Finally I got the idea:
Starting with $\frac{1}{1+x}\mapsto x$ then using Landens identity $\operatorname{Li}_2\left(\frac{x-1}{x}\right)=-\frac12\ln^2x-\operatorname{Li}_2(1-x)$ we obtain
$$\mathcal{I}=\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx=\int_{1/2}^1\frac{\ln^2x\operatorname{Li}_2\left(\frac{x-1}{x}\right)}{x(1-x)}dx$$
$$=-\frac12\underbrace{\int_{1/2}^1\frac{\ln^4x}{x(1-x)}dx}_{\mathcal{\large J}}-\underbrace{\int_{1/2}^1\frac{\ln^2x\operatorname{Li}_2(1-x)}{x}dx}_{ IBP}-\underbrace{\int_{1/2}^1\frac{\ln^2x\operatorname{Li}_2(1-x)}{1-x}dx}_{ IBP}$$
$$\text{Note for the third integral that} \int\frac{\ln x}{1-x}dx=\operatorname{Li}_2(1-x)$$
$$=-\frac12\mathcal{J}-\frac13\ln^32\operatorname{Li}_2(1/2)+\frac13\underbrace{\int_{1/2}^1\frac{\ln^4x}{1-x}dx}_{\frac1{1-x}=\frac1{x(1-x)}-\frac1x}-\frac12\ln2\operatorname{Li}_2^2(1/2)+\frac12\underbrace{\int_{1/2}^1\frac{\operatorname{Li}_2^2(1-x)}{x}dx}_{\mathcal{\large K}}$$
and the integral simplifies into
$$\mathcal{I}=\frac12\mathcal{K}-\frac16\mathcal{J}+\frac1{12}\ln^32\zeta(2)-\frac5{16}\ln2\zeta(4)-\frac1{40}\ln^52$$
where we substituted $\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$
The integral $\mathcal{J}$ is classical and can be done using the generalization
$$(-1)^n\int_{1/2}^1\frac{\ln^nx}{x(1-x)}dx=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$
so
$$\boxed{\mathcal{J}=24\zeta(5)-\frac{21}2\ln^22\zeta(3)+4\ln^32\zeta(2)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)}$$
where we substituted $\operatorname{Li}_3(1/2)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$
For the integral $\mathcal{K}$, we can just use the dilogarithm reflection formula: $$\operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)$$
upon expanding $\operatorname{Li}_2^2(1-x)$ we get
$$\mathcal{K}=\underbrace{\int_{1/2}^1\frac{\zeta^2(2)-2\zeta(2)\operatorname{Li}_2(x)}{x}dx}_{\mathcal{\large {K_1}}}-2\zeta(2)\underbrace{\int_{1/2}^1\frac{\ln x\ln(1-x)}{x}dx}_{\mathcal{\large {K_2}}}+\underbrace{\int_{1/2}^1\frac{\ln^2x\ln^2(1-x)}{x}dx}_{\mathcal{\large {K_3}}}\\+2\underbrace{\int_{1/2}^1\frac{\ln x\ln(1-x)\operatorname{Li}_2(x)}{x}dx}_{\mathcal{\large {K_4}}}+\underbrace{\int_{1/2}^1\frac{\operatorname{Li}_2^2(x)}{x}dx}_{\mathcal{\large {K_5}}}$$
$$\mathcal{K_1}=\frac52\ln2\zeta(4)-2\zeta(2)[\zeta(3)-\operatorname{Li}_3(1/2)]$$
$$\boxed{\mathcal{K_1}=\frac13\ln^32\zeta(2)-\frac14\zeta(2)\zeta(3)}$$
$$\mathcal{K_2}\overset{IBP}{=}-\ln2\operatorname{Li}_2(1/2)+\int_{1/2}^1\frac{\operatorname{Li}_2(x)}{x}dx$$
$$=-\ln2\operatorname{Li}_2(1/2)+\zeta(3)-\operatorname{Li}_3(1/2)$$
$$\boxed{\mathcal{K_2}=\frac18\zeta(3)+\frac13\ln^32}$$
$$\mathcal{K_3}=\int_0^{1/2}\frac{\ln^2x\ln^2(1-x)}{x}dx\overset{IBP}{=}\frac23{\int_{0}^{1/2}\frac{\ln^3x\ln(1-x)}{1-x}}dx=\frac23\color{blue}{\int_{1/2}^1\frac{\ln^3(1-x)\ln x}{x}dx}$$
I proved in this solution
$$\color{blue}{\int_{1/2}^1\frac{\ln^3(1-x)\ln x}{x}\ dx}=\frac3{16}\zeta(5)+\frac3{20}\ln^52-\frac14\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx+\frac12\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx$$
where
$$\int_{1/2}^1\frac{\ln^4x}{1-x}=\mathcal{J}-\int_{1/2}^1\frac{\ln^4x}{x}dx$$
$$=24\zeta(5)-\frac{21}2\ln^22\zeta(3)+4\ln^32\zeta(2)-\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)$$
and
$$\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx=\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^3x=6\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}$$
$$=6\sum_{n=1}^\infty\frac{H_n}{n^4}-6\zeta(5)=6\left(3\zeta(5)-\zeta(2)\zeta(3)\right)-6\zeta(5)=12\zeta(5)-6\zeta(2)\zeta(3)$$
combine the two integrals to get
$$\color{blue}{\int_{1/2}^1\frac{\ln^3(1-x)\ln x}{x}\ dx}\\=\frac3{16}\zeta(5)-3\zeta(2)\zeta(3)+\frac{21}8\ln^22\zeta(3)-\ln^32\zeta(2)+\frac25\ln^52+6\ln2\operatorname{Li}_4\left(\frac12\right)+6\operatorname{Li}_5\left(\frac12\right)$$
which gives
$$\boxed{\mathcal{K_3}=\frac1{8}\zeta(5)-2\zeta(2)\zeta(3)+\frac{7}4\ln^22\zeta(3)-\frac23\ln^32\zeta(2)+\frac4{15}\ln^52+4\ln2\operatorname{Li}_4\left(\frac12\right)+4\operatorname{Li}_5\left(\frac12\right)}$$
$$\mathcal{K_4}\overset{IBP}{=}-\frac12\ln2\operatorname{Li}_2^2(1/2)+\frac12\underbrace{\int_{1/2}^1\frac{\operatorname{Li}_2^2(x)}{x}dx}_{\mathcal{\large{K_5}}}$$
so
$$2\mathcal{K_4}+\mathcal{K_5}=-\ln2\operatorname{Li}_2^2(1/2)+2\int_{1/2}^1\frac{\operatorname{Li}_2^2(x)}{x}dx$$
where
$$\int_{1/2}^1\frac{\operatorname{Li}_2^2(x)}{x}dx=\int_0^{1}\frac{\operatorname{Li}_2^2(x)}{x}dx-\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}dx$$
we have
$$\int_0^{1}\frac{\operatorname{Li}_2^2(x)}{x}dx=\sum_{n=1}^\infty\frac1{n^2}\int_0^1 x^{n-1}\operatorname{Li}_2(x)dx=\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)$$
$$=\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{H_n}{n^4}=2\zeta(2)\zeta(3)-3\zeta(5)$$
and @Song proved here
$$\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}dx=\frac12\ln^32\zeta(2)-\frac78\ln^22\zeta(3)-\frac58\ln2\zeta(4)+\frac{27}{32}\zeta(5)+\frac78\zeta(2)\zeta(3)\\-\frac{7}{60}\ln^52-2\ln2\operatorname{Li}_4\left(\frac12\right)-2\operatorname{Li}_5\left(\frac12\right)$$
giving us
$$\int_{1/2}^1\frac{\operatorname{Li}_2^2(x)}{x}dx=-\frac12\ln^32\zeta(2)+\frac78\ln^22\zeta(3)+\frac58\ln2\zeta(4)-\frac{123}{32}\zeta(5)+\frac98\zeta(2)\zeta(3)\\+\frac{7}{60}\ln^52+2\ln2\operatorname{Li}_4\left(\frac12\right)+2\operatorname{Li}_5\left(\frac12\right)$$
consequently
$$\boxed{2\mathcal{K_4}+\mathcal{K_5}=-\frac12\ln^32\zeta(2)+\frac74\ln^22\zeta(3)+\frac{5}8\ln2\zeta(4)-\frac{123}{16}\zeta(5)+\frac94\zeta(2)\zeta(3)\\ \qquad\qquad\qquad-\frac{1}{60}\ln^52+4\ln2\operatorname{Li}_4\left(\frac12\right)+4\operatorname{Li}_5\left(\frac12\right)}$$
Finally combine the boxed results we obtain
$$\small{\mathcal{I}=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{125}{32}\zeta(5)-\frac{1}{8}\zeta(2)\zeta(3)+\frac{7}{4}\ln^22\zeta(3)-\frac2{3}\ln^32\zeta(2)+\frac{2}{15}\ln^52}$$
