How to evaluate $\int_0^y\frac{\ln x\ln^2(1-x)}{x}dx$

Thanks to @user97357329 for providing the key identity to solve the integral $C$,

$$\ln^3\left(\frac{x}{1-x}\right) =\ln^3(x)-3 \ln^2(x)\ln(1-x)+3 \ln(x)\ln^2(1-x)-\ln^3(1-x)$$

To have a less messy solution, I am going to evaluate the integral $C$ as an indefinite integral and the integral constant will be ignored as we will plug in the boundaries $0$ and $y$ eventually.

It follows from the identity above that

$$3\int\frac{\ln x\ln^2(1-x)}{x}dx$$

$$=\underbrace{\int\frac1{x}\ln^3\left(\frac{x}{1-x}\right)dx}_{\Large\mathcal{I}_1}+3\underbrace{\int\frac{\ln^2x\ln(1-x)}{x}dx}_{\Large\mathcal{I}_2}+\underbrace{\int\frac{\ln^3(1-x)}{x}dx}_{\Large\mathcal{I}_3}-\frac14\ln^4x$$

For $\mathcal{I}_1$, sub $\frac{x}{1-x}=t$

$$\mathcal{I}_1=\int\frac{\ln^3t}{t(1+t)}dt=\int\frac{\ln^3t}{t}dt-\int\frac{\ln^3t}{1+t}dt$$

$$=\frac14\ln^4t+\sum_{n=1}^\infty(-1)^n\int x^{n-1}\ln^3t\ dt$$

$$=\frac14\ln^4t+\sum_{n=1}^\infty(-1)^n\left(\frac{\ln^3t\ t^n}{n}-\frac{3\ln^2t\ t^n}{n^2}+\frac{6\ln t\ t^n}{n^3}-\frac{6t^n}{n^4}\right)$$

$$=\frac14\ln^4t-\ln^3t\ln(1+t)-3\ln^2t\operatorname{Li}_2(-t)+6\ln t\operatorname{Li}_3(-t)-6\operatorname{Li}_4(-t)$$

$$=\frac14\ln^4\left(\frac{x}{1-x}\right)+\ln^3\left(\frac{x}{1-x}\right)\ln(1-x)-3\ln^2\left(\frac{x}{1-x}\right)\operatorname{Li}_2\left(\frac{x}{x-1}\right)\\+6\ln \left(\frac{x}{1-x}\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)-6\operatorname{Li}_4\left(\frac{x}{x-1}\right)$$

$$\mathcal{I}_2=-\sum_{n-1}^\infty\frac{1}{n}\int x^{n-1}\ln^2x\ dx$$

$$=-\sum_{n=1}^\infty\frac1n\left(\frac{\ln^2x\ x^n}{n}-\frac{2\ln x\ x^n}{n^2}+\frac{2x^n}{n^3}\right)$$

$$=-\ln^2x\operatorname{Li}_2(x)+2\ln x\operatorname{Li}_3(x)-2\operatorname{Li}_4(x)$$

For $\mathcal{I}_3$, use $1-x=t$

$$\mathcal{I}_3=-\int\frac{\ln^3t}{1-t}dt=-\sum_{n=1}^\infty t^{n-1}\ln^3t\ dt$$

$$=-\sum_{n=1}^\infty\left(\frac{\ln^3t\ t^n}{n}-\frac{3\ln^2t\ t^n}{n^2}+\frac{6\ln t\ t^n}{n^3}-\frac{6t^n}{n^4}\right)$$

$$=\ln^3t\ln(1-t)+3\ln^2t\operatorname{Li}_2(t)-6\ln t\operatorname{Li}_3(t)+6\operatorname{Li}_4(t)$$

$$=\ln^3(1-x)\ln x+3\ln^2(1-x)\operatorname{Li}_2(1-x)-6\ln (1-x)\operatorname{Li}_3(1-x)+6\operatorname{Li}_4(1-x)$$

Combine the results of $\mathcal{I}_1$, $\mathcal{I}_2$ and $\mathcal{I}_3$, the terms $\frac14\ln^4x$ and $\ln x\ln^3(1-x)$ nicely cancel out

$$3\int\frac{\ln x\ln^2(1-x)}{x}dx$$ $$=\frac32\ln^2x\ln^2(1-x)-\ln^3x\ln(1-x)+\frac14\ln^4(1-x)+\ln^3\left(\frac{x}{1-x}\right)\ln(1-x)\\-3\ln^2\left(\frac{x}{1-x}\right)\operatorname{Li}_2\left(\frac{x}{x-1}\right)+6\ln \left(\frac{x}{1-x}\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)-6\operatorname{Li}_4\left(\frac{x}{x-1}\right)-3\ln^2x\operatorname{Li}_2(x)\\+6\ln x\operatorname{Li}_3(x)-6\operatorname{Li}_4(x)+3\ln^2(1-x)\operatorname{Li}_2(1-x)-6\ln (1-x)\operatorname{Li}_3(1-x)+6\operatorname{Li}_4(1-x)$$

Now plug in the boundaries $0$ and $y$ then divide by $3$ we obtain that

$$\int_0^y\frac{\ln x\ln^2(1-x)}{x}dx$$ $$=\frac12\ln^2y\ln^2(1-y)-\frac13\ln^3y\ln(1-y)+\frac1{12}\ln^4(1-y)+\frac13\ln^3\left(\frac{y}{1-y}\right)\ln(1-y)\\-\ln^2\left(\frac{y}{1-y}\right)\operatorname{Li}_2\left(\frac{y}{y-1}\right)+2\ln \left(\frac{y}{1-y}\right)\operatorname{Li}_3\left(\frac{y}{y-1}\right)-2\operatorname{Li}_4\left(\frac{y}{y-1}\right)-\ln^2y\operatorname{Li}_2(y)\\+2\ln y\operatorname{Li}_3(y)-2\operatorname{Li}_4(y)+\ln^2(1-y)\operatorname{Li}_2(1-y)-2\ln (1-y)\operatorname{Li}_3(1-y)+2\operatorname{Li}_4(1-y)-2\zeta(4)$$

Bonus:

In the question body we have

$$\frac12\int_0^y\frac{\ln x\ln^2(1-x)}{x}dx=\operatorname{Li}_4(y)-\ln y\operatorname{Li}_3(y)+\ln y\sum_{n=1}^\infty\frac{H_n}{n^2}y^n-\sum_{n=1}^\infty\frac{H_n} {n^3}y^n$$

Substitute

$$\sum_{n=1}^\infty\frac{H_{n}}{n^2}y^{n}=\operatorname{Li}_3(y)-\operatorname{Li}_3(1-y)+\ln(1-y)\operatorname{Li}_2(1-y)+\frac12\ln y\ln^2(1-y)+\zeta(3)$$

and $$\frac12\int_0^y\frac{\ln x\ln^2(1-x)}{x}dx$$ $$=\frac14\ln^2y\ln^2(1-y)-\frac16\ln^3y\ln(1-y)+\frac1{24}\ln^4(1-y)+\frac16\ln^3\left(\frac{y}{1-y}\right)\ln(1-y)\\-\frac12\ln^2\left(\frac{y}{1-y}\right)\operatorname{Li}_2\left(\frac{y}{y-1}\right)+\ln \left(\frac{y}{1-y}\right)\operatorname{Li}_3\left(\frac{y}{y-1}\right)-\operatorname{Li}_4\left(\frac{y}{y-1}\right)-\frac12\ln^2y\operatorname{Li}_2(y)\\+\ln y\operatorname{Li}_3(y)-\operatorname{Li}_4(y)+\frac12\ln^2(1-y)\operatorname{Li}_2(1-y)-\ln (1-y)\operatorname{Li}_3(1-y)+\operatorname{Li}_4(1-y)-\zeta(4)$$

we obtain that

$$\sum_{n=1}^\infty\frac{H_n}{n^3}y^n$$ $$=\zeta(4)-\frac1{24}\ln^4(1-y)+\frac16\ln^3y\ln(1-y)-\frac16\ln^3\left(\frac{y}{1-y}\right)\ln(1-y)+\frac14\ln^2y\ln^2(1-y)$$

$$-\frac12\ln^2(1-y)\operatorname{Li}_2(1-y)+\frac12\ln^2y\operatorname{Li}_2(y)+\ln (1-y)\operatorname{Li}_3(1-y)-\ln y\operatorname{Li}_3(y)$$

$$-\ln y\operatorname{Li}_3(1-y)+\ln y\ln(1-y)\operatorname{Li}_2(1-y)+\zeta(3)\ln y+2\operatorname{Li}_4(y)-\operatorname{Li}_4(1-y)$$

$$+\frac12\ln^2\left(\frac{y}{1-y}\right)\operatorname{Li}_2\left(\frac{y}{y-1}\right)-\ln \left(\frac{y}{1-y}\right)\operatorname{Li}_3\left(\frac{y}{y-1}\right)+\operatorname{Li}_4\left(\frac{y}{y-1}\right)$$

If we use Landen's identity

$$\operatorname{Li}_2(y)+\operatorname{Li}_2\left(\frac{y}{y-1}\right)=-\frac12\ln^2(1-y)$$

and

$$\operatorname{Li}_3(1-y)+\operatorname{Li}_3(y)+\operatorname{Li}_3\left(\frac{y}{y-1}\right)=\zeta(3)+\frac16\ln^3(1-y)-\frac12\ln^2y\ln(1-y)+\zeta(2)\ln y$$

the sum simplifies to

\begin{align} \sum_{n=1}^\infty\frac{H_n}{n^3}y^n&=\operatorname{Li}_4\left(\frac{y}{y-1}\right)-\frac12\operatorname{Li}_2^2\left(\frac{y}{y-1}\right)+2\operatorname{Li}_4(y)-\operatorname{Li}_4(1-y)-\ln(1-y)\operatorname{Li}_3(y)\\ &\quad +\frac12\ln^2(1-y)\operatorname{Li}_2(y)+\frac12\operatorname{Li}_2^2(y)+\frac16\ln^4(1-y)-\frac16\ln y\ln^3(1-y)\\ &\quad+\frac12\zeta(2)\ln^2(1-y)+\zeta(3)\ln(1-y)+\zeta(4) \end{align}