How to evaluate$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x$ I am trying evaluating this
$$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x}  \right ) }{(x-1)^2-k^2(x+1)^2}\ \text{d}x.$$
For $k=1$, there has
$$J(1)=\frac{\pi^4}{96}.$$
Maybe $J(k)$ doesn't have an explicit closed-form.


An integral relation:
$$\int_{0}^{\infty} \frac{\arctan^3x}{1+k^2x^2} \text{d}x
=-\frac{3}{2}J(k)+\frac{\pi^2}{8k} 
\left ( \operatorname{Li}_2\left ( \frac{1}{k}  \right ) 
-\operatorname{Li}_2\left ( -\frac{1}{k}  \right )  \right )
+\frac{\ln k}{8k}  \ln^3\left ( \frac{k-1}{k+1}  \right ),\qquad (k>1).$$



With some calculations, we followed that
$$\int_{0}^{\infty} \frac{\arctan^3x}{1+k^2x^2} \text{d}x
=\frac{\pi^4}{64k} 
+\frac{3}{4k}\left ( \operatorname{Li}_4\left ( \frac{k-1}{k+1}  \right )
-\operatorname{Li}_4\left (- \frac{k-1}{k+1}  \right )
  \right ) +\frac{3\pi^2}{8k} \operatorname{Li}_2\left (- \frac{k-1}{k+1}  \right ),\qquad(k>1).$$
Then the final result of $J(k)$ is
$${\color{Green}{J(k)
=\frac{\pi^2}{12k} 
\left ( \operatorname{Li}_2\left ( \frac{1}{k}  \right ) 
-\operatorname{Li}_2\left ( -\frac{1}{k}  \right )  \right )
+\frac{\ln k}{12k}  \ln^3\left ( \frac{k-1}{k+1}  \right )
-\frac{\pi^4}{96k} -\frac{1}{2k}\left ( \operatorname{Li}_4\left ( \frac{k-1}{k+1}  \right )
-\operatorname{Li}_4\left (- \frac{k-1}{k+1}  \right )
  \right ) -\frac{\pi^2}{4k} \operatorname{Li}_2\left (- \frac{k-1}{k+1}  \right ),\qquad (k>1).}}$$

 A: A solution idea of calculating the integral by Cornel Ioan Valean
The post is extremely short since I have no time, but once you know what to do, all is trivial. So, what do to? Observe that any $J(2n,k)$ is half the real part of the integral over the positive real line.
$$\int_{0}^{1} \frac{\log(x)^{2n}\log\left(\frac{1-x}{1+x}\right)}{(x-1)^2-k^2(x+1)^2}\text{d}x=\frac{1}{2}\Re\biggr\{\int_{0}^{\infty} \frac{\log(x)^{2n}\log\left ( \frac{1-x}{1+x}  \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x\biggr\},$$
which is easy to reach by also using a simple substiution.
Once you have reached this point, exploit multiple integrals involving PV integrals as in the calculations of $J(s)$ here https://math.stackexchange.com/q/3488566.
The rest is easy, boring work to do.
End of story
A: Since you have a relation with another integral, here are 2 Maclaurin series representations for a general anti derivative which you can then use to solve for J(x). Here is a Demo Graph and Integral Demo:
$$\int \frac{\arctan^3(x)}{k^2x^2+1}dx=\int \sum_{n=0}^\infty\frac{\left(\frac{d^n}{dx^n} \frac{\arctan^3(x)}{k^2x^2+1}\right)_{x=0}\ x^n}{n!}= \sum_{n=4,6,8,…}^\infty\frac{\left(\frac{d^{n-1}}{dx^{n-1}} \frac{\arctan^3(x)}{k^2x^2+1}\right)_{x=0}\ x^n}{n!}=C+\frac{x^4}{4}-\frac{(k^2+1)x^6}{6}+\frac{(15k^4+15k^2+14)x^8}{120}-\frac{(471k^6+4712k^4+ 441k^2+409)x^{10}}{4725}+\frac{211(4725k^8+4725k^6+4410k^4+4090k^2+3807)x^{12}}{11975040} $$
Here is a simpler series expansion related to the main inverse tangent expansion. The integration includes the Lerch Zeta function:
$$ \int\frac{\arctan^{3}(x)}{k^2 x^2+1}dx= \int \sum_{n=0}^\infty\frac{\left(\frac{d^n}{dx^n} \arctan^3(x)\right)_{x=0}\ }{n!} \int \frac{x^n}{k^2x^2+1}dx= \sum_{n=1}^\infty\frac{\left(\frac{d^{n-1}}{dx^{n-1}} \arctan^3(x)\right)_{x=0}\ }{(n-1)!}x^{n}\frac{1}{2}Φ\left(-k^2 x^2,1, \frac{1}{2}\right) $$
The nth derivative has a simple expansion.
From this you can possibly use it to solve for J(x). I have more to add.
