integration - Fourier transform in three dimensions getting out of hand

I have the following integral I wish to compute, it transforms a quantum position wave function into momentum space:

$$\phi(\mathbf p)=\int\frac{\mathrm d^3r}{(2\pi\hbar)^{3/2}}e^{-i\mathbf{p\cdot r}}\psi(\mathbf r)$$
Where $\psi(\mathbf r)=e^{-|\mathbf r|/a}/\sqrt{\pi a^3}$, the wave equation for the ground state of hydrogen.

So I thought about converting it to polar coordinates first then evaluate the integrals, but it turned out to be unexpectedly complicated.

Below is my attempt (note, since this is physics related, $\theta$ is swapped with $\varphi$ in polar coordinates):

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We first rewrite the integral in polar coordinates:
$$\phi(r_p, \varphi_p, \theta_p)=\frac{1}{(2a\hbar)^{3/2}\pi^2}\int_0^\pi\int_0^{2\pi}\int_0^\infty e^{-i\mathbf{p\cdot x}/h-r_x/a}\mathrm d r_x\mathrm d\varphi_x \mathrm d\theta_x$$

Where the dot product in spherical coordinates is:
$$\mathbf{p\cdot x}=r_xr_p[\sin\theta_p\sin\theta_x\cos(\varphi_p-\varphi_x)+\cos\theta_p\cos\theta_x]$$

Substitute and integrate, we get:
$$\begin{align*}\phi(r_p, \varphi_p, \theta_p)&=\frac{1}{(2a\hbar)^{3/2}\pi^2ir_p}\int_0^{2\pi}\int_0^\pi \frac{\mathrm d\theta_x \mathrm d\varphi_x}{\sin\theta_p\sin\theta_x\cos(\varphi_p-\varphi_x)+\cos\theta_p\cos\theta_x}\\ \textrm{Now let: } a &=\sin\theta_p\cos(\varphi_p-\varphi_x)\\ b &=\cos\theta_p \\ \phi&=\frac{1}{(2a\hbar)^{3/2}\pi^2ir_p}\int_0^\pi \frac{\ln(\sqrt{a^2+b^2}-a)-\ln(\sqrt{a^2+b^2}+a)}{\sqrt{a^2+b^2}}\mathrm d\varphi_x\\ \end{align*}$$

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The second integral is evaluated from wolfram alpha. From here I'm quite stuck. The final integral looks rather hopeless. Is there any tricks to Fourier transforms in space or did I make a mistake somewhere?

Edit: I think I accidentally substituted $2\pi$ instead of $\pi$, but the correct version is even more complicated.

Answer

Symmetry is your friend. Since $\psi$ is radially symmetric, so is $\phi$.
So wlog we can assume ${\bf p} = p \bf k$, and ${\bf p} . {\bf r} = p r \cos(\theta)$. Then

$$\eqalign{\phi(p {\bf k}) &= \dfrac{1}{(2 a \hbar)^{3/2} \pi^2} \int_0^{2\pi} d\varphi \int_{0}^\infty r^2 \; dr \int_{0}^{\pi} \sin(\theta)\; d\theta\; e^{ip r \cos(\theta)} e^{-r/a}\cr &= \dfrac{1}{\sqrt{2} (a \hbar)^{3/2} \pi} \int_0^\infty r^2\; dr \dfrac{2 \sin(pr)}{pr} e^{-r/a}\cr &= \dfrac{4 a^{3/2} }{\sqrt{2} \hbar^{3/2} \pi (a^2 p^2 + 1)^2}}$$