How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$ Before you think I haven't tried anything, please read.
I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$$
But I can't find a way to simplify it. Integration by parts is not valid since we face convergence issues. Subbing the $1-x$ term is also not quite helpful here. I also tried to use the sub $\frac{1}{1+x}$ but this is not useful either. Using algebraic identities isn't that useful either, since we run into similar difficulty integrals. Since these attempts do not lead to anything, how can I approach it?
 A: $$\int_0^1\frac{\ln^2(1-x)\ln^5(1+x)}{1+x}dx\overset{1+x\to x}{=}\int_1^2\frac{\ln^2(2-x)\ln^5x}{x}dx$$
$$=\ln^2(2)\int_1^2\frac{\ln^5x}{x}dx+2\ln(2)\int_1^2\frac{\ln(1-x/2)\ln^5x}{x}dx+\int_1^2\frac{\ln^2(1-x/2)\ln^5x}{x}dx$$
write $\ln(1-x/2)=-\sum_{n=1}^\infty\frac{x^n}{n2^n}$ for the first integral and write $\ln^2(1-x/2)=2\sum_{n=1}^\infty(\frac{H_n}{n2^n}-\frac{1}{n^22^n})x^n$ for the third integral
$$=\frac16\ln^8(2)+\sum_{n=1}^\infty\left(\frac{2H_n}{n}-\frac{2}{n^2}-\frac{2\ln(2)}{n}\right)\int_1^2 \frac{x^{n-1}\ln^5x}{2^n}dx$$
$$=\frac16\ln^8(2)+\sum_{n=1}^\infty\left(\frac{2H_n}{n}-\frac{2}{n^2}-\frac{2\ln(2)}{n}\right)$$
$$\left(\frac{\ln^5(2)}{n}-\frac{5\ln^4(2)}{n^2}+\frac{20\ln^3(2)}{n^3}-\frac{60\ln^2(2)}{n^4}+\frac{120\ln(2)}{n^5}-\frac{120}{n^6}+\frac{120}{n^62^n}\right)$$
$$=\frac16\ln^8(2)-2\ln^6(2)\zeta(2)+12\ln^5(2)\zeta(3)-\frac{85}{2}\ln^4(2)\zeta(4)+40\ln^3(2)[5\zeta(5)-\zeta(2)\zeta(3)]$$
$$-30\ln^2(2)[11\zeta(6)-2\zeta^2(3)]-240\ln(2)\left[\zeta(4)\zeta(3)+\zeta(2)\zeta(5)-4\zeta(7)+\text{Li}_7\left(\frac12\right)\right]$$
$$-300\zeta(8)+240\zeta(3)\zeta(5)-240\text{Li}_8\left(\frac12\right)+240\sum_{n=1}^\infty\frac{H_n}{n^72^n}\approx 0.113272.$$
There is no closed form for your integral as $\sum_{n=1}^\infty\frac{H_n}{n^72^n}$ has no closed form.

Remark:
The closed form of $\int_0^1\frac{\ln^2(1-x)\ln^a(1+x)}{1+x}dx$ can be expressed in terms of $\ln, \pi, \zeta$ and $\text{Li}_r$ plus $\sum_{n=1}^\infty\frac{H_n}{n^{a+2}2^n}$. What I know of is that there is no closed form for such series with $a>2$ and the case $a=2$ is calculated here.
