Dilogarithm integral $\int^x_0 \frac{\operatorname{Li}_2(1-t)\log(1-t)}{t}\, dt$

$$\begin{align} &\int^x_0 \frac{\operatorname{Li}_2(1-t)\log(1-t)}{t}\,dt=\\ &\frac{\operatorname{Li}_2^2(1-x)}2-2\operatorname{Li}_4\left(1-\frac1x\right)-2\operatorname{Li}_4(1-x)+2\operatorname{Li}_4(x)-\operatorname{Li}_2\left(1-\frac1x\right)\log^2\left(\frac1x-1\right)+\\ &\operatorname{Li}_2(x)\left(\log^2\left(\frac1x-1\right)+\log(1-x)\log(x)\right)+2\operatorname{Li}_3(x)\log\left(\frac1x\right)+\\&2\operatorname{Li}_3\left(1-\frac1x\right)\log\left(\frac1x-1\right)+2\operatorname{Li}_3(1-x)\left(\log\left(\frac1x-1\right)+\log(x)\right)-\\ &\frac14\log^4\left(\frac1x\right)+\frac13\log^3(1-x)\left(2\log(x)-\log\left(\frac1x\right)\right)-\\ &\log(1-x)\left(\log^3\left(\frac1x\right)+\frac13\pi^2\log\left(\frac1x\right)+\frac{\pi^2}6\log(x)\right)+\\&\log^2(1-x)\left(-\log^2\left(\frac1x\right)+\frac12\log^2(x)+\log(x)\log\left(\frac1x\right)-\frac{\pi^2}6\right)-\frac{11\pi^4}{360},\end{align}$$ that can be checked by taking derivatives from both sides.

See the corresponding indefinite integral at WolframAlpha.

The form of the integrand suggests the use of the reflection formula might come handy. We have: \begin{eqnarray} &&\int\limits_0^x Li_2(1-t) \frac{\log(1-t)}{t} dt = \int\limits_0^x \left(\frac{\pi^2}{6} - \log(t) \log(1-t) - Li_2(t) \right) \cdot \frac{\log(1-t)}{t} dt \\ &&=-\frac{\pi^2}{6} Li_2(x) + \frac{1}{2} Li_2(x)^2 - \int\limits_0^x \log(t) \frac{\log(1-t)^2}{t} dt \\ &&=-\frac{\pi^2}{6} Li_2(x) + \frac{1}{2} Li_2(x)^2 - \left(-2 S_{2,2}(x) + 2 \log(x) S_{1,2}(x)\right) \end{eqnarray} where $S_{p,q}(x)$ are the Nielsen generalized polylogarithms (see http://mathworld.wolfram.com/NielsenGeneralizedPolylogarithm.html for definition).

Here is a proof for your sum not the integral,

From here 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}

To get your integral, integrate by parts

$$\int_0^y\frac{\ln(1-x)\operatorname{Li}_2(1-x)}{x}dx=\operatorname{Li}_2(y)\operatorname{Li}_2(1-y)+\int_0^y\frac{\ln x\operatorname{Li}_2(x)}{1-x}dx$$

$$=\operatorname{Li}_2(y)\operatorname{Li}_2(1-y)+\sum_{n=1}^\infty \left(H_n^{(2)}-\frac1{n^2}\right)\int_0^y x^{n-1}\ln x\ dx$$

$$=\operatorname{Li}_2(y)\operatorname{Li}_2(1-y)+\sum_{n=1}^\infty \left(H_n^{(2)}-\frac1{n^2}\right)\left(\ln y\frac{y^n}{n}-\frac{y^n}{n^2}\right)$$

$$=\operatorname{Li}_2(y)\operatorname{Li}_2(1-y)-\ln y\operatorname{Li}_3(y)+\operatorname{Li}_4(y)+\ln y\sum_{n=1}^\infty\frac{H_n^{(2)}}{n}y^n-\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}y^n$$

by Cauchy product we have

$$\frac12\operatorname{Li}_2^2(y)=2\sum_{n=1}^\infty\frac{H_n}{n^3}y^n+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}y^n-3\operatorname{Li}_4(y)$$

which gives

$$\int_0^y\frac{\ln(1-x)\operatorname{Li}_2(1-x)}{x}dx$$ $$=\operatorname{Li}_2(y)\operatorname{Li}_2(1-y)-\frac12\operatorname{Li}_2^2(y)-\ln y\operatorname{Li}_3(y)-2\operatorname{Li}_4(y)+\ln y\sum_{n=1}^\infty\frac{H_n^{(2)}}{n}y^n+2\sum_{n=1}^\infty\frac{H_n}{n^3}y^n$$

where

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