Indefinite Integral $\int\sqrt[3]{\tan(x)}dx$

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

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