Closed form for $\int \sqrt[n]{\tan x}\ dx$

Question

I was solving $\displaystyle\int\sqrt[n]{\tan x}\ dx$.

Here's my method:

$$\begin{align}\int\sqrt[n]{\tan x}\ dx &= \int\frac{n \cdot t^n}{1 + (t)^{2n}}\tag{1}\ dt\\& = n \int\sum_{k=0}^\infty (-1)^k (t)^{2nk}\cdot t^n \ dt\tag{2} \\& = n\sum_{k=0}^\infty(-1)^k \int t^{2nk+n}\ dt \tag{3}\\& = n \sum_{k=0}^\infty(-1)^k \frac{t^{2nk +n+1}}{2nk +n+1} + C\tag{4}\\& =n \sum_{k=0}^\infty(-1)^k \frac{(\tan x)^{\frac{2nk +n+1}n}}{2nk +n+1} + C\tag{5} \\& = \boxed{n \sum_{k=0}^\infty(-1)^k \frac{(\tan x)^{2k +\frac{n+1}n}}{2nk +n+1} + C}\end{align}$$

Steps:

$(1)$ Substitution: $\tan{x} = t^n$

$(2)$ Taylor series: $\displaystyle\frac{1}{1+t} = \sum_{k=0}^\infty (-1)^k (t)^k\implies \frac{1}{1+t^{2n}} = \sum_{k=0}^\infty (-1)^k (t^{2n})^k$.

$(3)$ Interchanged integral and summation symbols.

$(4)$ Used power rule of integration.

$(5)$ Undone the substitution.

Source:

I was practicing integral calculus and came across $\displaystyle \int \sqrt{\tan x}\ dx$ and $\displaystyle \int \sqrt[3]{\tan x}\ dx $. Both of them were nice and I solved them. So I thought there would be definitely a general solution for $\displaystyle \int \sqrt[n]{\tan x}\ dx$ where $n\in \mathbb{Z}^+$.

My question:

Answers for $\int \sqrt{\tan x}\ dx$ and $\sqrt[3]{\tan {x}} $ were looking good, at least elementary (Having closed form). I expected the same for $\int \sqrt[n]{\tan x}\ dx$. Is there any closed form for $n \sum_{k=0}^\infty(-1)^k \frac{(\tan x)^{2k +\frac{n+1}n}}{2nk +n+1} + C$? And is my method right?

The closed form is here and using a transformation it is a Lerch transcendent or incomplete beta $\text B_z(a,0)$ expression. For rational $n$, there should be a closed form with a finite sum of logarithms. — Тyma Gaidash, Dec 01 '22 at 13:19
@TymaGaidash Thanks. I think beta function and Hypergeometric function are beyond the limits of my knowledge. — Utkarsh, Dec 01 '22 at 13:58
See https://math.stackexchange.com/questions/1998618/factorize-x2n1-to-evaluate-int-tan1-nxdx, which this question arguably duplicates. — Travis Willse, Dec 01 '22 at 22:03

Quanto · Accepted Answer · 2022-12-01T17:25:21.887

Apply the partial decomposition \begin{align} \frac{nt^n}{1+t^{2n}}= \sum_{k=1}^{n} (-1)^{k+1} \frac{t\sin \theta_k}{1-2t\cos \theta_k + t^2}, \>\>\>\>\>\theta_k=\frac{(2k-1)\pi}{2n} \end{align} and, with $t= \sqrt[n]{\tan x}$, integrate to obtain the close form below \begin{align} & \int\sqrt[n]{\tan x}\> dx= \int \frac{nt^n}{1+t^{2n}}dt\\ =& \sum_{k=0}^{n}{(-1)^{k+1}}\left[ {\sin \theta_k} \ln\sqrt{1- 2t\cos \theta_k + t^2} +\cos \theta_k \tan^{-1}\frac{t- \cos \theta_k}{\sin \theta_k} \right] \end{align}

Closed form for $\int \sqrt[n]{\tan x}\ dx$

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