<s>[INST] Is it possible to expand $\arctan x \ln(1-x^2)$ in series? I am not sure if its possible to find the Taylor series of $\arctan x \ln(1-x^2)$ so I am giving it a try:
By using the Cauchy product
$$\left(\sum_{n=1}^\infty a_n x^n\right)\left(\sum_{n=1}^\infty b_n x^n\right)=\sum_{n=1}^\infty \left(\sum_{k=1}^n a_k b_{n-k+1}\right)x^{n+1}$$
we have
$$\arctan x \ln(1-x^2)=\left(\sum_{n=1}^\infty \frac{(-1)^{n-1} x^{2n-1}}{2n-1}\right)\left(-\sum_{n=1}^\infty\frac{x^{2n}}{n}\right)$$
$$=\frac1x\left(\sum_{n=1}^\infty \underbrace{\frac{(-1)^{n}}{2n-1}}_{a_n}(x^2)^n\right)\left(\sum_{n=1}^\infty\underbrace{\frac{1}{n}}_{b_n}(x^2)^n\right)$$
$$=\frac1x\sum_{n=1}^\infty\left(\sum_{k=1}^n \frac{(-1)^k}{2k-1}\cdot\frac{1}{n-k+1}\right)(x^2)^{n+1}$$
$$\left\{\text{write $\frac{1}{(2k-1)(n-k+1)}=\frac{1}{2n+1}\left(\frac{2}{2k-1}+\frac{1}{n-k+1}\right)$}\right\}$$
$$=\sum_{n=1}^\infty \frac{2}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{2k-1}\right)x^{2n+1}+\sum_{n=1}^\infty \frac{1}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{n-k+1}\right)x^{2n+1}$$
$$\left\{\text{reverse the order of the terms in the second sum}\right\}$$
$$=\sum_{n=1}^\infty \frac{2}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{2k-1}\right)x^{2n+1}+\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{k}\right)x^{2n+1}$$
$$=\sum_{n=1}^\infty\frac{2}{2n+1}\left(\sum_{k=1}^n\frac{(-1)^k}{2k-1}\right)x^{2n+1}+\sum_{n=1}^\infty \frac{(-1)^{n}\overline{H}_n}{2n+1}x^{2n+1}.$$

Mathematica gives
$$\sum_{k=1}^n\frac{(-1)^k}{2k-1}=\int_0^1\frac{(-x^2)^n-1}{1+x^2}dx=-\frac{\pi}{4}+\frac{(-1)^n}{4}\left(\psi\left(\frac{2n+3}{4}\right)-\psi\left(\frac{2n+1}{4}\right)\right)$$
but this form makes the problem harder I guess. Any idea how to find a better form for $\sum_{k=1}^n\frac{(-1)^n}{2k-1}$ or how to find the whole $\sum_{n=1}^\infty\frac{2}{2n+1}\left(\sum_{k=1}^n\frac{(-1)^k}{2k-1}\right)x^{2n+1}$ or maybe a different way to find the Taylor series of $\arctan x \ln(1-x^2)$?
 [/INST] A bit tricky but it makes things simpler.
Rewrite
$$\tan ^{-1}(x)\log \left(1-x^2\right) = x \frac{\tan ^{-1}(x)}{x}\log \left(1-x^2\right)$$ and let $x=\sqrt t$
$$\frac{\tan ^{-1}(\sqrt t)}{\sqrt t}=\sum_{n=0}^\infty \frac{(-1)^n }{2 n+1}t^n$$
$$\log \left(1-t\right)=-\sum_{n=1}^\infty \frac {t^n} n$$
Where is see a problem is that the coefficients of the product
$$\frac{\tan ^{-1}(\sqrt t)}{\sqrt t}\,\log \left(1-t\right)=-\sum_{n=1}^\infty a_n\,t^n$$ being
$$\left\{1,\frac{1}{6},\frac{11}{30},\frac{121}{1260},\frac{281}{1260},\frac{929}{1
   3860},\frac{432683}{2702700},\frac{18527}{360360},\frac{84971}{680680},\frac{203
   75981}{488864376},\cdots\right\}$$ nothing has been identified by $OEIS$.
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