Evaluate an integral without using H(n)

Ignoring the title of the question, we can use the expansion for the polylogarithm to show that $\int_0^1 \frac{\ln(1-x)\mathrm{Li}_2(-x)}{x}\,dx \; = \; \sum_{k=1}^\infty \frac{(-1)^k}{k^2}\int_0^1 x^{k-1}\ln(1-x)\,dx \; = \; \sum_{k=1}^\infty (-1)^{k+1}\frac{H_k}{k^3}$ and this paper tells us that this is equal to $\frac12\int_0^1 \frac{(\ln x)^2\ln(1+x)}{x(1+x)}\,dx \; = \; -2\mathrm{Li}_2(\tfrac12) + \tfrac{11}{4}\zeta(4) + \tfrac12\zeta(2)(\ln2)^2 - \tfrac{1}{12}(\ln 2)^4 - \tfrac74\zeta(3)\ln2 = \boxed{0.859247}$