Nested sine integral!!!!

If we define $I_n = \int_0^{2\pi} \cos\big(n(x + \sin7x)\big)\,dx \hspace{2cm} n \in \mathbb{N} \cup \{0\}$ then $I_n \; = \; \mathfrak{Re}\left[\int_0^{2\pi} e^{in(x + \sin7x)}\,dx\right] \; = \; \mathfrak{Re}\left[\int_{|z|=1} z^n e^{\frac12n(z^7-z^{-7})}\,\frac{dz}{iz}\right] \; = \; \mathfrak{Re}\left[2\pi\mathrm{Res}_{z=0}z^{n-1}e^{\frac12n(z^7-z^{-7})}\right]$ from which it is clear that $I_n = 0$ whenever $n$ is not a multiple of $7$ . Of course $I_0 = 2\pi$ . Since $512\sin^{10}u \; = \; -\cos10u + 10\cos8u - 45\cos6u + 120\cos4u - 210\cos2u + 126$ we deduce that $\begin{aligned} \int_0^\pi \sin^{10}(x + \sin7x)\,dx & = \; \tfrac12\int_0^{2\pi}\sin^{10}(x + \sin7x)\,dx \; = \; \tfrac{1}{1024}\big[-I_{10} + 10I_8 - 45I_6 + 120I_4 - 210I_2 + 126I_0\big] \\ & = \; \tfrac{63}{512}I_0 \; = \; \tfrac{63}{256}\pi \end{aligned}$ making the answer $63 + 1 + 256 = \boxed{320}$ .