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This integral was also posted recently on other sites like AoPS, but since it didnt get an answer I thought no one minds if I post this here too. $$I=\int_{0}^{\infty} \frac{dx}{(1+x^2)(5-4\cos(2x))}$$ I have no successful attempt that lead actually to something relevant, but I would love to see a closed form.

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Of course it has a nice closed form. Since $$ \frac{1}{5-4\cos(2x)} = \frac{1}{3}+\frac{2}{3}\sum_{n\geq 1}\frac{\cos(2nx)}{2^n}\tag{1} $$ and $$ \int_{0}^{+\infty}\frac{\cos(2nx)}{1+x^2}\,dx = \frac{\pi}{2}e^{-2n}\tag{2}$$ we have $$ \int_{0}^{+\infty}\frac{dx}{(1+x^2)(5-4\cos(2x))}=\frac{\pi}{6}+\frac{2}{3}\sum_{n\geq 1}\frac{\pi}{2^{n+1}}e^{-2n}=\color{red}{\frac{\pi}{6}\cdot\frac{2e^2+1}{2e^2-1}}.\tag{3} $$ The same technique can be used for computing $ \int_{0}^{+\infty}\frac{dx}{(1+x^2)^m (5-4\cos(2x))^n}$, too.

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    $\begingroup$ You are correct about the possible generalization and the last series is just a geometric series. $\endgroup$ – Jack D'Aurizio Apr 20 '18 at 8:12

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