Evaluating $\int \tan^{1/3}(\theta) d \theta$ I know $\int \tan^{1/3}\theta  d \theta$ is a non integrable but wolfram alpha does it easily. And can anyone explain how a fuction is non integrable? My argument is that there has to be a function which represents the area under a graph. It's just that we do not know it. Please shed a bit light on it. Also, how can we solve $\int \tan^{1/3}\theta  d \theta$?
 A: Set $x=\left(\tan\theta\right)^{\frac23}$, then $xdx=\frac23\left(1+x^3\right)\left(\tan\theta\right)^{\frac13}d\theta$, and therefore
$$\int\left(\tan\theta\right)^{\frac13}d\theta=\frac32\int\frac{x\,dx}{x^3+1}.$$
The rest should be easy.
A: I know a way which can solve such these kinds of integral so examine it for $\sqrt[3]{\tan(x)}$. If you have $$\int\sin^p(x)\cos^q(x)dx$$ where $p,~q$ are rationals so, by taking  $t=\sin(t)$ you'll have $$\int t^p(1-t^2)^{q-1}dt~$$ Now consult this method (however I think it is an old way) for the latter integral. If the conditions, illustrated in this method, are well satisfied so the integral can be expressed as elementary functions so we can find the anti-derivative. Unless, I think we can not solve the integral by routine approaches. 
A: I will just leave here the exact solution.\begin{align}
\int (\tan x)^\frac13\,dx=&\frac14 \bigg[-2\sqrt{3}  \tan^{-1}\left(\sqrt{3}-2(\tan x)^\frac13 \right)-2\sqrt{3}\tan^{-1}\left(\sqrt{3}+2(\tan x)^\frac13  \right)\\
&-2\log\left(\tan^\frac23x+1\right)+\log\left(\tan^\frac23x -\sqrt{3}(\tan x)^\frac13+1   \right)\\
&\hspace{5mm}+\log\left(\tan^\frac23x+\sqrt{3}(\tan x)^\frac13+1      \right)    \bigg].
\end{align}
A: By enforcing the substitutions $x=\arctan(t)$, then $t=u^3$, we get:
$$ \int\left(\tan x\right)^{1/3}\,dx = \int\frac{t^{1/3} dt}{1+t^2}=\int\frac{3u^3}{1+u^6}\,du $$
and the last integral can be computed through a partial fraction decomposition. We have:
$$ \int\frac{3u^3}{1+u^6}\,du =C+\frac{\sqrt{3}}{2}\arctan\left(\frac{2u^2-1}{\sqrt{3}}\right)+\frac{1}{4}\log\left(\frac{1+u^6}{(1+u^2)^3}\right),$$
not that simple.
