tanの平方根の積分

\begin{align*} \int\sqrt{\tan x}dx & =2\int\frac{t^{2}}{t^{4}+1}dt\cmt{t=\sqrt{\tan x}}\\ & =\frac{1}{\sqrt{2}}\int t^{2}\left(-\frac{t-\sqrt{2}}{t^{2}-\sqrt{2}t+1}+\frac{t+\sqrt{2}}{t^{2}+\sqrt{2}t+1}\right)dt\\ & =\frac{1}{\sqrt{2}}\int\left(-t+\frac{t}{t^{2}-\sqrt{2}t+1}+t-\frac{t}{t^{2}+\sqrt{2}t+1}\right)dt\\ & =\frac{1}{\sqrt{2}}\int\left(\frac{t}{t^{2}-\sqrt{2}t+1}-\frac{t}{t^{2}+\sqrt{2}t+1}\right)dt\\ & =\frac{1}{\sqrt{2}}\int\left(\frac{\frac{1}{2}\left(t^{2}-\sqrt{2}t+1\right)'}{t^{2}-\sqrt{2}t+1}+\frac{\sqrt{2}}{2}\frac{1}{t^{2}-\sqrt{2}t+1}-\frac{\frac{1}{2}\left(t^{2}+\sqrt{2}t+1\right)'}{t^{2}+\sqrt{2}t+1}+\frac{\sqrt{2}}{2}\frac{1}{t^{2}+\sqrt{2}t+1}\right)dt\\ & =\frac{1}{\sqrt{2}}\int\left(\frac{\frac{1}{2}\left(t^{2}-\sqrt{2}t+1\right)'}{t^{2}-\sqrt{2}t+1}+\frac{\sqrt{2}}{2}\frac{1}{\left(t-\frac{\sqrt{2}}{2}\right)^{2}+\frac{1}{2}}-\frac{\frac{1}{2}\left(t^{2}+\sqrt{2}t+1\right)'}{t^{2}+\sqrt{2}t+1}+\frac{\sqrt{2}}{2}\frac{1}{\left(t+\frac{\sqrt{2}}{2}\right)^{2}+\frac{1}{2}}\right)dt\\ & =\frac{1}{\sqrt{2}}\left(\frac{1}{2}\log\left(t^{2}-\sqrt{2}t+1\right)+\frac{\sqrt{2}}{2}\sqrt{2}\tan^{\circ}\left(\sqrt{2}t-1\right)-\frac{1}{2}\log\left(t^{2}+\sqrt{2}t+1\right)+\frac{\sqrt{2}}{2}\sqrt{2}\tan^{\circ}\left(\sqrt{2}t+1\right)\right)+C\\ & =\frac{\sqrt{2}}{4}\log\left(t^{2}-\sqrt{2}t+1\right)-\frac{\sqrt{2}}{4}\log\left(t^{2}+\sqrt{2}t+1\right)+\frac{\sqrt{2}}{2}\tan^{\circ}\left(\sqrt{2}t-1\right)+\frac{\sqrt{2}}{2}\tan^{\circ}\left(\sqrt{2}t+1\right)+C\\ & =\frac{\sqrt{2}}{4}\log\left(\tan x-\sqrt{2\tan x}+1\right)-\frac{\sqrt{2}}{4}\log\left(\tan x+\sqrt{2\tan x}+1\right)+\frac{\sqrt{2}}{2}\tan^{\circ}\left(\sqrt{2\tan x}-1\right)+\frac{\sqrt{2}}{2}\tan^{\circ}\left(\sqrt{2\tan x}+1\right)+C \end{align*}