Closed form 0f $I=\int _{ 0 }^{ 1 }{ \frac { \ln { x } { \left( \ln { \left( 1-{ x }^{ 2 } \right) } \right) }^{ 3 } }{ 1-x } dx }$ While solving a problem, I got stuck at an integral. The integral is as follows:

Find the closed form of: $$I=\int _{ 0 }^{ 1 }{ \frac { \ln { x } { \left( \ln { \left( 1-{ x }^{ 2 } \right)  }  \right)  }^{ 3 } }{ 1-x } dx } $$

I tried using power series but it failed. I tried various subtitutions which came to be of no use. Please help.
 A: The closed form is  $$I=21\zeta({3})\log^22-\pi^2\log^32-\frac{13}{4}\pi^2\zeta({3})+48\zeta({5})-\frac{1}{4}\pi^4\log2$$
A: We have $$ I=\int_{0}^{1}\frac{\log\left(x\right)\log^{3}\left(1-x^{2}\right)}{1-x}dx=\int_{0}^{1}\frac{\log\left(x\right)\log^{3}\left(1-x^{2}\right)}{1-x^{2}}dx+\int_{0}^{1}\frac{x\log\left(x\right)\log^{3}\left(1-x^{2}\right)}{1-x^{2}}dx
 $$ and so if we put $x=\sqrt{y}$ we get $$I=\frac{1}{4}\int_{0}^{1}\frac{y^{-1/2}\log\left(y\right)\log^{3}\left(1-y\right)}{1-y}dy+\frac{1}{4}\int_{0}^{1}\frac{\log\left(y\right)\log^{3}\left(1-y\right)}{1-y}dy$$ and recalling the definition of beta function $$B\left(a,b\right)=\int_{0}^{1}x^{a-1}\left(1-x\right)^{b-1}dx
 $$ we have $$\frac{\partial^{h+k}B}{\partial a^{h}\partial b^{k}}\left(a,b\right)=\int_{0}^{1}x^{a-1}\log^{h}\left(x\right)\left(1-x\right)^{b-1}\log^{k}\left(1-x\right)dx
 $$ hence $$I=\frac{1}{4}\frac{\partial^{4}B}{\partial a\partial b^{3}}\left(\frac{1}{2},0^{+}\right)+\frac{1}{4}\frac{\partial^{4}B}{\partial a\partial b^{3}}\left(1,0^{+}\right).$$ For the computation of the limit, we can use the asymptotic $\Gamma(x)=\frac{1}{x}+O(1)$ when $x\rightarrow 0$ and the relations between the polygamma terms and zeta.
