\begin{align}
&{\rm I}\left(\varphi,\psi\right)
\equiv
\int_{0}^{\pi}{\rm d}\theta\ \
{\rm f}\left(\alpha\cos\left(\theta - \psi\right)\right)
\int_{-\infty}^{\infty}{\rm d}x\ \ x\,
{\rm e}^{-{\rm i}x\cos\left(\theta - \varphi\right)}
\\[3mm]&=
\int_{0}^{\pi}{\rm d}\theta\ \
{\rm f}\left(\alpha\cos\left(\theta - \psi\right)\right)\left\lbrack%
{2\pi \over {\rm i}\sin\left(\theta - \psi\right)}\,
{\partial \over \partial\theta}\
\overbrace{\quad%
\int_{-\infty}^{\infty}{{\rm d}x \over 2\pi}\,
{\rm e}^{-{\rm i}x\cos\left(\theta - \varphi\right)}\quad}
^{\delta\left(\cos\left(\theta - \varphi\right)\right)}
\right\rbrack
\\[3mm]&=
2\pi{\rm i}\int_{0}^{\pi}{\rm d}\theta\,
\delta\left(\cos\left(\theta - \varphi\right)\right)\,
{\partial \over \partial\theta}\left\lbrack%
{{\rm f}\left(\alpha\cos\left(\theta - \psi\right)\right)
\over
\sin\left(\theta - \psi\right)}
\right\rbrack
=
2\pi{\rm i}\int_{0}^{\pi}\!\!\!{\rm d}\theta\,
\delta\left(\cos\left(\theta - \varphi\right)\right)\,
{\partial{\rm F}\left(\alpha,\theta - \psi\right) \over \partial\theta}
\end{align}
$\displaystyle{%
\mbox{where}\quad {\rm F}\left(\alpha,\phi\right)
\equiv
{{\rm f}\left(\alpha\cos\left(\phi\right)\right)
\over
\sin\left(\phi\right)}}
$
\begin{align}
-------&----------------------------------
\\[3mm]
{\rm I}\left(\varphi,\psi\right)
&=
-2\pi{\rm i}\,{\partial \over \partial\psi}
\int_{0}^{\pi}{\rm d}\theta\,
{\rm F}\left(\alpha,\theta - \psi\right)
\delta\left(\cos\left(\theta - \varphi\right)\right)
\\[3mm]&=
-2\pi{\rm i}\,{\partial \over \partial\psi}
\int_{0}^{\pi}{\rm d}\theta\,
{\rm F}\left(\alpha,\theta - \psi\right)
\sum_{\ell = -\infty}^{\infty}
{\delta\left(\theta - \theta_{\ell}\left(\varphi\right)\right)
\over
\left\vert
-\sin\left(\theta_{\ell}\left(\varphi\right) - \varphi\right)
\right\vert}
\end{align}
where
$$
\theta_{\ell}\left(\varphi\right) \equiv \varphi + \left(2\ell + 1\right)\pi/2.
\quad\mbox{Notice that}\quad
\left\vert%
\begin{array}{rcl}
\sin\left(\theta_{\ell}\left(\varphi\right) - \varphi\right)
& = &
\left(-1\right)^{\ell}
\\
\sin\left(\theta_{\ell}\left(\varphi\right) - \psi\right)
& = &
\left(-1\right)^{\ell}\cos\left(\varphi - \psi\right)
\\
\cos\left(\theta_{\ell}\left(\varphi\right) - \psi\right)
& = &
\left(-1\right)^{\ell + 1}\sin\left(\varphi - \psi\right)
\end{array}\right.
$$
Then
\begin{align}
{\rm I}\left(\varphi,\psi\right)
&=
-2\pi{\rm i}\sum_{\ell = -\infty}^{\infty}
{\partial{\rm F}\left(\alpha,\theta_{\ell}\left(\varphi\right) - \psi\right)
\over
\partial\psi}
\int_{0}^{\pi}{\rm d}\theta\,
\delta\left(\theta - \theta_{\ell}\left(\varphi\right)\right)
\\[3mm]&=
-2\pi{\rm i}\!\!\!\!\!\!\!\!\!\!
\sum_{%
\ell = -\infty
\atop
{0\ <\ \theta_{\ell}\left(\varphi\right)\ <\ \pi\vphantom{\Huge A^{A}}}}
^{\infty}\!\!\!\!\!\!\!\!\!\!
{\partial{\rm F}\left(\alpha,\theta_{\ell}\left(\varphi\right) - \psi\right)
\over
\partial\psi}
=
-2\pi{\rm i}\!\!\!\!\!\!\!\!\!\!
\sum_{%
\ell = -\infty
\atop
{\left\vert\ell + \varphi/\pi\right\vert\ <\ 1/2\vphantom{\Huge A}}}
^{\infty}\!\!\!\!\!\!\!\!\!\!
{\partial{\rm F}\left(\alpha,\varphi - \psi + \ell\pi + \pi/2\right)
\over
\partial\psi}
\end{align}
$$
\left\lbrace%
\begin{array}{rcl}
\cos\left(\phi + \ell\pi + {\pi \over 2}\right)
& = &
-\sin\left(\ell\pi + \phi\right)
=
\left(-1\right)^{\ell + 1}\sin\left(\phi\right)
\\
\sin\left(\phi + \ell\pi + {\pi \over 2}\right)
& = &
\phantom{-}\cos\left(\ell\pi + \phi\right)
=
\left(-1\right)^{\ell\phantom{+1}}\cos\left(\phi\right)
\end{array}\right.
$$
\begin{align}
---&--------------------------------------
\\
{\rm I}\left(\varphi,\psi\right)
&=
\color{#ff0000}{-2\pi{\rm i}\!\!\!\!\!\!\!\!\!\!
\sum_{%
\ell = -\infty
\atop
{\left\vert\ell + \varphi/\pi\right\vert\ <\ 1/2\vphantom{\Huge A}}}
^{\infty}\!\!\!\!\!\!\!\!\!\!\left(-1\right)^{\ell}\,
{\partial \over \partial\psi}\left\lbrack%
{{\rm f}\left(\left(-1\right)^{\ell + 1}\alpha
\sin\left(\varphi - \psi\right)\right)
\over
\cos\left(\varphi - \psi\right)}
\right\rbrack
=
{\rm I}_{1}\left(\phi,\psi\right) + {\rm I}_{2}\left(\phi,\psi\right)}
\\[5mm]
{\rm I}_{1}\left(\varphi,\psi\right)
&=
\color{#ff0000}{-2\pi{\rm i}\alpha\!\!\!\!\!\!\!\!\!\!
\sum_{%
\ell = -\infty
\atop
{\left\vert\ell + \varphi/\pi\right\vert\ <\ 1/2\vphantom{\Huge A}}}
^{\infty}\!\!\!\!\!\!\!\!\!\!
{\rm f}'\left(\vphantom{\LARGE A}\left(-1\right)^{\ell + 1}\alpha
\sin\left(\varphi - \psi\right)\right)\sec\left(\varphi - \psi\right)}
\\[3mm]
{\rm I}_{2}\left(\varphi,\psi\right)
&=
\color{#ff0000}{2\pi{\rm i}\!\!\!\!\!\!\!\!\!\!
\sum_{%
\ell = -\infty
\atop
{\left\vert\ell + \varphi/\pi\right\vert\ <\ 1/2\vphantom{\Huge A}}}
^{\infty}\!\!\!\!\!\!\!\!\!\!\left(-1\right)^{\ell}
{\rm f}\left(\vphantom{\LARGE A}\left(-1\right)^{\ell + 1}\alpha
\sin\left(\varphi - \psi\right)\right)\sec\left(\varphi - \psi\right)
\tan\left(\varphi - \psi\right)}
\end{align}
By using the Maclaurin series of $\exp(x)$ and $\cos(x)$ both integrals boil down to the computation of
$$ \int_{0}^{\pi/2}\frac{\left(\cos x\right)^n}{a^2 \sin^2(x)+b^2\cos^2(x)}\,dx=\int_{0}^{\pi/2}\frac{\left(\sin x\right)^n}{a^2 \sin^2(x)+b^2\cos^2(x)}\,dx $$
which can be tackled thrugh the tangent half-angle substitution and the residue theorem.
Best Answer
This seems to be the Lambertian reflectance for parallelly polarised light passing from medium 0 to medium 1, so let us introduce $n = \frac{n_1}{n_0} > 1$ and write your integral as \begin{align} r_\text{p} (n) &= \int \limits_0^{\pi/2} \left(\frac{n^2 \cos(f) - \sqrt{n^2 - \sin^2(f)}}{n^2 \cos(f) + \sqrt{n^2-\sin^2(f)}}\right)^2 2 \sin(f) \cos(f) \, \mathrm{d} f \\ &\!\!\!\!\!\!\stackrel{\sin^2(f) = t}{=} \int \limits_0^1 \left(\frac{1 - \frac{n^2 \sqrt{1-t}}{\sqrt{n^2-t}}}{1 + \frac{n^2 \sqrt{1-t}}{\sqrt{n^2-t}}}\right)^2 \, \mathrm{d} t \stackrel{\frac{n \sqrt{1-t}}{\sqrt{n^2-t}} = u}{=} 2 n^2 (n^2 - 1) \int \limits_0^1 \frac{u (1-nu)^2}{(1+nu)^2(n+u)^2(n-u)^2} \, \mathrm{d} u \\ &= 2 n^2 (n^2 - 1) \int \limits_0^1 \left[\frac{8 n^3(n^4+1)}{(n^4-1)^3(1+nu)} - \frac{4n^3}{(n^4-1)^2 (1+nu)^2} \right. \\ &\phantom{= 2 n^2 (n^2 - 1) \int \limits_0^1 \left[\vphantom{\frac{8 n^3(n^4+1)}{(n^4-1)^3(1+nu)}}\right.} \left. - \, \frac{n^2-1}{(n^2+1)^3 (n-u)} + \frac{(n^2-1)^2}{4(n^2+1)^2n(n-u)^2} \right. \\ &\phantom{= 2 n^2 (n^2 - 1) \int \limits_0^1 \left[\vphantom{\frac{8 n^3(n^4+1)}{(n^4-1)^3(1+nu)}}\right.} \!\left. - \, \frac{n^2+1}{(n^2-1)^3(n+u)} - \frac{(n^2+1)^2}{4(n^2-1)^2 n (n+u)^2}\right] \mathrm{d} u \, . \end{align} The remaining integrals are elementary and after some simplification we end up with $$ r_\text{p} (n) = 1 - \frac{4 n^3 (n^2+2n-1)}{(n^2-1)(n^2+1)^2} + \frac{16 n^4 (n^4+1)}{(n^2-1)^2(n^2+1)^3} \ln(n) - \frac{4 n^2 (n^2-1)^2}{(n^2+1)^3} \operatorname{arcoth} (n) \, .$$