Integral Stuff 3

Integating by parts twice, and then putting $x = \cot\theta$ , we obtain $\begin{aligned} \int_0^\infty \left(\tfrac12\pi - \tan^{-1}x\right)^3\,dx & = \; \int_0^\infty \big(\tan^{-1}x^{-1}\big)^3\,dx \; = \; \left[ x\big(\tan^{-1}x^{-1}\big)^3\right]_0^\infty + 3 \int_0^\infty\frac{\big(\tan^{-1}x^{-1}\big)^2 x}{x^2+1}\,dx \\ & = \; 3\int_0^\infty\frac{\big(\tan^{-1}x^{-1}\big)^2 x}{x^2+1}\,dx \; = \; 3\left[\tfrac12\big(\tan^{-1}x^{-1}\big)^2\,\ln(x^2+1)\right]_0^\infty + 3\int_0^\infty \frac{\tan^{-1}x^{-1}\, \ln(x^2+1)}{x^2+1}\,dx \\ & = \; 3\int_0^\infty \frac{\tan^{-1}x^{-1}\, \ln(x^2+1)}{x^2+1}\,dx \; = \; 6\int_0^{\frac12\pi}\theta\ln(\mathrm{cosec}\,\theta)\,d\theta \\ & = \; -6\int_0^{\frac12\pi}\theta\ln(\sin\theta)\,d\theta \end{aligned}$ Using Euler's integral $\int_0^{\frac12\pi}\theta \ln(\sin\theta)\,d\theta \; = \; \tfrac{7}{16}\zeta(3) - \tfrac18\pi^2\ln2$ we obtain $\int_0^\infty \left(\tfrac12\pi - \tan^{-1}x\right)^3\,dx \; = \; \tfrac34\pi^2\ln2 - \tfrac{21}{8}\zeta(3) \; = \; \tfrac38\big(\pi^2\ln4 - 7\zeta(3)\big)$ making the answer $2+4+7+3=\boxed{16}$ .