. q=\frac{1}{4}
  11. \frac{d}{dx}\left(-\tanh^{-1}\left(\frac{5x-7}{2\sqrt{x^{3}-7x+10}}\right)-\tanh^{-1}\left(\frac{14\sqrt{x^{3}-7x+10}}{37x+31}\right)+\tanh^{-1}\left(\frac{2\sqrt{x^{3}-7x+10}}{x^{2}-4x+7}\right)\right)
  12. \frac{77x^{3}-177x^{2}+259x+801}{(x-3)(x-1)(49x+111)\sqrt{x^{3}-7x+10}}
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 Author| 青青子衿 Posted at 2025-3-5 14:17:11
Last edited by 青青子衿 at 2025-4-2 15:07:28
青青子衿 发表于 2025-2-22 00:20
\begin{align*}
&\qquad\quad{\large{\int}}\frac{77x^{3}-177x^{2}+259x+801}{(x-3)(x-1)(49x+111)\sqrt{x^{3}-7x+10}}{\mathrm{d}}x\\
\\
&=\operatorname{artanh}\left(\frac{x^{2}-4x+7}{2\sqrt{x^{3}-7x+10}}\right)-\operatorname{artanh}\left(\frac{5x-7}{2\sqrt{x^{3}-7x+10}}\right)\\
&\qquad\qquad-\operatorname{artanh}\left(\frac{37x+31}{14\sqrt{x^{3}-7x+10}}\right)+C
\end{align*}

\begin{align*}
\Xi_j&=\sqrt{\xi_j^3+A\xi_j+B}\qquad\qquad(j=1,2,3)\\
\Delta _j&=R(\xi_j,\Xi _j)\\
&=-\tfrac{3A^{4}+32AB^{2}+40A^{2}B\xi_{j}+44A^{3}\xi_{j}^{2}-48B^{2}\xi_{j}^{2}-176AB\xi_{j}^{3}+10A^{2}\xi_{j}^{4}-312B\xi_{j}^{5}-100A\xi_{j}^{6}-21\xi_{j}^{8}}{24 \left(A^3+8 B^2+4 A B \xi_{j}+5 A^2 \xi_{j}^2-20 B \xi_{j}^3-5 A \xi_{j}^4-\xi_{j}^6\right)\cdot\Xi_{j}}
\end{align*}

\begin{align*}
S_{1}&=\alpha_{1,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{1,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{1,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\

&=\alpha_{1,1}\Delta_{1}+\alpha_{1,2}\Delta_{2}+\alpha_{1,3}\Delta_{3}+\frac{\alpha_{1,1}\Xi_{1}}{\xi_{1}-x}+\frac{\alpha_{1,2}\Xi_{2}}{\xi_{2}-x}+\frac{\alpha_{1,3}\Xi_{3}}{\xi_{3}-x}\\
&=\frac{s_{1,0}+s_{1,1}x}{a_{0}+a_{1}x+a_{2}x^2+x^3}\\

S_{2}&=\alpha_{2,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{2,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{2,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\

&=\alpha_{2,1}\Delta_{1}+\alpha_{2,2}\Delta_{2}+\alpha_{2,3}\Delta_{3}+\frac{\alpha_{2,1}\Xi_{1}}{\xi_{1}-x}+\frac{\alpha_{2,2}\Xi_{2}}{\xi_{2}-x}+\frac{\alpha_{2,3}\Xi_{3}}{\xi_{3}-x}\\
&=\frac{s_{2,0}+s_{2,2}x^2}{a_{0}+a_{1}x+a_{2}x^2+x^3}\\

S_{3}&=\alpha_{3,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{3,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{3,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\

&=\alpha_{3,1}\Delta_{1}+\alpha_{3,2}\Delta_{2}+\alpha_{3,3}\Delta_{3}+\frac{\alpha_{3,1}\Xi_{1}}{\xi_{1}-x}+\frac{\alpha_{3,2}\Xi_{2}}{\xi_{2}-x}+\frac{\alpha_{3,3}\Xi_{3}}{\xi_{3}-x}\\
&=\frac{s_{3,0}+s_{3,3}x^3}{a_{0}+a_{1}x+a_{2}x^2+x^3}\\

\end{align*}

当然,不仅这些代数数有限制,而且给定的线性有理函数的系数也有限制.
先给定三个线性有理分式
\begin{align*}
\boldsymbol{\beta}^{\mathrm{T}}=\left(\Delta _1+\frac{\Xi _1}{\xi _1-x},\,\Delta _2+\frac{\Xi _2}{\xi _2-x},\,\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)
\end{align*}
其中
\begin{gather*}
\begin{split}
\Xi_j&=\sqrt{\xi_j^3+A\xi_j+B}\qquad\qquad(j=1,2,3)\\
\Delta _j&=R(\xi_j,\Xi _j)\\
&=-\tfrac{3A^{4}+32AB^{2}+40A^{2}B\xi_{j}+44A^{3}\xi_{j}^{2}-48B^{2}\xi_{j}^{2}-176AB\xi_{j}^{3}+10A^{2}\xi_{j}^{4}-312B\xi_{j}^{5}-100A\xi_{j}^{6}-21\xi_{j}^{8}}{24 \left(A^3+8 B^2+4 A B \xi_{j}+5 A^2 \xi_{j}^2-20 B \xi_{j}^3-5 A \xi_{j}^4-\xi_{j}^6\right)\cdot\Xi_{j}}\\
\\
\end{split}\\
(\xi_1-x)(\xi_2-x)(\xi_3-x)\in\mathbb{Q}[x]
\end{gather*}
设待求的代数数向量为
\begin{align*}
\boldsymbol{\alpha}_{l}^{\mathrm{T}}=\left(\alpha _{l,1},\alpha _{l,2},\alpha _{l,3}\right)
\end{align*}
待求的代数数要求是Q-共轭的,换句话说,这些代数数是同一个不可约整系数多项式的不同根,当给定的线性有理函数的系数分别在同构的三次域时,我们有一些试探的方法,不是一个严谨的算法.
具体例子如下
已知$\xi_1,\xi_2,\xi_3$是有理系数多项式(碰巧也是整系数)$f(x)=x^3+6 x^2+72 x+148$的根,作如下记号
\begin{align*}
\omega&=\frac{-1+i\sqrt{3}}{2},\\
\xi _1&= -2+2\cdot10^{1/3}-10^{2/3},\\
\xi _2&=-2 + 2\cdot10^{1/3}\omega-10^{2/3}\omega^2,\\
\xi _3&=-2 + 2\cdot10^{1/3}\omega^2-10^{2/3}\omega.\\
\end{align*}
取定$(A,B)=(-12,-11)$,利用$\Xi_{j}$的公式,可以得到
\begin{align*}
\Xi _1&=3 \left(5-2\cdot10^{1/3}\right),\\
\Xi _2&=3 \left(5-2\cdot10^{1/3}\omega\right),\\
\Xi _3&=3 \left(5-2\cdot10^{1/3}\omega^2\right).
\end{align*}
利用$\Delta _j$的公式,可以得到
\begin{align*}
\Delta _1&=\frac{1}{3} \left(3+10^{1/3}-10^{2/3}\right),\\
\Delta _2&=\frac{1}{3} \left(3+10^{1/3}\omega-10^{2/3}\omega^2\right),\\
\Delta _3&=\frac{1}{3} \left(3+10^{1/3}\omega^2-10^{2/3}\omega\right).
\end{align*}
接下来寻找另一个多项式,使得其分裂域与$\xi_{j}$极小多项式的分裂域同构
\begin{align*}
\alpha _{0,1}&=1+2\cdot10^{1/3}+3\cdot10^{2/3},\\
\alpha _{0,2}&=1+2\cdot10^{1/3}\omega+3\cdot10^{2/3}\omega^2,\\
\alpha _{0,3}&=1+2\cdot10^{1/3}\omega^2+3\cdot10^{2/3}\omega.
\end{align*}
这些$\alpha_{j}$的极小多项式为$x^3-3 x^2-177 x-2601$,用SageMath验证是同构的
x = polygen(QQ, 'x')
K = NumberField(148 + 72*x + 6*x^2 + x^3, 'a')
L = NumberField(-2601 - 177*x - 3*x^2 + x^3, 'b')
K.is_isomorphic(L, True)
于是
\begin{align*}
\boldsymbol{\alpha}_{0}^{\mathrm{T}}\boldsymbol{\beta}
&=\alpha_{0,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{0,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{0,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\
&=\frac{2 x^3+4 x^2+3 x-404}{x^3+6 x^2+72 x+148}
\end{align*}
另外,凑出了三组结果
\begin{align*}
\boldsymbol{\alpha}_{1}^{\mathrm{T}}\boldsymbol{\beta}&=\frac{x-8}{x^3+6 x^2+72 x+148}\\
\boldsymbol{\alpha}_{2}^{\mathrm{T}}\boldsymbol{\beta}&=\frac{x^2+26}{x^3+6 x^2+72 x+148}\\
\boldsymbol{\alpha}_{3}^{\mathrm{T}}\boldsymbol{\beta}&=\frac{x^3-242}{x^3+6 x^2+72 x+148}
\end{align*}

.




### Russian Translation:

Дано:
\begin{align*}
\Xi_j&=\sqrt{\xi_j^3+A\xi_j+B}\qquad\qquad(j=1,2,3)\\
\Delta _j&=R(\xi_j,\Xi _j)\\
&=-\tfrac{3A^{4}+32AB^{2}+40A^{2}B\xi_{j}+44A^{3}\xi_{j}^{2}-48B^{2}\xi_{j}^{2}-176AB\xi_{j}^{3}+10A^{2}\xi_{j}^{4}-312B\xi_{j}^{5}-100A\xi_{j}^{6}-21\xi_{j}^{8}}{24 \left(A^3+8 B^2+4 A B \xi_{j}+5 A^2 \xi_{j}^2-20 B \xi_{j}^3-5 A \xi_{j}^4-\xi_{j}^6\right)\cdot\Xi_{j}}
\end{align*}

\begin{align*}
S_{1}&=\alpha_{1,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{1,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{1,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\

&=\alpha_{1,1}\Delta_{1}+\alpha_{1,2}\Delta_{2}+\alpha_{1,3}\Delta_{3}+\frac{\alpha_{1,1}\Xi_{1}}{\xi_{1}-x}+\frac{\alpha_{1,2}\Xi_{2}}{\xi_{2}-x}+\frac{\alpha_{1,3}\Xi_{3}}{\xi_{3}-x}\\
&=\frac{s_{1,0}+s_{1,1}x}{a_{0}+a_{1}x+a_{2}x^2+x^3}\\

S_{2}&=\alpha_{2,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{2,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{2,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\

&=\alpha_{2,1}\Delta_{1}+\alpha_{2,2}\Delta_{2}+\alpha_{2,3}\Delta_{3}+\frac{\alpha_{2,1}\Xi_{1}}{\xi_{1}-x}+\frac{\alpha_{2,2}\Xi_{2}}{\xi_{2}-x}+\frac{\alpha_{2,3}\Xi_{3}}{\xi_{3}-x}\\
&=\frac{s_{2,0}+s_{2,2}x^2}{a_{0}+a_{1}x+a_{2}x^2+x^3}\\

S_{3}&=\alpha_{3,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{3,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{3,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\

&=\alpha_{3,1}\Delta_{1}+\alpha_{3,2}\Delta_{2}+\alpha_{3,3}\Delta_{3}+\frac{\alpha_{3,1}\Xi_{1}}{\xi_{1}-x}+\frac{\alpha_{3,2}\Xi_{2}}{\xi_{2}-x}+\frac{\alpha_{3,3}\Xi_{3}}{\xi_{3}-x}\\
&=\frac{s_{3,0}+s_{3,3}x^3}{a_{0}+a_{1}x+a_{2}x^2+x^3}\\

\end{align*}

Конечно, не только эти алгебраические числа имеют ограничения, но и коэффициенты заданных линейных рациональных функций также ограничены.  
Сначала заданы три линейные рациональные дроби:
\begin{align*}
\boldsymbol{\beta}^{\mathrm{T}}=\left(\Delta _1+\frac{\Xi _1}{\xi _1-x},\,\Delta _2+\frac{\Xi _2}{\xi _2-x},\,\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)
\end{align*}
где
\begin{gather*}
\begin{split}
\Xi_j&=\sqrt{\xi_j^3+A\xi_j+B}\qquad\qquad(j=1,2,3)\\
\Delta _j&=R(\xi_j,\Xi _j)\\
&=-\tfrac{3A^{4}+32AB^{2}+40A^{2}B\xi_{j}+44A^{3}\xi_{j}^{2}-48B^{2}\xi_{j}^{2}-176AB\xi_{j}^{3}+10A^{2}\xi_{j}^{4}-312B\xi_{j}^{5}-100A\xi_{j}^{6}-21\xi_{j}^{8}}{24 \left(A^3+8 B^2+4 A B \xi_{j}+5 A^2 \xi_{j}^2-20 B \xi_{j}^3-5 A \xi_{j}^4-\xi_{j}^6\right)\cdot\Xi_{j}}\\
\\
\end{split}\\
(\xi_1-x)(\xi_2-x)(\xi_3-x)\in\mathbb{Q}[x]
\end{gather*}
Пусть искомый вектор алгебраических чисел равен
\begin{align*}
\boldsymbol{\alpha}_{l}^{\mathrm{T}}=\left(\alpha _{l,1},\alpha _{l,2},\alpha _{l,3}\right)
\end{align*}
Требуется, чтобы искомые алгебраические числа были Q-сопряженными, другими словами, эти алгебраические числа являются различными корнями одного и того же неприводимого многочлена с целыми коэффициентами. Когда коэффициенты заданных линейных рациональных функций находятся в изоморфных кубических полях, у нас есть некоторые эвристические методы, но это не строгий алгоритм.  
Конкретный пример следующий:  
Известно, что $\xi_1,\xi_2,\xi_3$ являются корнями многочлена с рациональными (и даже целыми) коэффициентами $f(x)=x^3+6 x^2+72 x+148$. Введем обозначения:
\begin{align*}
\omega&=\frac{-1+i\sqrt{3}}{2},\\
\xi _1&= -2+2\cdot10^{1/3}-10^{2/3},\\
\xi _2&=-2 + 2\cdot10^{1/3}\omega-10^{2/3}\omega^2,\\
\xi _3&=-2 + 2\cdot10^{1/3}\omega^2-10^{2/3}\omega.\\
\end{align*}
Выбрав $(A,B)=(-12,-11)$ и используя формулу для $\Xi_{j}$, получаем:
\begin{align*}
\Xi _1&=3 \left(5-2\cdot10^{1/3}\right),\\
\Xi _2&=3 \left(5-2\cdot10^{1/3}\omega\right),\\
\Xi _3&=3 \left(5-2\cdot10^{1/3}\omega^2\right).
\end{align*}
Используя формулу для $\Delta _j$, получаем:
\begin{align*}
\Delta _1&=\frac{1}{3} \left(3+10^{1/3}-10^{2/3}\right),\\
\Delta _2&=\frac{1}{3} \left(3+10^{1/3}\omega-10^{2/3}\omega^2\right),\\
\Delta _3&=\frac{1}{3} \left(3+10^{1/3}\omega^2-10^{2/3}\omega\right).
\end{align*}
Далее ищем другой многочлен, чье поле разложения изоморфно полю разложения минимального многочлена для $\xi_{j}$:
\begin{align*}
\alpha _{0,1}&=1+2\cdot10^{1/3}+3\cdot10^{2/3},\\
\alpha _{0,2}&=1+2\cdot10^{1/3}\omega+3\cdot10^{2/3}\omega^2,\\
\alpha _{0,3}&=1+2\cdot10^{1/3}\omega^2+3\cdot10^{2/3}\omega.
\end{align*}
Минимальный многочлен для этих $\alpha_{j}$ равен $x^3-3 x^2-177 x-2601$. Проверим изоморфизм с помощью SageMath:
x = polygen(QQ, 'x')
K = NumberField(148 + 72*x + 6*x^2 + x^3, 'a')
L = NumberField(-2601 - 177*x - 3*x^2 + x^3, 'b')
K.is_isomorphic(L, True)
Тогда:
\begin{align*}
\boldsymbol{\alpha}_{0}^{\mathrm{T}}\boldsymbol{\beta}
&=\alpha_{0,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{0,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{0,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\
&=\frac{2 x^3+4 x^2+3 x-404}{x^3+6 x^2+72 x+148}
\end{align*}
Кроме того, подобраны три группы результатов:
\begin{align*}
\boldsymbol{\alpha}_{1}^{\mathrm{T}}\boldsymbol{\beta}&=\frac{x-8}{x^3+6 x^2+72 x+148}\\
\boldsymbol{\alpha}_{2}^{\mathrm{T}}\boldsymbol{\beta}&=\frac{x^2+26}{x^3+6 x^2+72 x+148}\\
\boldsymbol{\alpha}_{3}^{\mathrm{T}}\boldsymbol{\beta}&=\frac{x^3-242}{x^3+6 x^2+72 x+148}
\end{align*}

---

### English Translation:

Given:
\begin{align*}
\Xi_j&=\sqrt{\xi_j^3+A\xi_j+B}\qquad\qquad(j=1,2,3)\\
\Delta _j&=R(\xi_j,\Xi _j)\\
&=-\tfrac{3A^{4}+32AB^{2}+40A^{2}B\xi_{j}+44A^{3}\xi_{j}^{2}-48B^{2}\xi_{j}^{2}-176AB\xi_{j}^{3}+10A^{2}\xi_{j}^{4}-312B\xi_{j}^{5}-100A\xi_{j}^{6}-21\xi_{j}^{8}}{24 \left(A^3+8 B^2+4 A B \xi_{j}+5 A^2 \xi_{j}^2-20 B \xi_{j}^3-5 A \xi_{j}^4-\xi_{j}^6\right)\cdot\Xi_{j}}
\end{align*}

\begin{align*}
S_{1}&=\alpha_{1,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{1,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{1,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\

&=\alpha_{1,1}\Delta_{1}+\alpha_{1,2}\Delta_{2}+\alpha_{1,3}\Delta_{3}+\frac{\alpha_{1,1}\Xi_{1}}{\xi_{1}-x}+\frac{\alpha_{1,2}\Xi_{2}}{\xi_{2}-x}+\frac{\alpha_{1,3}\Xi_{3}}{\xi_{3}-x}\\
&=\frac{s_{1,0}+s_{1,1}x}{a_{0}+a_{1}x+a_{2}x^2+x^3}\\

S_{2}&=\alpha_{2,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{2,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{2,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\

&=\alpha_{2,1}\Delta_{1}+\alpha_{2,2}\Delta_{2}+\alpha_{2,3}\Delta_{3}+\frac{\alpha_{2,1}\Xi_{1}}{\xi_{1}-x}+\frac{\alpha_{2,2}\Xi_{2}}{\xi_{2}-x}+\frac{\alpha_{2,3}\Xi_{3}}{\xi_{3}-x}\\
&=\frac{s_{2,0}+s_{2,2}x^2}{a_{0}+a_{1}x+a_{2}x^2+x^3}\\

S_{3}&=\alpha_{3,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{3,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{3,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\

&=\alpha_{3,1}\Delta_{1}+\alpha_{3,2}\Delta_{2}+\alpha_{3,3}\Delta_{3}+\frac{\alpha_{3,1}\Xi_{1}}{\xi_{1}-x}+\frac{\alpha_{3,2}\Xi_{2}}{\xi_{2}-x}+\frac{\alpha_{3,3}\Xi_{3}}{\xi_{3}-x}\\
&=\frac{s_{3,0}+s_{3,3}x^3}{a_{0}+a_{1}x+a_{2}x^2+x^3}\\

\end{align*}

Of course, not only are these algebraic numbers constrained, but the coefficients of the given linear rational functions are also restricted.  
First, three linear rational fractions are given:
\begin{align*}
\boldsymbol{\beta}^{\mathrm{T}}=\left(\Delta _1+\frac{\Xi _1}{\xi _1-x},\,\Delta _2+\frac{\Xi _2}{\xi _2-x},\,\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)
\end{align*}
where
\begin{gather*}
\begin{split}
\Xi_j&=\sqrt{\xi_j^3+A\xi_j+B}\qquad\qquad(j=1,2,3)\\
\Delta _j&=R(\xi_j,\Xi _j)\\
&=-\tfrac{3A^{4}+32AB^{2}+40A^{2}B\xi_{j}+44A^{3}\xi_{j}^{2}-48B^{2}\xi_{j}^{2}-176AB\xi_{j}^{3}+10A^{2}\xi_{j}^{4}-312B\xi_{j}^{5}-100A\xi_{j}^{6}-21\xi_{j}^{8}}{24 \left(A^3+8 B^2+4 A B \xi_{j}+5 A^2 \xi_{j}^2-20 B \xi_{j}^3-5 A \xi_{j}^4-\xi_{j}^6\right)\cdot\Xi_{j}}\\
\\
\end{split}\\
(\xi_1-x)(\xi_2-x)(\xi_3-x)\in\mathbb{Q}[x]
\end{gather*}
Let the sought vector of algebraic numbers be
\begin{align*}
\boldsymbol{\alpha}_{l}^{\mathrm{T}}=\left(\alpha _{l,1},\alpha _{l,2},\alpha _{l,3}\right)
\end{align*}
The sought algebraic numbers must be Q-conjugate, meaning these algebraic numbers are different roots of the same irreducible integer-coefficient polynomial. When the coefficients of the given linear rational functions lie in isomorphic cubic fields, we have some heuristic methods, but this is not a rigorous algorithm.  
A specific example is as follows:  
It is known that $\xi_1,\xi_2,\xi_3$ are roots of the rational (and even integer) coefficient polynomial $f(x)=x^3+6 x^2+72 x+148$. We introduce the notation:
\begin{align*}
\omega&=\frac{-1+i\sqrt{3}}{2},\\
\xi _1&= -2+2\cdot10^{1/3}-10^{2/3},\\
\xi _2&=-2 + 2\cdot10^{1/3}\omega-10^{2/3}\omega^2,\\
\xi _3&=-2 + 2\cdot10^{1/3}\omega^2-10^{2/3}\omega.\\
\end{align*}
Choosing $(A,B)=(-12,-11)$ and using the formula for $\Xi_{j}$, we obtain:
\begin{align*}
\Xi _1&=3 \left(5-2\cdot10^{1/3}\right),\\
\Xi _2&=3 \left(5-2\cdot10^{1/3}\omega\right),\\
\Xi _3&=3 \left(5-2\cdot10^{1/3}\omega^2\right).
\end{align*}
Using the formula for $\Delta _j$, we obtain:
\begin{align*}
\Delta _1&=\frac{1}{3} \left(3+10^{1/3}-10^{2/3}\right),\\
\Delta _2&=\frac{1}{3} \left(3+10^{1/3}\omega-10^{2/3}\omega^2\right),\\
\Delta _3&=\frac{1}{3} \left(3+10^{1/3}\omega^2-10^{2/3}\omega\right).
\end{align*}
Next, we search for another polynomial whose splitting field is isomorphic to the splitting field of the minimal polynomial of $\xi_{j}$:
\begin{align*}
\alpha _{0,1}&=1+2\cdot10^{1/3}+3\cdot10^{2/3},\\
\alpha _{0,2}&=1+2\cdot10^{1/3}\omega+3\cdot10^{2/3}\omega^2,\\
\alpha _{0,3}&=1+2\cdot10^{1/3}\omega^2+3\cdot10^{2/3}\omega.
\end{align*}
The minimal polynomial for these $\alpha_{j}$ is $x^3-3 x^2-177 x-2601$. We verify the isomorphism using SageMath:
x = polygen(QQ, 'x')
K = NumberField(148 + 72*x + 6*x^2 + x^3, 'a')
L = NumberField(-2601 - 177*x - 3*x^2 + x^3, 'b')
K.is_isomorphic(L, True)
Then:
\begin{align*}
\boldsymbol{\alpha}_{0}^{\mathrm{T}}\boldsymbol{\beta}
&=\alpha_{0,1}\left(\Delta _1+\frac{\Xi _1}{\xi _1-x}\right)+\alpha_{0,2}\left(\Delta _2+\frac{\Xi _2}{\xi _2-x}\right)+\alpha_{0,3}\left(\Delta _3+\frac{\Xi _3}{\xi _3-x}\right)\\
&=\frac{2 x^3+4 x^2+3 x-404}{x^3+6 x^2+72 x+148}
\end{align*}
Additionally, three sets of results have been derived:
\begin{align*}
\boldsymbol{\alpha}_{1}^{\mathrm{T}}\boldsymbol{\beta}&=\frac{x-8}{x^3+6 x^2+72 x+148}\\
\boldsymbol{\alpha}_{2}^{\mathrm{T}}\boldsymbol{\beta}&=\frac{x^2+26}{x^3+6 x^2+72 x+148}\\
\boldsymbol{\alpha}_{3}^{\mathrm{T}}\boldsymbol{\beta}&=\frac{x^3-242}{x^3+6 x^2+72 x+148}
\end{align*}


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 Author| 青青子衿 Posted at 2025-4-12 23:30:56
Last edited by 青青子衿 at ereyesterday 18:06
\begin{align*}
\int_{x}^{+\infty}\frac{{\mathrm{d}}t}{(t^{3}+at+b)^{2/3}}
=
3\int_{4(3x^{2}+3x(x^{3}+ax+b)^{1/3}+3(x^{3}+ax+b)^{2/3}+a)}^{+\infty}\frac{{\mathrm{d}}t}{\sqrt{t^{3}-64a^{3}-432b^{2}}}
\end{align*}


\begin{align*}
\int_{0}^{1}\frac{{\mathrm{d}}t}{\sqrt[4]{t(t+4)(t^{2}+12t+48)}}=-\frac{\pi}{12}-\frac{1}{6}\arctan\left(\tfrac{3(2867+75\sqrt{305})\sqrt[4]{305}}{19520}\right)+\frac{1}{6}\operatorname{arctanh}\left(\tfrac{(375+47\sqrt{305})\sqrt[4]{305}}{4998}\right)
\end{align*}

\begin{align*}
\int_{0}^{\frac{1}{4}}\frac{{\mathrm{d}}t}{\sqrt[4]{t(t+1)(t^{2}+3t+3)}}=\frac{2}{3}\operatorname{arctanh}\left(\frac{\sqrt[4]{305}}{5}\right)-\frac{2}{3}\arctan\left(\frac{\sqrt[4]{305}}{5}\right)
\end{align*}

\begin{align*}
{\Large\int}\frac{\mathrm{d}x}{\sqrt[4]{{x^{4}-4x^{2}+4({\small1-\sqrt{3}}\>\!)x-{\small4+2\sqrt{3}}}} }
\end{align*}

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