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
$$\int_{0}^{\infty} \frac{\ln \left(ax^{2}+bx+1\right)}{1+x^{2}} d x?$$

Alternatively, we can tackle the problem using contour integration on the upper semi-circle.
$$I=\int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}} d x= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}} d x =\operatorname{Re} \left(\int_{-\infty}^{\infty} \frac{\ln (\sqrt{a} x+i \sqrt{c})}{1+x^{2}} d x \right)$$
Now let’s take the contour integration on the last integral along anti-clockwise direction of the path
$$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where }  \gamma_{1}(t)=t+i 0(-R \leq t \leq R)  \textrm{ and } \gamma_{2}(t)=R e^{i t}  (0<t<\pi) ,$$
$$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{\ln (\sqrt{a} x+i \sqrt{c})}{1+x^{2}} d x  &=\int_{-\infty}^{\infty} \frac{\ln (\sqrt{a} x+i \sqrt{c})}{1+x^{2}} d x \\
&=\oint_{\gamma} \frac{\ln (\sqrt{a} z+i \sqrt{c})}{1+z} d z \\
&=2 \pi i \operatorname{Res}\left(\frac{\ln (\sqrt{a} z+i \sqrt{c})}{1+z^{2}}, z=i\right)\\
&=2 \pi i \cdot \frac{\ln (\sqrt{a} + \sqrt{c})i}{2 i}\\
&=\pi  \ln (\sqrt{a}+\sqrt{c})+\frac{\pi}{2}i 
\end{aligned}
$$
Hence we conclude that
$$
\boxed{I=\pi \ln (\sqrt{a}+\sqrt{c})}
$$
