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 $${\color{Green}{J(k) =\frac{\pi^2}{12k} \left ( \operatorname{Li}_2\left ( \frac{1}{k} \right ) -\operatorname{Li}_2\left ( -\frac{1}{k} \right ) \right ) +\frac{\ln k}{12k} \ln^3\left ( \frac{k-1}{k+1} \right ) -\frac{\pi^4}{96k} -\frac{1}{2k}\left ( \operatorname{Li}_4\left ( \frac{k-1}{k+1} \right ) -\operatorname{Li}_4\left (- \frac{k-1}{k+1} \right ) \right ) -\frac{\pi^2}{4k} \operatorname{Li}_2\left (- \frac{k-1}{k+1} \right ),\qquad (k>1).}}$$