二重嵌套根式的正切函数不定积分

青青子衿

\begin{align*}
\int\sqrt{1+\tan x}\,\mathrm{d}x
&\xrightarrow[]{\quad\,t\,=\,\tan x\quad}
\int\dfrac{\sqrt{1+t}}{\,\,1+t^2}\mathrm{d}t\\
&\xrightarrow[]{\quad\,u\,=\,\sqrt{1+t\,}\quad}
\int\dfrac{2u^2}{1+\left(u^2-1\right)^2}\mathrm{d}u
\end{align*}
\begin{gather*}
\int\dfrac{2u^2}{1+\left(u^2-1\right)^2}\mathrm{d}u\\
=
\frac{1}{\sqrt{2\left(\sqrt{2}\color{blue}{-}1\right)}}\arctan\left(\frac{u\color{blue}{-}\frac{\sqrt{2}}{u}}{\sqrt{2\left(\sqrt{2}\color{blue}{-}1\right)}}\right)-\frac{1}{\sqrt{2\left(\sqrt{2}\color{gold}{+}1\right)}}\operatorname{arccoth}\left(\frac{u\color{gold}{+}\frac{\sqrt{2}}{u}}{\sqrt{2\left(\sqrt{2}\color{gold}{+}1\right)}}\right)+C
\end{gather*}

1. \int_0^{\frac{\pi}{2}}\sqrt{1+\tan x}\,\mathrm{d}x\\
2. =\frac{\pi}{2}\sqrt{\frac{\sqrt{2}+1}{2}}
3. +\sqrt{\frac{\sqrt{2}+1}{2}}\arctan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\,\right)
4. +\sqrt{\frac{\sqrt{2}-1}{2}}\operatorname{arccoth}\left(\sqrt{\frac{\sqrt{2}+1}{2}}\,\right)

复制代码

\begin{align*}
\int\sqrt{1+\sqrt{1+\tan x}}\,\mathrm{d}x
&\xrightarrow[]{\quad\,t\,=\,\tan x\quad}
\int\dfrac{\sqrt{1+\sqrt{1+t}\,}}{\,\,1+t^2}\mathrm{d}t\\
&\xrightarrow[]{\quad\,u\,=\,\sqrt{1+t\,}\quad}
\int\dfrac{2u\sqrt{1+u}}{1+\left(u^2-1\right)^2}\mathrm{d}u\\
&\xrightarrow[]{\quad\,v\,=\,\sqrt{1+u\,}\quad}
\int\dfrac{4v^2\left(v^2-1\right)}{1+\left(\left(v^2-1\right)^2-1\right)^2}\mathrm{d}v
\end{align*}

\begin{array}{}
\bigstar\circ\bullet\bigodot{\normalsize\boldsymbol{\bigcirc}}\bigcirc
\end{array}

Infinity

结果很复杂

1. int = Integrate[Sqrt[1 + Sqrt[1 + t]]/(1 + t^2), t];
2. Simplify[Re[ComplexExpand[int]] //. (f : Re | Im)[x__] :>
3.    ComplexExpand[f[x]], t > -1]//ToRadicals

复制代码