How to evaluate $\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx$ I want to evaluate: $$\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx,$$ but I don't see how can I achieve so.
My attempts so far have been rewriting the integral using algebraic identities such as: $$ab=\frac{1}{2}a^2+\frac{1}{2}b^2-\frac{1}{2}\left(a-b\right)^2=\frac{1}{4}\left(a+b\right)^2-\frac{1}{4}\left(a-b\right)^2,$$ that yield other integrals like:
$$\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln ^2\left(1-x\right)}{1+x}\:dx,\:\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln ^2\left(\frac{1-x}{1+x}\right)}{1+x}\:dx.$$Normally the beta function and expanding terms into series is used to deal with these kind of integrals yet neither can be used because of the upper bound.
What else can be done in order to compute the main integral? Thanks.
 A: An idea by Cornel Ioan Valean
Exploit the result
$$f(a)=\int_0^{1/2} \frac{\log (x) \log (1-x)}{1-a x} \textrm{d}x$$
$$\small =\frac{1}{2 a}\log ^3(2)-\frac{3 }{4 a}\zeta (3)-\frac{\log ^2(2) }{2 a}\log \left(1-\frac{a}{2}\right)-\frac{\log ^2(2) }{2 a}\log \left(\frac{a-2}{a-1}\right)+\frac{\log(2)}{a}\text{Li}_2\left(\frac{a}{2}\right)$$
$$\small+\frac{\log (2) }{a}\text{Li}_2\left(\frac{a}{2 (a-1)}\right)+\frac{1}{a}\text{Li}_3\left(\frac{a}{2}\right)+\frac{1}{a}\text{Li}_3\left(\frac{a}{2 (a-1)}\right)-\frac{1}{a}\text{Li}_3\left(\frac{a}{a-1}\right)-\frac{\text{Li}_3(a-1)}{a}.$$
This is a modified form of Lemma 2 in the paper  A special way of extracting the real part of the Trilogarithm,  $\displaystyle  \operatorname{Li}_3\left(\frac{1\pm i}{2}\right)$ easily obtained by the means presented in the paper https://www.researchgate.net/publication/337868999_A_special_way_of_extracting_the_real_part_of_the_Trilogarithm_Li_31i2
A nice fact: maybe good to mention also this integral variant
$$\int_0^{1/2}\frac{\log(1-x)\log(x)\log(1+x)}{x}\textrm{d}x=\int_{-1}^0 f(a) da.$$
