Evaluation of $\int\sqrt[4]{\tan x}dx$ Evaluation of $\displaystyle \int\sqrt[4]{\tan x}dx$
$\bf{My\; Try}::$ Let $\tan x = t^4\;\;,$ Then $\sec^2 xdx = 4t^3dt$. So $\displaystyle dx = \frac{4t^3}{1+t^8}dt$
So Integral Convert into $\displaystyle 4\int\frac{t^4}{1+t^8}dt = 2\int \frac{(t^4+1)+(t^4-1)}{t^8+1}dt$
So Integral is $\displaystyle 2\int\frac{t^4+1}{t^8+1}dt+2\int\frac{t^4-1}{t^4+1}dt$
Now How can I solve after that
Help me
Thanks

Transforming the integral into an integral of rational function by letting $y^{4}=\tan x$, then
$$
I=4 \int \frac{y^{4}}{1+y^{8}} d y
$$
By my post,
$$ \displaystyle \int \frac{x^{4}}{1+x^{8}} d x =\frac{1}{4 \sqrt{2}}\left[\frac{1}{2 \sqrt{2+\sqrt{2}}} \ln \left|\frac{x^{2}-\sqrt{2+\sqrt{2}} x+1}{x^{2}+\sqrt{2+\sqrt{2}} x+1}\right|+\frac{1}{\sqrt{2-\sqrt{2}}} \tan^{-1}\left(\frac{x^{2}-1}{\sqrt{2-\sqrt{2}} x}\right)\\ \quad  -  \frac{1}{2 \sqrt{2-\sqrt{2}}} \ln \left|\frac{x^{2}-\sqrt{2-\sqrt{2}} x+1}{x^{2}+\sqrt{2-\sqrt{2} x}+1}\right|-\frac{1}{\sqrt{2+\sqrt{2}}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2+\sqrt{2}} x}\right)\right]+C$$
Now we can conclude that
$$ \displaystyle \int \sqrt[4]{\tan x} d x =\frac{1}{ \sqrt{2}}\left[\frac{1}{2 \sqrt{2+\sqrt{2}}} \ln \left|\frac{\sqrt{\tan x} -\sqrt{2+\sqrt{2}} \sqrt[4]{\tan x} +1}{\sqrt{\tan x} +\sqrt{2+\sqrt{2}} \sqrt[4]{\tan x} +1}\right|+\frac{1}{\sqrt{2-\sqrt{2}}} \tan^{-1}\left(\frac{\sqrt{\tan x}-1}{\sqrt{2-\sqrt{2}} \sqrt[4]{\tan x}}\right)\\ -  \frac{1}{2 \sqrt{2-\sqrt{2}}} \ln \left|\frac{\sqrt{\tan x}-\sqrt{2-\sqrt{2}} \sqrt[4]{\tan x} +1}{\sqrt{\tan x} +\sqrt{2-\sqrt{2} \sqrt[4]{\tan x}}+1}\right|-\frac{1}{\sqrt{2+\sqrt{2}}} \tan ^{-1}\left(\frac{\sqrt{\tan x}-1}{\sqrt{2+\sqrt{2}} \sqrt[4]{\tan x}}\right)\right]+C$$
:|D Wish you enjoy the solution!
