Don't expand

$\large{J=\displaystyle \int^{\pi}_{0} \frac{dx}{a- \cos x}}$

using $\displaystyle \int^{t}_{0} f(x).dx=\displaystyle \int^{t}_{0} f(t-x).dx$

and adding both, we obtain

$\large{2J=\displaystyle \int^{\pi}_{0} \frac{2a}{a^{2}-\cos^{2} (x)}.dx}$

$\large{J=a \displaystyle \int^{\pi}_{0} \frac{sec^{2}(x).dx}{a^{2}sec^{2}(x)-1}}$

$\large{J=2a \displaystyle \int^{\frac{\pi}{2}}_{0} \frac{sec^{2}(x).dx}{a^{2}sec^{2}(x)-1}}$

let $tanx=t$

$\large{J=2a \displaystyle \int^{\infty}_{0} \frac{dt}{a^{2}t^{2}+a^{2}-1}}$

$\large{J=2a \displaystyle \int^{\infty}_{0} \frac{dt}{(at)^{2}+(\sqrt{a^{2}-1})^{2}}}$

$\huge{J=\frac { 2a\left( { \arctan { \frac { at }{ \sqrt { { a }^{ 2 }-1 } } } }_{ 0 }^{ \infty } \right) }{ a\sqrt { { a }^{ 2 }-1 } } }$

$\large{J=\frac{\pi}{\sqrt{a^2-1}}}$

let $\large{I(a)=\displaystyle \int^{\pi}_{0} \frac{dx}{a- \cos x}=\frac{\pi}{\sqrt{a^{2}-1}}}$

on differentiating $wrt$ $a$

$\large{I^{\prime}(a)=\displaystyle \int^{\pi}_{0} \frac{dx}{(a- \cos x)^{2}}=\frac{\pi a}{(a^2-1)^{\frac{3}{2}}}}$

again differentiating

$\large{I^{\prime \prime}(a)=\displaystyle \int^{\pi}_{0} \frac{-2 dx}{(a- \cos x)^{3}}=\frac{- \pi \sqrt{a^2-1} (2a^2+1)}{(a^2-1)^{3}}}$

$\huge{\displaystyle \int^{\pi}_{0} \frac{dx}{(a- \cos x)^{3}}=\frac{\pi \sqrt{a^2-1}(2a^2+1)}{2(a^2-1)^{3}}}$

substitute $a=\sqrt{17}$ , we get

$\frac{35 \pi}{2048}$