Formula for $\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ b} \ln \left(a^2+b^2+a b \sqrt{2}\right) $ In my post, I had proved that $$
\int_0^{\infty} \frac{\ln \left(x^2+a^2\right)}{b^2+x^2} d x=\frac{ \pi}{b} \ln (a+b) \tag*{(*)} 
$$
To go further, I guess that
$$\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ |b|} \ln \left(a^2+b^2+|a|| b| \sqrt{2}\right) $$

Proof:
For $a,b>0$,
Using $\ln \left(a^2+b^2\right)=2 Re(\ln (a+b i))$, we can reduce the power $4$ to $2$.
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x
& = 2\int_{0}^{\infty} \frac {Re\left[\ln \left(x^2+a ^2i\right)\right]}{b^2+x^2} d x \\
& =2  Re\left(\int_0^{\infty} \frac{\ln \left(x^2+\left[\left(\frac{1+i}{\sqrt{2}}\right) a\right]^2\right)}{b^2+x^2} d x\right)
\end{aligned}
$$
Using (*), we have $$
\begin{aligned}\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x&=2 Re\left[\frac{\pi}{b} \ln \left(\frac{1+i}{\sqrt{2}} a+b\right)\right] \\&=\frac{2 \pi}{b} R e\left[\ln \left(\frac{a}{\sqrt{2}}+b+\frac{a}{\sqrt{2}}i\right)\right] \\&= \boxed{\frac{\pi}{b} \ln \left(a^2+b^2+a b \sqrt{2}\right)}\end{aligned}
$$
In general, for any $a, b \in \mathbb{R} \backslash\{0\}$, replacing $a$ and $b$ by $|a|$ and $|b|$ yields
$$\boxed{\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ |b|} \ln \left(a^2+b^2+|a|| b| \sqrt{2}\right) }$$

For example, $$
\int_0^{\infty} \frac{\ln \left(x^4+16\right)}{9+x^2} d x= \frac{\pi}{3} \ln (13+6 \sqrt{2})
$$

Comments and alternative methods are highly appreciated.

An option using a semicircular contour in the upper half-plane: due to its even integrand, if $a,\,b>0$ your integral is$$\begin{align}\Re\int_{\Bbb R}\frac{\ln(x^2-a^2+aix\sqrt{2})}{b^2+x^2}dx=\Re\left[2\pi i\lim_{z\to bi}\frac{\ln(z^2-a^2+aiz\sqrt{2})}{z+bi}\right]=\frac{\pi}{b}\ln(a^2+b^2+ab\sqrt{2}).\end{align}$$The trick is to multiply the linear factors of $x^4+a^4$ whose roots have negative imaginary parts, so the integrand is meromorphic inside the contour. In the above calculation, the pole makes the logarithm's argument negative, so it's an especially simple $\Re\ln z=\ln|z|$ calculation.
