Inspired from Hot Integral-14

This follows by applying a collection of standard results. Firstly $f$ can be expressed in terms of Bessel functions: $f(x) \; = \; \int_0^\infty \cos\big(x \cosh t\big)\,dt \; = \; -\tfrac12\pi Y_0(x) \hspace{2cm} x > 0$ while $\int_0^\infty \cos(2ax) Y_0(x)^2\,dx \; = \; \frac{2}{\pi a}K\left(\sqrt{1-a^{-2}}\right) \hspace{2cm} a > 1$ where $K$ is the complete elliptic integral of the first kind. Hence $\int_0^\infty \cos(2\sqrt{2}x)f(x)^2\,dx \; = \; \frac{\pi}{2\sqrt{2}} K\big(\tfrac{1}{\sqrt{2}}\big) \; = \; \frac{\pi}{2\sqrt{2}} \times \frac{1}{4\sqrt{\pi}}\Gamma\big(\tfrac14\big)^2 \; = \; \frac{\sqrt{2\pi}}{16}\Gamma\big(\tfrac14\big)^2$ using a known special value of the function $K$ . This makes the answer $2 + 16 + 1 + 4 + 2 = \boxed{25}$ .