On the integral $\iint_{\partial B_1(0)}\frac{d\Omega}{\left(1+a\cdot r\right)\left(1+b\cdot r\right)}$ Let $a, b$ be arbitrary vectors in $\mathbb{R}^3$ and define $\partial B_1(0):=\{r\in\mathbb{R}^3:|r|=1\}$ as the boundary of the three-dimensional unit sphere.
I am trying to evaluate the following integral:
$$I = \iint_{\partial B_1(0)}\frac{d\Omega}{\left(1+a\cdot r\right)\left(1+b\cdot r\right)}$$
So far, I've said that we can, given a vector $a$ and a fixed vector $b$, write the following:
$$a = c b + x$$
where $c = \frac{a\cdot b}{|b|}$ and $x\in\mathbb{R}^3$ such that $x$ orthogonal to $b$.
Then, if we choose our coordinate system so that the positive $z$-axis runs parallel to the vector $b$, then, considering spherical coordinates, the angle between $r$ and $b$ is given by $\theta$, and we have:
\begin{align}
I = \int_0^\pi d\theta\sin\theta\int_0^{2\pi}d\phi\frac{1}{\left(1+|a|\cos\theta\right)\left(1+c\cos\theta + x\cdot r\right)}
\end{align}
Now, I know that, since $x\cdot b = 0$, $x$ must lie in the $x-y-$plane, so we can express it as $(\rho\cos\zeta, \rho\sin\zeta, 0)$ for some modulus $\rho$ and some polar angle $\zeta$. Thus:
$$x\cdot r = \rho\sin\theta\left(\cos\zeta\cos\phi+\sin\zeta\sin\phi\right) = \rho\sin\theta\cos\left(\phi-\zeta\right)$$
Letting $\phi-\zeta=:\xi$, we have $d\xi = d\phi$ and thus, since all periods of the cosine function are equal, we have:
\begin{align}
I&=\int_0^\pi d\theta\frac{\sin\theta}{1+|a|\cos\theta}\int_{-\zeta}^{2\pi-\zeta}\frac{d\xi}{1+c\cos\theta+\rho\sin\theta\cos\xi}\\
&=\int_{-1}^1 \frac{du}{1+|a|u}\int_{0}^{2\pi}\frac{d\xi}{1+cu+\rho\sqrt{1-u^2}\cos\xi}
\end{align}
And this is where I'm stumped. I've been thinking of maybe converting the second integral into a contour integral over the unit circle, which reduces to evaluating
$$\oint_{|z|=1}\frac{dz}{z^2+\alpha z + 1}$$
where $\alpha = 2\frac{1+cu}{\rho\sqrt{1-u^2}}$, but this then leads to a very complicated dependancy of the residues on $u$, so I was maybe wondering if there was a simpler way. Is this even the right way to attack the integral $I$? Or is $I$ perhaps so complicated that it admits no simpler solution?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

lets $\ds{\mathbf{A} \equiv \pars{\mathbf{a} + \mathbf{b}}/2}$ and
  $\ds{\mathbf{B} \equiv \pars{\mathbf{a} - \mathbf{b}}/2}$. With $\ds{\mrm{\mathbf{R}}\pars{u} \equiv \mathbf{A} +  \mathbf{B}u}$:

\begin{align}
I & \equiv
\iint_{\partial B_1\pars{0}}{\dd\Omega \over
\pars{1 + \mathbf{a}\cdot\mathbf{r}}\pars{1 + \mathbf{b}\cdot\mathbf{r}}}
\\[5mm] & =
\iint_{\partial B_1\pars{0}}
2\int_{-1}^{1}{\dd u \over
\bracks{\pars{1 + \mathbf{a}\cdot\mathbf{r}}\pars{1 + u} +
\pars{1 + \mathbf{b}\cdot\mathbf{r}}\pars{1 - u}}^{\, 2}}
\,\dd\Omega
\\[5mm] & =
{1 \over 2}\int_{-1}^{1}\iint_{\partial B_1\pars{0}}{\dd\Omega \over
\bracks{1 + \mrm{\mathbf{R}}\pars{u}\cdot\mathbf{r}}^{\, 2}}\,\dd u =
{1 \over 2}\int_{-1}^{1}\int_{0}^{2\pi}\int_{0}^{\pi}{\sin\pars{\theta}\,\dd\theta\,\dd\phi \over
\bracks{\rule{0pt}{5mm}%
1 + \verts{\mrm{\mathbf{R}}\pars{u}}\cos\pars{\theta}}^{\, 2}}\,\dd u
\\[5mm] & =
\pi\int_{-1}^{1}\bracks{{1 \over 1 - \verts{\mrm{\mathbf{R}}\pars{u}}} -
{1 \over 1 + \verts{\mrm{\mathbf{R}}\pars{u}}}}\,\dd u
\\[5mm] & =
{\pi \over B}\int_{-B}^{B}{\dd u \over
1 - \root{u^{2} + 2\mathbf{A}\cdot\hat{\mathbf{B}}\, u + A^{2}}} -
{\pi \over B}\int_{-B}^{B}{\dd u \over
1 + \root{u^{2} + 2\mathbf{A}\cdot\hat{\mathbf{B}}\, u + A^{2}}}
\end{align}

where I used the notation
  $\ds{P \equiv \verts{\mathbf{P}}}$.
  Both integrals can be evaluated by means of an
  Euler Substitution.

