"cot" in the act!

This one needs a couple of technical integrals: $\begin{aligned} I_n & = \; \int_0^\infty \frac{u^n\,du}{u^{10} + 1} \; = \; \tfrac{1}{10}\pi \,\mathrm{cosec}\left(\tfrac{(n+1)\pi}{10}\right) \hspace{2cm} 0 \le n \le 8 \\ K_\alpha & = \; \int_1^\infty \frac{u^\alpha-1}{u-1} \; = \; -\tfrac12H_{-\frac12(1+\alpha)} - \ln2 \hspace{2cm} -1 < \alpha < 1 \end{aligned}$ The first of these can be proved by fairly straightforward contour integration, while the second is easily derivable from standard integrals about the generalised Harmonic number function.

If we define $F(t) \; = \; \int_0^\pi \cot^{-1}\big(t \cot^5x\big)\, \cot x\,dx \; = \; 2\int_0^{\frac12\pi} \cot^{-1}\big(t\cot^5x\big)\,\cot x\,dx$ then $F(t) \to 0$ as $t \to \infty$ and $F'(t) \; = \; -2\int_0^{\frac12\pi} \frac{\cot^6x\,dx}{t^2\cot^{10}x + 1} \; = \; -2\int_0^{\frac12\pi} \frac{\tan^4x\,dx}{\tan^{10}x + t^2} \; = \; -2\int_0^\infty \frac{u^4\,du}{(u^2+1)(u^{10}+t^2)}$ Expanding this last integrand by partial fractions we see that $\begin{aligned} F'(t) & = \; -\frac{2}{t^2-1}\int_0^\infty \left\{ \frac{1}{u^2+1} - \frac{u^8 - u^6 + u^4 - t^2u^2 + t^2}{u^{10} + t^2}\right\}\,du \\ & = \; -\frac{2}{t^2-1}\left[ \tfrac12\pi - t^{-\frac15}I_8 + t^{-\frac35}I_6 - t^{-1}I_4 + t^{\frac35}I_2 - t^{\frac15}I_0\right] \\ & = \; \frac{\pi}{5(t^2-1)}\left[ t^{-\frac15}\mathrm{cosec}\left(\tfrac{9\pi}{10}\right) - t^{-\frac35}\mathrm{cosec}\left(\tfrac{7\pi}{10}\right) + t^{-1} - t^{\frac35} \mathrm{cosec}\left(\tfrac{3\pi}{10}\right) + t^{\frac15}\mathrm{cosec}\left(\tfrac{\pi}{10}\right) - 5\right] \\ & = \; \frac{\pi}{5(t^2-1)}\left[(1 +\sqrt{5})\big(t^{\frac15} + t^{-\frac15}\big) + (1 - \sqrt{5})\big(t^{\frac35} + t^{-\frac35}\big) + t^{-1} - 5\right] \\ & = \; \frac{\pi}{5(t^2-1)}\left[(1 + \sqrt{5})\big(t^{\frac15} - 1 + t^{-\frac15} - 1\big) + (1 - \sqrt{5})\big(t^{\frac35} - 1 + t^{-\frac35} - 1\big)\right] - \frac{\pi}{5t(t+1)} \end{aligned}$ so the desired integral is $\begin{aligned} F(1) & = \; -\int_1^\infty F'(t)\,dt \; = \; -\tfrac15\pi\left[(1+\sqrt{5})\left(K_{\frac15} + K_{-\frac15}\right) + (1 - \sqrt{5})\left(K_{\frac35} + K_{-\frac35}\right)\right] + \tfrac15\pi\ln2 \\ & = \; \tfrac{1}{10}\pi\left[(1 + \sqrt{5})\left(H_{-\frac35} + H_{-\frac25}\right) + (1 - \sqrt{5})\left(H_{-\frac45} + H_{-\frac15}\right)\right] + \pi\ln2 \\ & = \; \tfrac{\pi}{5}\ln\left[\tfrac{64}{125}(25 + 11\sqrt{5})\right] \; = \; \tfrac{\pi}{5}\Big[4\ln2 - 3\ln5 + \ln(25 + 11\sqrt{5})\Big] \end{aligned}$ giving the answer $5 + 4 + 2 + 3 + 25 + 11 = \boxed{50}$ .

(N.B. RMM notified)

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It is worth noting that this integral can be much more elegantly expressed, since $\tfrac{64}{125}(25 + 11\sqrt{5}) \; = \; \left(\frac{5 + \sqrt{5}}{5}\right)^5$ and hence $F(1) \; = \; \pi \ln\left(\frac{5+\sqrt{5}}{5}\right)$