# What is the indefinite integral of $\sqrt[2n] {\tan \left( x\right) }$

I have already solved the following integrals $\sqrt {\tan \left( x\right) }$ and $\sqrt[4] {\tan \left( x\right) }$ (the last one with some help) so I want to know if it's possible to have a solution for the general case.

• I have not much idea about it, but I think it's not possible
– ashi
Apr 12 '16 at 4:05

Let us consider $$I_n=\int \sqrt[2 n]{\tan (x)}\,dx$$ Change variable $$\sqrt[2 n]{\tan (x)}=u\implies x=\tan ^{-1}\left(u^{2 n}\right)\implies dx=\frac{2 n u^{2 n-1}}{u^{4 n}+1}\,du$$ All of this makes $$I_n=2n\int\frac{ u^{2 n}}{u^{4 n}+1}\,du$$ For a specific value of $n$,we could consider that $$u^{4n}+1=\prod_{i=1}^{4n}(u-r_i)$$ and use partial fraction decompositions to get something like $$\frac{ u^{2 n}}{u^{4 n}+1}=\sum_{i=1}^{4n}\frac {a_i}{u-r_ i}$$ and integrate each term which will result in a complex logarithm. Recombining them, we should arrive to a sum of logarithms and arctangents. I do not see how to get a general expression except using the Gaussian hypergeometric function which would give $$I_n=\frac{2 n }{2 n+1}u^{2 n+1}\, _2F_1\left(1,\frac{1}{4} \left(\frac{1}{n}+2\right);\frac{1}{4} \left(\frac{1}{n}+6\right);-u^{4 n}\right)$$