Who's up to the challenge? 53

Note that $\int_0^1 \frac{x^{\alpha-1} (\ln x)^2}{1-x}\,dx \; = \;\displaystyle \lim_{\beta\to0}\frac{\partial^2B}{\partial\alpha^2}(\alpha,\beta) \; =\; -\psi^{(2)}(\alpha)$ for $\alpha > 0$ , so that $\int_0^1 \frac{x^n (\ln x)^2}{1-x}\,dx \; = \; -\psi^{(2)}(n+1) \; = \; 2\big[\zeta(3) - H^{(3)}_n\big]$ for all integers $n \ge 1$ . Hence $\begin{array}{rcl} \displaystyle \int_0^1 \frac{(\ln x)^2 \mathrm{Li}_3(x)}{1-x}\,dx & = & \displaystyle \sum_{n=1}^\infty \int_0^1 \frac{x^n (\ln x)^2}{n^3 (1-x)}\,dx \; =\; 2\sum_{n=1}^\infty \frac{\zeta(3) - H^{(3)}_n}{n^3} \\ & = & \displaystyle 2\left[\zeta(3)^2 - \sum_{n=1}^\infty \frac{H^{(3)}_n}{n^3}\right] \; = \; 2\left[ \zeta(3)^2 - \tfrac12\big(\zeta(3)^2 + \zeta(6)\big)\right] \\ & = & \displaystyle \zeta(3)^2 - \zeta(6) \end{array}$ making the answer $2+3+6 = \boxed{11}$ .