Integrate :$\int\frac{1}{\sqrt[3]{\tan(x)}}\, \mathrm dx$ How to integrate 
$$\int\dfrac{1}{\sqrt[3]{\tan(x)}}\, \mathrm dx?$$

Sub $u=\tan{x}$ and get
$$\int du \frac{u^{-1/3}}{1+u^2}$$
Then sub $u=y^3$ to get
$$3 \int dy \frac{y}{1+y^6} = \frac{3}{2} \int \frac{dv}{1+v^3} = \frac12 \int \frac{dv}{1+v} + \int \frac{dv}{1-v+v^2} - \frac12 \int dv \frac{v}{1-v+v^2} $$
Each of these integrals may be evaluated in turn.
$$\int \frac{dv}{1+v} = \log{(1+v)}$$
$$\int \frac{dv}{1-v+v^2} = \int \frac{dv}{(v-1/2)^2+3/4} = \frac{2}{\sqrt{3}} \arctan{\frac{2 v-1}{\sqrt{3}}}$$
$$\int dv \frac{v}{1-v+v^2} = \int dv \frac{v-1/2}{(v-1/2)^2+3/4} +\frac12 \int \frac{dv}{(v-1/2)^2+3/4} = \\\frac12 \log{[(v-1/2)^2+3/4]} + \frac{1}{\sqrt{3}} \arctan{\frac{2 v-1}{\sqrt{3}}}$$
I get, putting this all together,
$$\int dx \, (\tan{x} )^{-1/3} = \frac12 \log{\frac{1+v}{1-v+v^2}} + \frac{\sqrt{3}}{2} \arctan{\frac{2 v-1}{\sqrt{3}}} +C$$
where $v=(\tan{x})^{-2/3}$.
