How can I evaluate $\displaystyle \int _0^1\frac{\ln ^2\left(x\right)\ln ^3\left(\frac{1-x}{1+x}\right)}{1-x}\:dx$ in an elegant (simple?) way? this integral seems very challenging, by simply expanding the binomial we have $$\int _0^1\frac{\ln ^2\left(x\right)\ln ^3\left(1-x\right)}{1-x}\:dx-3\int _0^1\frac{\ln ^2\left(x\right)\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1-x}\:dx$$ $$+3\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)\ln ^2\left(1+x\right)}{1-x}\:dx-\int _0^1\frac{\ln ^2\left(x\right)\ln ^3\left(1+x\right)}{1-x}\:dx$$ which is clearly not helpful.
By using the sub $t=\frac{1-x}{1+x}$ I also obtain $$\int _0^1\frac{\ln ^2\left(\frac{1-t}{1+t}\right)\ln ^3\left(t\right)}{t}\:dt-\int _0^1\frac{\ln ^2\left(\frac{1-t}{1+t}\right)\ln ^3\left(t\right)}{1+t}\:dt$$ for that first integral $\displaystyle \ln ^2\left(\frac{1-x}{1+x}\right)=-2\sum _{n=1}^{\infty }\frac{H_n-2H_{2n}}{n}x^{2n}$ can be useful yet not so much for the $2$nd one.
The integral in question can also be converted into $$\int _0^1\frac{\ln ^2\left(x\right)\ln ^3\left(\frac{\left(1-x\right)^2}{1-x^2}\right)}{1-x}\:dx$$ $$=8\int _0^1\frac{\ln ^2\left(x\right)\ln ^3\left(1-x\right)}{1-x}\:dx-12\int _0^1\frac{\ln ^2\left(x\right)\ln ^2\left(1-x\right)\ln \left(1-x^2\right)}{1-x}\:dx$$ $$+6\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)\ln ^2\left(1-x^2\right)}{1-x}\:dx-\int _0^1\frac{\ln ^2\left(x\right)\ln ^3\left(1-x^2\right)}{1-x}\:dx$$ The first and last integral can be done by using the beta function but the others are as challenging.