Evaluating $\int_0^{\frac{\pi}{2}} \frac{K_0 \left( \frac{x}{\sqrt{B(\phi)}} \right)}{B(\phi)} \, \sin^2 \phi \, \mathrm{d} \phi$

Question

I am currently working on the evaluation of the following definite integral, which emerged during the application of an inverse Fourier transform: $$ f(x) = \int_0^{\frac{\pi}{2}} \frac{K_0 \left( \frac{x}{\sqrt{B(\phi)}} \right)}{B(\phi)} \, \sin^2 \phi \, \mathrm{d} \phi, \tag{1} $$

where $x \in \mathbb{R}$ and $B = 1 + A \sin^2\phi \cos^2 \phi$ with $A \in [-4, \infty)$. In this expression, $K_0$ represents the modified Bessel function of the second kind with a zeroth-order. It can readily be checked that $B(\phi) \ge 0, \forall \phi \in [0,\pi/2]$.

My initial approach was to employ a change of variables by substituting $u = x/\sqrt{B(\phi)}$. However, this method introduced a challenge, as it rendered the upper and lower limits of the integral equal. I am seeking assistance from this forum to explore the possibility of evaluating this integral analytically.

EDIT:

The above integral arose after applying the inverse Fourier transform of: $$ \int_0^{2\pi} \int_0^\infty \frac{k J_0(xk)}{1+B(\phi) k^2} \, \sin^2\phi \, \mathrm{d}k \, \mathrm{d}\phi \, , \tag{2} $$ with $k$ denoting the wavenumber. It can be verified that the integral over the range of 0 to $2\pi$ is exactly four times the integral over the range of 0 to $\pi/2$.

By evaluating the integral (2) with respect to $\phi$ first I get $$ \int_0^\infty \frac{2\pi \, k J_0(xk)}{\sqrt{(1+k^2) \left( 4+(A+4)k^2 \right)}} \, \mathrm{d} k \, . \tag{3} $$