I am trying to find a closed-form for the Maclaurin series of $\frac{\ln(1-x)\ln(1+x)}{1+x^2}$. I am not sure if it's possible but here is what I did:
We know that:
$$\ln(1-x)\ln(1+x)=\sum_{n=1}^\infty\left(\frac{H_n-H_{2n}}{n}-\frac{1}{2n^2}\right)x^{2n}\tag1$$
$$\frac1{1+x^2}=\sum_{n=1}^\infty (-1)^{n-1}x^{2n-2}$$
And by the Cauchy product of power series
$$\left(\sum_{n=1}^\infty a_nx^n\right)\left(\sum_{n=1}^\infty b_nx^n\right)=\sum_{n=1}^\infty x^{n+1}\left(\sum_{k=1}^n a_k\ b_{n-k+1}\right)$$
we have
$$\frac{\ln(1-x)\ln(1+x)}{1+x^2}=\sum_{n=1}^\infty(-1)^nx^{2n}\color{blue}{\left(\sum_{k=1}^n(-1)^k\left(\frac{H_k-H_{2k}}{k}-\frac1{2k^2}\right)\right)},$$
and I have no idea how to deal with the blue sum, any idea? Or maybe a different approach?
The identity of $(1)$ can be found in the book Almost Impossible Integrals, Sums and Series.