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 $\int\sqrt[3]{\tan(x)}~dx$?

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 $\tan x \ge 0$.
