How to evaluate $\int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}} d x$ for positive constants $a$ and $c$? First of all, we deal with the simple case.
$$
\begin{aligned}
 \int_{0}^{\infty} \frac{\ln \left(ax^{2}+1\right)}{1+x^{2}} d x \stackrel{x=\tan \theta}{=}  & \int_{0}^{\frac{\pi}{2}} \ln \left(1+a \tan ^{2} \theta\right) d \theta \\
=& \int_{0}^{\frac{\pi}{2}} \ln \left(\frac{\cos ^{2} \theta+a \sin ^{2} \theta}{\cos ^{2} \theta}\right) d \theta \\
=& \int_{0}^{\frac{\pi}{2}} \ln \left(\cos ^{2} \theta+a \sin ^{2} \theta\right) d \theta-2 \int_{0}^{\frac{\pi}{2}} \ln (\cos \theta) d \theta \\
\end{aligned}
$$
Using the result in my post and $2 \int_{0}^{\frac{\pi}{2}} \ln (\cos \theta) d \theta =-\pi \ln 2$ yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln \left(1+ax^{2}\right)}{1+x^{2}} d x &=\pi \ln \left(\frac{1+\sqrt{a}}{2}\right)+\pi \ln 2 =\pi \ln (1+\sqrt{a})
\end{aligned} \tag*{(*)}
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}}d x \\
=& \int_{0}^{\infty} \frac{\left.\ln \left[c(\frac{a}{c} x^{2}+1\right)\right]}{1+x^{2}} d x \\
=& \int_{0}^{\infty} \frac{\ln c}{1+x^{2}} d x+\int_{0}^{\infty} \frac{\ln \left(\frac{a}{c} x^{2}+1\right)}{1+x^{2}}
\end{aligned}
Replacing $a$ by $\frac{a}{c}$ in $(*)$ yields
$$ \boxed{\int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}}d x =\pi \ln (\sqrt{a}+\sqrt{c})}$$
My question: Can we go further to evaluate a more general integral
$$\int_{0}^{\infty} \frac{\ln \left(ax^{2}+bx+1\right)}{1+x^{2}} d x?$$

$$ \frac{\log \left(ax^{2}+bx+1\right)}{1+x^{2}}= \frac {\log(a)}{1+x^{2}}+\frac i 2\Bigg[\frac 1 {x+i} -\frac 1 {x-i} \Bigg]\Bigg[\log(x+\alpha)+\log(x+\beta)\Bigg]$$ where $(-\alpha,-\beta)$ are the complex roots of the quadratic.
$$\int \frac {\log(x+\gamma)}{x+i}\,dx=\text{Li}_2\left(\frac{x+\gamma }{\gamma -i}\right)+\log \left(-\frac{x+i}{\gamma -i}\right) \log (\gamma +x)$$
$$\int \frac {\log(x+\gamma)}{x-i}\,dx=\text{Li}_2\left(\frac{x+\gamma }{\gamma +i}\right)+\log \left(-\frac{x-i}{\gamma
   +i}\right) \log (\gamma +x)$$
