How to integrate $\int (\tan x)^{1/ 6} \,\text{d}x$? How do I compute the following integral

$$
I=\int (\tan x)^{1/ 6} \,\text{d}x
$$

 A: The final answer maybe complicated as mentioned in the comments. But here are the steps that you may take to get there.
First use  the substitution $\tan x=u$ and then use another substitution $u=t^6$ to get
$$\begin{align}
I &= \int (\tan x)^{\frac 16} dx \\
&= \int \frac{u^{\frac 16}}{1+u^2} du \\
&= \int \frac{5t^6}{1+t^{12}} dt \\
&= 5 \int \frac{t^6}{1+t^{12}} dt
\end{align}$$
For tackling down the last integral you should use partial fractions. You may wonder that the denominator has no real roots. The answer is that the partial fraction decomposition works for complex roots as well.
A: Given you said you "won't even try" to solve this integral, try Mathematica:
$$\frac{-2 \left(\sqrt{3}-1\right) \tan ^{-1}\left(\frac{-2 \sqrt{2} \sqrt[6]{\tan
   (x)}+\sqrt{3}-1}{1+\sqrt{3}}\right)+4 \tan ^{-1}\left(1-\sqrt{2} \sqrt[6]{\tan
   (x)}\right)-4 \tan ^{-1}\left(\sqrt{2} \sqrt[6]{\tan (x)}+1\right)+2
   \left(\sqrt{3}-1\right) \tan ^{-1}\left(\frac{2 \sqrt{2} \sqrt[6]{\tan
   (x)}+\sqrt{3}-1}{1+\sqrt{3}}\right)-2 \left(1+\sqrt{3}\right) \tan ^{-1}\left(-2
   \sqrt{2+\sqrt{3}} \sqrt[6]{\tan (x)}+\sqrt{3}+2\right)+2 \left(1+\sqrt{3}\right) \tan
   ^{-1}\left(\left(\sqrt{2}+\sqrt{6}\right) \sqrt[6]{\tan (x)}+\sqrt{3}+2\right)-2 \log
   \left(\sqrt[3]{\tan (x)}-\sqrt{2} \sqrt[6]{\tan (x)}+1\right)+2 \log
   \left(\sqrt[3]{\tan (x)}+\sqrt{2} \sqrt[6]{\tan (x)}+1\right)-\left(1+\sqrt{3}\right)
   \log \left(2 \sqrt[3]{\tan (x)}+\sqrt{2} \left(\sqrt{3}-1\right) \sqrt[6]{\tan
   (x)}+2\right)+\left(\sqrt{3}-1\right) \log \left(2 \sqrt[3]{\tan (x)}-2
   \sqrt{2+\sqrt{3}} \sqrt[6]{\tan (x)}+2\right)+\left(1+\sqrt{3}\right) \log \left(2
   \sqrt[3]{\tan (x)}+\left(\sqrt{2}-\sqrt{6}\right) \sqrt[6]{\tan
   (x)}+2\right)-\left(\sqrt{3}-1\right) \log \left(2 \sqrt[3]{\tan
   (x)}+\left(\sqrt{2}+\sqrt{6}\right) \sqrt[6]{\tan (x)}+2\right)}{4 \sqrt{2}}$$
...just in case you need the answer.
A: HINT (the integral will become very very big):
$$\int\left(\tan(x)\right)^{\frac{1}{6}}\space\text{d}x=\int\sqrt[6]{\tan(x)}\space\text{d}x=$$

Substitute $u=\tan(x)$ and $\frac{\text{d}u}{\text{d}x}=\sec^2(x)$:

$$\int\frac{\sqrt[6]{u}}{1+u^2}\space\text{d}u=$$

Substitute $s=\sqrt[6]{u}$ and $\frac{\text{d}s}{\text{d}u}=\frac{1}{6\sqrt[6]{u^5}}$:

$$\int\frac{5s^6}{1+s^{12}}\space\text{d}s=5\int\frac{s^6}{1+s^{12}}\space\text{d}s=$$
$$5\int\left[\frac{s^6+s^2}{3(s^8-s^4+1)}-\frac{s^2}{3(s^4+1)}\right]\space\text{d}s=$$
$$5\left[\int\frac{s^6+s^2}{3(s^8-s^4+1)}\space\text{d}s-\int\frac{s^2}{3(s^4+1)}\space\text{d}s\right]=$$
$$5\left[\frac{1}{3}\int\frac{s^6+s^2}{s^8-s^4+1}\space\text{d}s-\frac{1}{3}\int\frac{s^2}{s^4+1}\space\text{d}s\right]=$$
$$5\left[\frac{1}{3}\int\frac{s^6+s^2}{s^8-s^4+1}\space\text{d}s-\frac{1}{3}\int\left[-\frac{s}{2\sqrt{2}\left(-s^2+\sqrt{2}s-1\right)}-\frac{s}{2\sqrt{2}\left(s^2+\sqrt{2}s+1\right)}\right]\space\text{d}s\right]$$
