Integral $I=\int_0^1 \frac{\log x \log (1+x) \log(1-x) \log(1+x^2)\log(1-x^2)}{x^{3/2}}dx$

Question

Hi I am trying to integrate and obtain a closed form result for $$ I:=\int_0^1 \frac{\log x \log (1+x) \log(1-x) \log(1+x^2)\log(1-x^2)}{x^{3/2}}dx. $$ Here is what I tried (but I do not think this is a smart way because of all the sums): writing $$ I=\int_0^1 \frac{dx\log x}{x^{3/2}} \sum_{n=1}^\infty\frac{ (-1)^{n}x^n}{n}\sum_{m=1}^\infty \frac{x^m}{m}\sum_{l=1}^\infty \frac{(-1)^lx^{2l}}{l}\sum_{p=1}^\infty \frac{(-1)^p(-x^2)^{p}}{p}. $$ Now we can write $$ \sum_{n=1}^\infty\frac{ (-1)^{n}}{n}\sum_{m=1}^\infty \frac{1}{m}\sum_{l=1}^\infty \frac{(-1)^l}{l}\sum_{p=1}^\infty \frac{1}{p} \int_0^1 x^{n+m+2l+2p-3/2}\log x \, dx. $$ We can simplify this as $$ \sum_{n=1}^\infty \frac{ (-1)^{n}}{n}v\sum_{m=1}^\infty \frac{1}{m}\sum_{l=1}^\infty \frac{(-1)^l}{l}\sum_{p=1}^\infty \frac{1}{p}\left[ \frac{-1}{\big( n+m+2l+2p -\frac{1}{2} \big)^2}\right]. $$ I am stuck as to what to do From here, note above I used $$ -\sum_{n=1}^\infty \frac{x^n}{n}=\log(1-x),\quad \log(1+x)=-\sum_{n=1}^\infty \frac{(-1)^{n}x^n}{n},\quad \int_0^1 x^n \log x \, dx =\frac{-1}{(n+1)^2}. $$