$1/\sqrt[3]{x^3-3 x^2-3 x-1}$不定积分

hbghlyj

blog.wolfram.com/2021/08/18/new-methods-for-c … algebraic-integrals/
For example, the following integral possesses an elementary (albeit enormous) solution:$$\int \frac{d x}{\sqrt[3]{x^3-3 x^2-3 x-1}}$$

ljh25252

$$
\begin{aligned}
\int\frac{\mathrm{d}x}{\sqrt[3]{x^3-3x^2-3x-1}}&=\frac{3}{2}\int\frac{1+2x}{3x\sqrt[3]{x^3-3x^2-3x-1}}\mathrm{d}x-\frac{1}{2}\int\frac{1}{x\sqrt[3]{x^3-3x^2-3x-1}}\mathrm{d}x\\
&=\frac{3}{2}\int\frac{\mathrm{d}\left(\frac{x-1}{\sqrt[3]{x^3-3x^2-3x-1}}\right)}{1-\left(\frac{x-1}{\sqrt[3]{x^3-3x^2-3x-1}}\right)^3}-\frac{1}{2}\int\frac{\mathrm{d}\left(\frac{-x-1}{\sqrt[3]{x^3-3x^2-3x-1}}\right)}{1-\left(\frac{-x-1}{\sqrt[3]{x^3-3x^2-3x-1}}\right)^3}\\
&=\frac{3}{2}\int\frac{\mathrm{d}p}{1-p^3}-\frac{1}{2}\int\frac{\mathrm{d}q}{1-q^3}\\
&\color{gray}{p=\frac{x-1}{\sqrt[3]{x^3-3x^2-3x-1}}}\quad \color{gray}{q=\frac{-x-1}{\sqrt[3]{x^3-3x^2-3x-1}}}
\\
\color{gray}{\int\frac{\mathrm{d}z}{1-z^3}}&\color{gray}{=\frac{1}{\sqrt{3}}\arctan\frac{1+2z}{\sqrt{z}}+\frac{1}{3}\ln\frac{\sqrt{1+z+z^2}}{1-z}}\\
\end{aligned}
$$

青青子衿

ljh25252 发表于 2024-12-26 17:48
\begin{align*}
\int\frac{\mathrm{d}x}{\sqrt[3]{x^3-3x^2-3x-1}}&=\frac{3}{2}\int\frac{1+2x}{3x\sqrt[3]{x^3-3x^2-3x-1}}\mathrm{d}x-\frac{1}{2}\int\frac{1}{x\sqrt[3]{x^3-3x^2-3x-1}}\mathrm{d}x\\
\end{align*}

下面这个好像也是初等的，有办法解决吗？
\begin{align*}
\int\frac{\mathrm{d}x}{\sqrt[3]{x^3-3x^2-6x-1}}
\end{align*}

青青子衿

青青子衿 发表于 2025-4-10 21:15
下面这个好像也是初等的，有办法解决吗？
\begin{align*}

\int\frac{\mathrm{d}x}{\sqrt[3]{x^3-3x^2-6x-1}}

\end{align*}

凑出来了

\begin{align*}
\int\frac{\mathrm{d}x}{\sqrt[3]{x^3-3x^2-3x-1}}
&=\frac{3}{2}{\large\int}\frac{\mathrm{d}\left(\frac{x-1}{\sqrt[3]{x^3-3 x^2-3 x-1}}\right)}{1-\left(\frac{x-1}{\sqrt[3]{x^3-3 x^2-3 x-1}}\right)^3}\\
&\qquad\qquad
+
\frac{1}{2}{\large\int}\frac{\mathrm{d}\left(\frac{x+1}{\sqrt[3]{x^3-3 x^2-3 x-1}}\right)}{1+\left(\frac{x+1}{\sqrt[3]{x^3-3 x^2-3 x-1}}\right)^3}\\

\int\frac{\mathrm{d}x}{\sqrt[3]{x^3-3x^2-6x-1}}
&=\frac{4}{3}{\large\int}\frac{\mathrm{d}\left(\frac{x-1}{\sqrt[3]{x^3-3 x^2-6 x-1}}\right)}{1-\left(\frac{x-1}{\sqrt[3]{x^3-3 x^2-6 x-1}}\right)^3}\\
&\qquad\qquad+
\frac{1}{3}{\large\int}\frac{\mathrm{d}\left(\frac{x+2}{\sqrt[3]{x^3-3 x^2-6 x-1}}\right)}{1-\left(\frac{x+2}{\sqrt[3]{x^3-3 x^2-6 x-1}}\right)^3}\\
&\qquad\qquad\qquad\quad+\frac{2}{3}{\large\int}\frac{\mathrm{d}\left(\frac{2 x+1}{\sqrt[3]{x^3-3 x^2-6 x-1}}\right)}{1+\left(\frac{2 x+1}{\sqrt[3]{x^3-3 x^2-6 x-1}}\right)^3}\\
\end{align*}

1. int(convert(1/(x^3 - 3*x^2 - 3*x - 1)^(1/3), RootOf), x)
2. int(convert(1/(x^3 - 3*x^2 - 6*x - 1)^(1/3), RootOf), x)

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1. 3/2*D[(-1 + x)/(-1 - 3 x - 3 x^2 + x^3)^(1/3), x]/(
2.    1 - ((-1 + x)/(-1 - 3 x - 3 x^2 + x^3)^(1/3))^3) +
3.   1/2*D[(1 + x)/(-1 - 3 x - 3 x^2 + x^3)^(1/3), x]/(
4.    1 + ((1 + x)/(-1 - 3 x - 3 x^2 + x^3)^(1/3))^3) // Factor
5. 4/3*D[(-1 + x)/(-1 - 6 x - 3 x^2 + x^3)^(1/3), x]/(
6.    1 - ((-1 + x)/(-1 - 6 x - 3 x^2 + x^3)^(1/3))^3) +
7.   1/3*D[(2 + x)/(-1 - 6 x - 3 x^2 + x^3)^(1/3), x]/(
8.    1 - ((2 + x)/(-1 - 6 x - 3 x^2 + x^3)^(1/3))^3) +
9.   2/3*D[(1 + 2 x)/(-1 - 6 x - 3 x^2 + x^3)^(1/3), x]/(
10.    1 + ((1 + 2 x)/(-1 - 6 x - 3 x^2 + x^3)^(1/3))^3) // Factor

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