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

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    $\begingroup$ $ \int \sqrt{\tanh (x)} \, dx = \tanh ^{-1}\left(\sqrt{\tanh (x)}\right)-\tan ^{-1}\left(\sqrt{\tanh (x)}\right)$ $\endgroup$ – alfC Sep 25 '13 at 7:24
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    $\begingroup$ Here's something along the same lines, at least for the cube root. math.stackexchange.com/questions/479865/… $\endgroup$ – Ron Gordon Sep 25 '13 at 11:35
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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

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

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    $\begingroup$ In fact OP want to ask whether $\int\dfrac{u^3}{u^6+1}~du$ and $\int\dfrac{u^4}{u^8+1}~du$ have some trickly approaches beside directly taking partial functions. Given that for example $\int\dfrac{u^2}{u^4+1}~du$ has (for example in math.stackexchange.com/questions/425603). $\endgroup$ – Harry Peter Mar 19 '17 at 8:05

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