Formula for $\int_0^\infty \frac{\log(1+x^2)}{\sqrt{(a^2+x^2)(b^2+x^2)}}dx$ Is it possible to express the following integral in terms of known special functions?
$$I(a,b)=\int_0^\infty \frac{\log(1+x^2)}{\sqrt{(a^2+x^2)(b^2+x^2)}}dx$$


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*I have managed to solve the special case when $b=1$ and $a>1$. $$I(a,1)=\frac{\pi}{2a}K\left(\frac{1}{a} \right)+\frac{\log(a^2-1)}{2a}K\left( \sqrt{1-\frac{1}{a^2}}\right) $$
where $K(k)$ is the complete elliptic integral of the first kind. (Please refer to this answer for it's derivation) Unfortunately, I was not able to solve any other case.

*Is it possible to express $I(a,b)$ in terms of complete elliptic integrals?


Update
Using the substitution $x\mapsto1/x$, it can be shown that
$$I(a,b)=\frac{1}{ab}I\left(\frac{1}{a},\frac{1}{b}\right)+\frac{\log(ab)}{b}K\left( \frac{\sqrt{b^2-a^2}}{b}\right) \quad b>a$$

Assume $a > b$, define $m$ and $s$ such that $\frac{b^2}{a^2} = 1 - m = \frac{1-3s}{2}$. Substitute $x^2$ by $\frac{b^2}{p-\frac{1-s}{2}}$, we can rewrite $I(a,b)$ as
$$I(a,b) = -\frac{1}{a}\int_\infty^{\frac{1-s}{2}}
\frac{\log\left[\frac{p-\left(\frac{1-s}{2}-b^2\right)}{p-\left(\frac{1-s}{2}\right)}\right]}{
\sqrt{4p^3 - g_2 p - g_3}} dz
\quad\text{ where }\quad
\begin{cases}g_2 = 1+3s^2\\g_3 = s(s^2-1)\end{cases}$$
Let $\wp(z), \zeta(z), \sigma(z)$ be the Weierstrass elliptic, zeta and sigma functions associated with the ODE:
$$\wp'(z)^2 = 4\wp(z)^3 - g_2 \wp(z) - g_3 =
4 \left(\wp(z) - \frac{1-s}{2}\right)\left(\wp(z) + \frac{1+s}{2}\right)\left(\wp(z) - s\right)$$
The double poles of $\wp(z)$ lies on a rectangular lattice with half period
$$\omega = K(\sqrt{m}),\quad \omega' = iK'(\sqrt{m}) = iK(\sqrt{1-m})$$
where $K(k)$ is the complete elliptic integral of the first kind. Furthermore, $\wp(\omega) = \frac{1-s}{2}$ and $\wp(\omega') = -\frac{1+s}{2}$.
Choose $\rho$ such that $\wp(\omega \pm \rho ) = \frac{1-s}{2} - b^2$. 
When $b$ increases from $0$ to $\infty$, $\wp(\omega\pm\rho)$ decreases from $\wp(\omega) = \frac{1-s}{2}$ to $-\infty$. For concreteness, we will choose $\rho$ such that as $b$
varies from $0$ to $\infty$. $\omega + \rho$ will move along the polygonal path joining
$\omega, \omega+\omega', \omega'$ and $0$. i.e. for small $b$, $\rho$ is purely imaginary.
In terms of all these, we can simplify $I(a,b)$ as
$$\begin{align}
I(a,b) 
&= \frac{1}{a}\int_0^\omega\log\left[
\frac{\wp(z)-\wp(\omega+\rho)}{\wp(z)-\wp(\omega)}
\right]dz
= \frac{1}{2a}\int_{-\omega}^\omega\log\left[
\frac{\wp(z)-\wp(\omega+\rho)}{\wp(z)-\wp(\omega)}\right]dz\\
&= \frac{1}{2a}\int_{-\omega}^\omega\log\left[C
\frac{\sigma(z+\omega+\rho)\sigma(z+\omega-\rho)}{\sigma(z+\omega)^2}
\right]
\end{align}
$$
where $\displaystyle\;C = \frac{\sigma(\omega)^2}{\sigma(\omega+\rho)\sigma(\omega-\rho)}$.
Using the same tricks as this answer, we can evaluate the $z$ dependent part and get
$$\begin{align}
I(a,b) 
&= \frac{1}{2a}\left(
2\omega\log C
+ \varphi_{+}(\omega+\rho) + \varphi_{-}(\omega-\rho)-\varphi_{+}(\omega)-\varphi_{-}(\omega)\right)\\
&= \frac{1}{2a}\left(
2\omega\log C
+ \zeta(\omega)((\omega+\rho)^2 + (\omega-\rho)^2 - 2\omega^2) -\pi i((\omega + \rho) - (\omega - \rho))\right)\\
&= \frac{1}{a}\left( \omega \log C + \zeta(\omega) \rho^2 - \pi i \rho\right)
\end{align}$$
Using various identities for the elliptic functions:
$$\begin{align}
\wp(z) - \wp(u) &= -\frac{\sigma(z+u)\sigma(z-u)}{\sigma(z)^2\sigma(u)^2}\\
\wp(\omega) + \wp(\rho) + \wp(\omega+\rho) &= \frac14\left(\frac{\wp'(\omega)-\wp'(\omega+\rho)}{\wp(\omega)-\wp(\omega+\rho)}\right)^2
\end{align}
$$
We find
$$C = -\frac{1}{\sigma(\rho)^2(\wp(\omega)-\wp(\rho))}
\quad\text{ and }\quad
\wp(\omega) - \wp(\rho) = \frac{1}{a^2}$$
This finally gives us
$$I(a,b) = \frac{1}{a}\left\{ \omega\log\left[ -\frac{a^2}{\sigma(\rho)^2}\right] + \zeta(\omega) \rho^2 - \pi i\rho \right\}
\quad\text{ with }\quad \rho = \wp^{-1}\left(\frac{a^2+b^2-3}{3a^2}\right)
$$
As a double check, let us study two special cases $b = 1$ and $a = 1$.
Case I: $b = 1$.
When $b = 1$, $\wp(\rho) = s = \wp(\omega'\pm\omega) \implies \rho = \omega'-\omega$. 
Let $\eta = \zeta(\omega)$ and $\eta' = \zeta(\omega')$. The addition formula for sigma function 
$$\wp'(z+\rho) = -\frac{\sigma(2z+2\rho)}{\sigma(z+\rho)^4}
= e^{2(\eta'-\eta)(2z + \omega'-\omega)}\frac{\sigma(2z)}{\sigma(z+\rho)^4}$$
implies
$$\sigma(\rho)^4 = \frac{2}{\wp''(\rho)}e^{2(\eta'-\eta)(\omega'-\omega)}
= -\frac{a^4}{b^2(a^2-b^2)}e^{2(\eta'-\eta)(\omega'-\omega)}
$$
Up to a sign factor in the intermediate imaginary pieces, this allow us to fix $I(a,b)$ to the form
$$\frac{1}{a}\left\{\omega\left(\log(b\sqrt{a^2-b^2}) \stackrel{?}{\pm} \frac{\pi i}{2}\right)
-\omega ( (\eta'-\eta)(\omega'-\omega) )
+ \eta(\omega'-\omega)^2 - \pi i(\omega'-\omega)
\right\}$$
Using the identity $\eta\omega'-\omega\eta' = \frac{\pi i}{2}$, the mess after
the logarithm can be simplified:
$$\begin{align}
& -\omega ( (\eta'-\eta)(\omega'-\omega) ) + \eta(\omega'-\omega)^2 - \pi i(\omega'-\omega)\\
=& (\omega'-\omega)( -\omega(\eta'-\eta) + \eta(\omega'-\omega) - \pi i)\\
=& -\frac{\pi i}{2}(\omega' - \omega)
\end{align}
$$
We know $I(a,b)$ is a real number, the imaginary part $\frac{\pi i}{2}\omega$
in above expression will get cancelled by the imaginary piece associated with the logarithm. 
As a result, we find when $b = 1$,
$$\begin{align}
I(a,1) 
&= \frac{1}{a}\left\{\omega\log(b\sqrt{a^2-b^2}) - \frac{\pi i}{2}\omega'\right\}\\
&= \frac{1}{a}\left\{\log(b\sqrt{a^2-b^2})K\left(\sqrt{1-\frac{b^2}{a^2}}\right) + \frac{\pi}{2}K\left(\frac{b}{a}\right)\right\}
\end{align}$$
reproducing what is known.
Case II: $a = 1$.
When $a = 1$, $\wp(\rho) = -\frac{1+s}{2} = \wp(\pm\omega') \implies \rho = \omega'$. Once again, by the addition formula for sigma function, we have
$$\sigma(\rho)^4 = \frac{2}{\wp''(\rho)}e^{2\eta'\omega'} = \frac{a^2}{a^2-b^2}e^{2\eta'\omega'}$$
This allow us to conclude $I(a,b)$ has the form
$$\frac{1}{a}\left\{\omega\log(a\sqrt{a^2-b^2})
- \omega\eta'\omega' + \eta\omega'^2 - \pi i\omega'\right\}
= \frac{1}{a}\left\{\omega\log(a\sqrt{a^2-b^2}) -\frac{\pi i}{2}\omega'\right\}
$$
and hence $I(1,b)$ can be casted to a form very similar to what we have for $b = 1$.
$$I(1,b) = 
\frac{1}{a}\left\{\log(a\sqrt{a^2-b^2})K\left(\sqrt{1-\frac{b^2}{a^2}}\right) + \frac{\pi}{2}K\left(\frac{b}{a}\right)\right\}
$$
