Integrals $\int \frac{1}{\operatorname{arctanh}(x)} \, dx$ and $\int \frac{1}{\operatorname{arccoth}(x)} \, dx$

Question

Do we know anything about this integrals?

$$ \begin{align} I_1(x) = \int \frac{1}{\operatorname{artanh}(x)} \, dx \\ I_2(x) = \int \frac{1}{\operatorname{arcoth}(x)} \, dx \end{align}$$

Similar integrals.

$$ \begin{align} \int \operatorname{artanh}(x) \, dx & = x \operatorname{artanh}(x) + \frac12 \ln(1-x^2) + C, \\ \int \operatorname{arcoth}(x) \, dx & = x \operatorname{arcoth}(x) + \frac12 \ln(1-x^2) + C, \\ \int \frac{1}{\operatorname{arsinh}(x)} \, dx & = \operatorname{Chi}(\operatorname{arsinh}(x)) + C, \\ \int \frac{1}{\operatorname{arcosh}(x)} \, dx & = \operatorname{Shi}(\operatorname{arcosh}(x)) + C, \\ \end{align}$$

where $\operatorname{Chi}$ is the hyperbolic cosine integral, and $\operatorname{Shi}$ is the hyperbolic sine integral.

I found nothing with Maple or Mathematica. As I see some kind of "hyperbolic tangent integral" is not defined.

Answer 1

Approach $1$:

For $\int\dfrac{1}{\text{arcoth }x}dx$ ,

Let $u=\text{arcoth }x$ ,

Then $\int\dfrac{1}{\text{arcoth }x}dx$

$=\int\dfrac{1}{u}d(\coth u)$

$=\dfrac{\coth u}{u}-\int\coth u~d\left(\dfrac{1}{u}\right)$

$=\dfrac{\coth u}{u}+\int\dfrac{\coth u}{u^2}du$

$=\dfrac{\coth u}{u}+\int\dfrac{e^{2u}+1}{u^2(e^{2u}-1)}du$

$=\dfrac{\coth u}{u}-\int\dfrac{1}{u^2(e^{-2u}-1)}du+\int\dfrac{1}{u^2(e^{2u}-1)}du$

$=\dfrac{\coth u}{u}-\int\dfrac{1}{u^2}\sum\limits_{n=0}^\infty\dfrac{B_n(-2u)^{n-1}}{n!}du+\int\dfrac{1}{u^2}\sum\limits_{n=0}^\infty\dfrac{B_n(2u)^{n-1}}{n!}du$ (with the formula in http://en.wikipedia.org/wiki/Bernoulli_number#Generating_function)

$=\dfrac{\coth u}{u}+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{n-1}B_nu^{n-3}}{n!}du+\int\sum\limits_{n=0}^\infty\dfrac{2^{n-1}B_nu^{n-3}}{n!}du$

$=\dfrac{\coth u}{u}+\int\sum\limits_{n=0}^\infty\dfrac{4^nB_{2n}u^{2n-3}}{(2n)!}du$

$=\dfrac{\coth u}{u}+\int\left(\dfrac{1}{u^3}+\dfrac{1}{3u}+\sum\limits_{n=2}^\infty\dfrac{4^nB_{2n}u^{2n-3}}{(2n)!}\right)du$

$=\dfrac{\coth u}{u}-\dfrac{1}{2u^2}+\dfrac{\ln u}{3}+\sum\limits_{n=2}^\infty\dfrac{2^{2n-1}B_{2n}u^{2n-2}}{(2n)!(n-1)}+C$

$=\dfrac{x}{\text{arcoth }x}-\dfrac{1}{2~\text{arcoth}^2x}+\dfrac{\ln\text{arcoth }x}{3}+\sum\limits_{n=2}^\infty\dfrac{2^{2n-1}B_{2n}\text{arcoth}^{2n-2}x}{(2n)!(n-1)}+C$

For $\int\dfrac{1}{\text{artanh }x}dx$ ,

Let $u=\text{artanh }x$ ,

Then $\int\dfrac{1}{\text{artanh }x}dx$

$=\int\dfrac{1}{u}d(\tanh u)$

$=\dfrac{\tanh u}{u}-\int\tanh u~d\left(\dfrac{1}{u}\right)$

$=\dfrac{\tanh u}{u}+\int\dfrac{\tanh u}{u^2}du$

$=\dfrac{\tanh u}{u}+\int\dfrac{1}{u^2}\sum\limits_{n=1}^\infty\dfrac{4^n(4^n-1)B_{2n}u^{2n-1}}{(2n)!}du$ (with the formula in http://en.wikipedia.org/wiki/Hyperbolic_function#Taylor_series_expressions)

$=\dfrac{\tanh u}{u}+\int\sum\limits_{n=1}^\infty\dfrac{4^n(4^n-1)B_{2n}u^{2n-3}}{(2n)!}du$

$=\dfrac{\tanh u}{u}+\int\left(\dfrac{1}{u}+\sum\limits_{n=2}^\infty\dfrac{4^n(4^n-1)B_{2n}u^{2n-3}}{(2n)!}\right)du$

$=\dfrac{\tanh u}{u}+\ln u+\sum\limits_{n=2}^\infty\dfrac{2^{2n-1}(4^n-1)B_{2n}u^{2n-2}}{(2n)!(n-1)}+C$

$=\dfrac{x}{\text{artanh }x}+\ln\text{artanh }x+\sum\limits_{n=2}^\infty\dfrac{2^{2n-1}(4^n-1)B_{2n}\text{artanh}^{2n-2}x}{(2n)!(n-1)}+C$

Approach $2$:

For $\int\dfrac{1}{\text{arcoth }x}dx$ ,

Let $u=\text{arcoth }x$ ,

Then $\int\dfrac{1}{\text{arcoth }x}dx$

$=\int\dfrac{1}{u}d(\coth u)$

$=\dfrac{\coth u}{u}-\int\coth u~d\left(\dfrac{1}{u}\right)$

$=\dfrac{\coth u}{u}+\int\dfrac{\coth u}{u^2}du$

$=\dfrac{\coth u}{u}+\int\dfrac{1}{u^3}+\int\sum\limits_{n=1}^\infty\dfrac{2}{u(u^2+n^2\pi^2)}du$ (use Mittag-Leffler Expansion of hyperbolic cotangent)

$=\dfrac{\coth u}{u}-\dfrac{1}{2u^2}+\int\sum\limits_{n=1}^\infty\dfrac{2}{n^2\pi^2u}du-\int\sum\limits_{n=1}^\infty\dfrac{2u}{n^2\pi^2(n^2\pi^2+u^2)}du$

$=\dfrac{\coth u}{u}-\dfrac{1}{2u^2}+\sum\limits_{n=1}^\infty\dfrac{2\ln u}{n^2\pi^2}-\sum\limits_{n=1}^\infty\dfrac{\ln(n^2\pi^2+u^2)}{n^2\pi^2}+C$

$=\dfrac{\ln\text{arcoth }x}{3}+\dfrac{x}{\text{arcoth }x}-\dfrac{1}{2(\text{arcoth }x)^2}-\sum\limits_{n=1}^\infty\dfrac{\ln(n^2\pi^2+(\text{arcoth }x)^2)}{n^2\pi^2}+C$

For $\int\dfrac{1}{\text{artanh }x}dx$ ,

Let $u=\text{artanh }x$ ,

Then $\int\dfrac{1}{\text{artanh }x}dx$

$=\int\dfrac{1}{u}d(\tanh u)$

$=\dfrac{\tanh u}{u}-\int\tanh u~d\left(\dfrac{1}{u}\right)$

$=\dfrac{\tanh u}{u}+\int\dfrac{\tanh u}{u^2}du$

$=\dfrac{\tanh u}{u}+\int\sum\limits_{n=0}^\infty\dfrac{8}{u((2n+1)^2\pi^2+4u^2)}du$ (use Mittag-Leffler Expansion of hyperbolic tangent)

$=\dfrac{\tanh u}{u}+\int\sum\limits_{n=0}^\infty\dfrac{8}{(2n+1)^2\pi^2u}du-\int\sum\limits_{n=0}^\infty\dfrac{32u}{(2n+1)^2\pi^2((2n+1)^2\pi^2+4u^2)}du$

$=\dfrac{\tanh u}{u}+\sum\limits_{n=0}^\infty\dfrac{8\ln u}{(2n+1)^2\pi^2}-\sum\limits_{n=0}^\infty\dfrac{4\ln((2n+1)^2\pi^2+4u^2)}{(2n+1)^2\pi^2}+C$

$=\ln\text{artanh }x+\dfrac{x}{\text{artanh }x}-\sum\limits_{n=0}^\infty\dfrac{4\ln((2n+1)^2\pi^2+4(\text{artanh }x)^2)}{(2n+1)^2\pi^2}+C$