$$\int_0^1\frac{\ln^2(1-x)\ln^5(1+x)}{1+x}dx\overset{1+x\to x}{=}\int_1^2\frac{\ln^2(2-x)\ln^5x}{x}dx$$
$$=\ln^2(2)\int_1^2\frac{\ln^5x}{x}dx+2\ln(2)\int_1^2\frac{\ln(1-x/2)\ln^5x}{x}dx+\int_1^2\frac{\ln^2(1-x/2)\ln^5x}{x}dx$$
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.
$$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x=\int _0^1\frac{1}{\sqrt{1-x^2}}\left(\int_0^1\frac{x^2\ln(y)}{1+x^2 y}\mathrm{d}y\right)\mathrm{d}x$$
$$=\int _0^1\ln(y)\left(\int_0^1\frac{x^2}{(1+x^2y)\sqrt{1-x^2} }\mathrm{d}x\right)\mathrm{d}y$$
$$=\int _0^1\ln(y)\left(\frac{\pi }{2y}\left(1-\frac{1}{\sqrt{1+y}}\right)\right)\mathrm{d}y$$
$$=\int _0^1\ln(y)\left(\frac{\pi }{2y}\left(-\sum_{n=1}^\infty\frac{2n\choose n}{4^n}(-y)^n\right)\right)\mathrm{d}y$$
$$=-\frac{\pi}{2}\sum_{n=1}^\infty(-1)^n\frac{2n\choose n}{4^n}\int_0^1 y^{n-1}\ln(y)\mathrm{d}y$$
$$=\frac{\pi}{2}\sum_{n=1}^\infty(-1)^n\frac{2n\choose n}{4^n n^2}.\qquad (*)$$
To get this sum, first replace $x$ by $-x$ in
$$\sum_{n=1}^\infty\frac{{2n\choose n}}{4^n}\frac{x^n}{n}=-2\ln\left(\frac{1+\sqrt{1-x}}{2}\right)$$
then divide both sides by $x$ and integrate from $0$ to $1$,
$$\sum_{n=1}^\infty(-1)^n\frac{2n\choose n}{4^n n^2}=-2\int_0^1\frac{\ln\left(\frac{1+\sqrt{1+x}}{2}\right)}{x}\mathrm{d}x$$
$$\overset{\frac{1+\sqrt{1+x}}{2}=t}{=}2\int_1^{\frac{1+\sqrt{2}}{2}}\frac{\ln(t)}{1-t}\mathrm{d}t-2\int_1^{\frac{1+\sqrt{2}}{2}}\frac{\ln(t)}{t}\mathrm{d}t$$
$$=2\operatorname{Li}_2(1-t)-\ln^2(t)\bigg|_1^{\frac{1+\sqrt{2}}{2}}$$
$$=2 \operatorname{Li}_2\left(\frac{1-\sqrt{2}}{2}\right)-\ln^2 \left(\frac{1+\sqrt{2}}{2}\right).$$
Addendum:
You can get the summation representation in $(*)$ by expanding $\operatorname{Li}_2(-x^2)$ in series as you mentioned in your question body, then using $\int_0^1\frac{x^{2n}}{\sqrt{1-x^2}}\mathrm{d}x=\frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac12+n\right)}{n\Gamma(n)}=\frac{\pi}{2}\frac{{2n\choose n}}{4^n}$ which follows from using the beta function and the Legendre duplication formula:
$$\Gamma\left(\frac12+n\right)=\frac{2\sqrt{\pi}\,\Gamma(2n)}{4^n\Gamma(n)}=\sqrt{\pi}n\Gamma(n)\frac{2n\choose n}{4^n}.$$
Best Answer
Lets start with the following integral
$$\int_0^1\frac{\text{Li}_2(-x)\ln(1-x)}{1+x}dx=\sum_{n=0}^\infty (-1)^nH_n^{(2)}\int_0^1 x^{n}\ln(1-x)dx$$
$$=-\sum_{n=0}^\infty \frac{(-1)^nH_n^{(2)}H_{n+1}}{n+1}=\sum_{n=1}^\infty \frac{(-1)^nH_{n-1}^{(2)}H_{n}}{n}$$
$$=\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}H_n}{n}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$
where
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$
The other sum was already evaluated by Cornel in this solution and I am writing it in more details:
First relation:
From here we have
$$\int_0^1x^{n-1}\ln^3(1-x)\ dx=-\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$
Multiply both sides by $(-1)^{n-1}$ then $\sum_{n=1}^\infty$ we have
$$\sum_{n=1}^\infty (-1)^n\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}=\int_0^1\ln^3(1-x)\sum_{n=1}^\infty (-x)^{n-1}dx=\int_0^1\frac{\ln^3(1-x)}{1+x}dx$$
$$=\int_0^1\frac{\ln^3x}{2-x}dx=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln^3xdx=-6\sum_{n=1}^\infty\frac{1}{2^nn^4}=-6\text{Li}_4\left(\frac12\right)\tag1$$
Second relation:
From here we have
$$\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)x^n=-\frac{\ln^3(1-x)}{1-x}$$
Replace $x$ by $-x$ then divide both sides by $x$ and $\int_0^1$ we get
$$\sum_{n=1}^\infty (-1)^n\frac{H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}}{n}=-\int_0^1\frac{\ln^3(1+x)}{x(1+x)}dx$$ $$\overset{x=\frac{1-y}{y}}{=}\int_{1/2}^1\frac{\ln^3x}{1-x}dx=\sum_{n=1}^\infty \int_{1/2}^1 x^{n-1}\ln^3xdx$$
$$=\sum_{n=1}^\infty\left(\frac{6}{n^42^n}-\frac{6}{n^4}+\frac{6\ln2}{n^32^n}+\frac{3\ln^22}{n^22^n}+\frac{\ln^32}{n2^n}\right)$$
$$=6\text{Li}_4\left(\frac12\right)-6\zeta(4)+6\ln2\text{Li}_3\left(\frac12\right)+3\ln^22\text{Li}_2\left(\frac12\right)+\ln^42$$
$$=6\text{Li}_4\left(\frac12\right)-6\zeta(4)+\frac{21}{4}\ln2\zeta(3)-\frac32\ln^22\zeta(2)+\frac12\ln^42\tag2$$
Thus, $(1)-(2)$ gives
$$\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n}=-2\text{Li}_4\left(\frac12\right)+\zeta(4)-\frac{7}{8}\ln2\zeta(3)+\frac14\ln^22\zeta(2)-\frac1{12}\ln^42$$
Combining the two sums we finally get
$$\int_0^1\frac{\text{Li}_2(-x)\ln(1-x)}{1+x}dx=-4 \text{Li}_4\left(\frac{1}{2}\right)+\frac{15}{4}\zeta(4)-\frac{21}{8}\ln2\zeta(3)+\frac34\ln^22\zeta(2)-\frac16\ln^42$$
Bonus:
By writing $\text{Li}_2(-x)=\int_0^1\frac{x\ln y}{1+xy}dy$ then changing the order of integration we have
$$\int_0^1\frac{\text{Li}_2(-x)\ln(1-x)}{1+x}dx=\int_0^1\ln y\left(\int_0^1\frac{x\ln(1-x)}{(1+x)(1+xy)}dx\right)dy$$
$$=\int_0^1 \ln y\left(\frac{\zeta(2)-\ln^22}{2(1-y)}-\frac{\text{Li}_2\left(\frac{y}{1+y}\right)}{y(1-y)}\right)dy$$
$$=-\frac12(\zeta(2)-\ln^22)\zeta(2)-\int_0^1\frac{\ln y}{y(1-y)}\text{Li}_2\left(\frac{y}{1+y}\right)dx$$
Or
$$\int_0^1\frac{\text{Li}_2(-x)\ln(1-x)}{1+x}dx+\int_0^1\frac{\ln x}{x(1-x)}\text{Li}_2\left(\frac{x}{1+x}\right)dx=\frac12\ln^22\zeta(2)-\frac54\zeta(4)$$