Indefinite Integral $\int\sqrt[3]{\tan(x)}~dx$ For calculating $\int\sqrt{\tan(x)}~dx$, I used this easy method
$$\begin{align}\int\sqrt{\tan(x)}~dx&=\frac{1}{2}\int\left(\sqrt{\tan(x)}+\sqrt{\cot(x)}\right)dx+\frac{1}{2}\int\left(\sqrt{\tan(x)}-\sqrt{\cot(x)}\right)dx\\&=\frac{1}{2}\int\frac{\sin(x)+\cos(x)}{\sqrt{\sin(x)\cos(x)}}~dx-\frac{1}{2}\int\frac{\cos(x)-\sin(x)}{\sqrt{\sin(x)\cos(x)}}~dx\\&=\frac{\sqrt{2}}{2}\int\frac{du}{\sqrt{1-u^2}}-\frac{\sqrt{2}}{2}\int\frac{dv}{\sqrt{v^2-1}}.\end{align}$$$$u=\sin(x)-\cos(x), v=\sin(x)+\cos(x)$$
Does there exist an easy method for $\int\sqrt[3]{\tan(x)}~dx$?
 A: If you assume $\tan(x)=u^3$, then

$$\int (\tan(x))^{1/3}dx = 3\,\int \!{\frac {{u}^{3}}{{u}^{6}+1}}{du}. $$

For the other one, you can assume $ \tan(x)=u^4 $ to get

$$\int (\tan(x))^{1/4}dx  = 4\,\int \!{\frac {{u}^{4}}{{u}^{8}+1}}{du}. $$

Now, you can use some integration techniques to evaluate the integrals. Note that, for the integral you already did, you can assume $\tan(x)=u^2$ to get

$$ = \int (\tan(x))^{1/2} dx =  2\,\int \!{\frac {{u}^{2}}{{u}^{4}+1}}{du}. $$

Note: When you use these substitutions you need the identity

$$ \sec^2(x) = 1+\tan^2(x). $$

A: $\displaystyle \int\sqrt[3]{tanx}dx$
$\displaystyle tanx=u^{\frac{3}{2}}\Rightarrow (1+tan^{2}x)dx=\frac{3}{2}u^{\frac{1}{2}}du$
$\displaystyle\sqrt[3]{tanx}dx=\frac{3}{2}\frac{udu}{1+u^{3}}$
$\displaystyle\frac{u}{u^{3}+1}=\frac{1}{3}(\frac{u}{u+1}-\frac{u(u-2)}{u^{2}-u+1})$
$\displaystyle\frac{u}{u+1}=1-\frac{1}{u+1}$
$\displaystyle\frac{u(u-2)}{u^{2}-u+1}=1-\frac{2u-1}{2(u^{2}-u+1)}-\frac{3}{2(u^{2}-u+1)}$
$\displaystyle\frac{1}{u^{2}-u+1}=\frac{\frac{4}{3}}{(\frac{2}{\sqrt{3}}(u-1))^{2}+1}$
$\displaystyle\frac{3}{2}\frac{u}{u^{3}+1}=\frac{2u-1}{4(u^{2}-u+1)}+\frac{1}{(\frac{2}{\sqrt{3}}(u-1))^{2}+1}-\frac{1}{2(u+1)}$
$\displaystyle\frac{3}{2}\int\frac{udu}{u^{3}+1}=\frac{1}{4}ln\left| u^{2}-u+1 \right|+\frac{\sqrt{3}}{2}$
$\displaystyle\arctan{(\frac{2}{\sqrt{3}}(u-1))}-\frac{1}{2}ln\left| u+1 \right|+c$
$\displaystyle \int\sqrt[3]{tanx}dx=\frac{1}{4}ln\left|\sqrt[3]{ tan^{4}x}-\sqrt[3]{tan^{2}(x)}+1 \right|+\frac{\sqrt{3}}{2}$
$\displaystyle\arctan{(\frac{2}{\sqrt{3}}(\sqrt[3]{tan^{2}(x)}-1))}-\frac{1}
{2}ln\left|\sqrt[3]{tan^{2}(x)}+1 \right|+c$
A: Maple says,
$$
\int \!(\tan \left( x \right))^{1/3} \,{\rm d}x={\frac {1}{4}\ln 
 \left(  \left( \tan \left( x \right)  \right) ^{{\frac{4}{3}}}-
 \left( \tan \left( x \right)  \right) ^{{\frac{2}{3}}}+1 \right) }+{
\frac {\sqrt {3}}{2}\arctan \left( {\frac {\sqrt {3}}{3} \left( 2\,
 \left( \tan \left( x \right)  \right) ^{2/3}-1 \right) } \right) }-{
\frac {1}{2}\ln  \left(  \left( \tan \left( x \right)  \right) ^{{
\frac{2}{3}}}+1 \right) }
$$
But note: this assumes the principal cube root, so it is real only for $\tan x \ge 0$.
