Who's up to the challenge? 51

If we define $F(x) \; =\; \sum_{n\ge1} \frac{H_n x^n}{n} \; = \; -\int_0^x \frac{\ln(1-u)}{u(1-u)}\,du \; = \; \tfrac12\ln^2(1-x) + \mathrm{Li}_2(x)$ for $|x| < 1$ , so that (by Abel's Lemma) $\begin{array}{rcl} F(i)\; = \; \lim_{y\to1-}F(yi) & = & \displaystyle \tfrac12\ln^2(1-i) + \mathrm{Li}_2(i) \; =\; \tfrac12\big[\tfrac12\ln2 - \tfrac14\pi i\big]^2 + iG - \tfrac{1}{48}\pi^2 \\ & = & \displaystyle \tfrac18(\ln2)^2 - \tfrac{5}{96}\pi^2 + i\big(G - \tfrac18\pi\ln2\big) \end{array}$ then since $\begin{array}{rcl} \displaystyle \int_0^1 x^{2n}(1-x)\ln(1-x)\,dx & = & \displaystyle \sum_{m \ge 1} \frac{1}{m}\int_0^1 x^{2n+m}(x-1)\,dx \\ & = & \displaystyle \sum_{m \ge 1} \left(\frac{1}{m(m+2n+2)} - \frac{1}{m(m+2n+1)}\right) \\ & = & \displaystyle \frac{H_{2n+2}}{2n+2} - \frac{H_{2n+1}}{2n+1} \end{array}$ for all $n\ge 0$ , we see that $\begin{array}{rcl} \displaystyle \int_0^1 \frac{(1-x)\ln(1-x)}{1+x^2}\,dx & = & \displaystyle \sum_{n\ge0}(-1)^n \left( \frac{H_{2n+2}}{2n+2} - \frac{H_{2n+1}}{2n+1} \right) \\ & = & \displaystyle -\sum_{n \ge 0} \frac{H_{2n+2}i^{2n+2}}{2n+2} + i\sum_{n\ge0}\frac{H_{2n+1}i^{2n+1}}{2n+1} \\ & = & \displaystyle -\mathrm{Re}\big[F(i)\big] - \mathrm{Im}\big[F(i)\big] \\ & = & \displaystyle -G + \tfrac{5}{96}\pi^2 - \tfrac18(\ln2)^2 + \tfrac18\pi \ln2 \end{array}$ making the answer $5+2+96+ 2 + 2 + 8 + 2 + 8 = \boxed{125}$ .