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.
 A: To make it more general, consider the  case of
$$I_n=\int_0^{\infty} \frac{\log \left(x^n+a^n\right)}{x^2+b^2} \,d x$$ where $a$ and $b$ are positive. Let $x=a t$ to get
$$I_n=\frac{\pi   \log (a)}{2 b}n+\frac 1a\int_0^{\infty} \frac{\log \left(t^n+1\right)}{t^2+c^2} \,d x  \qquad\text{with} \qquad c=\frac ba$$ Using the roots of unity, this is the summation of integrals looking like
$$J=\int_0^{\infty} \frac{\log \left(t+r\right)}{t^2+c^2} \,d x $$The antiderivative exists (polylogarithms appear, for sure) and, in terms of the Lerch transcendent function
$$4c^2\,J=r \,\Phi \left(-\frac{r^2}{c^2},2,\frac{1}{2}\right)+c \left(\pi  \log \left(c^2+r^2\right)+4 \log
   \left(\frac{c}{r}\right) \tan
   ^{-1}\left(\frac{r}{c}\right)\right) $$ Summing over the roots, it is "just" a matter of simplifications.
A: Note that $x^4+a^4= (x^2+\sqrt2 ax+a^2) (x^2-\sqrt2 ax+a^2)$
\begin{align}
&\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{x^2+b^2} d x \\
 =&\int_{-\infty}^{\infty} \frac{\ln (x^2+\sqrt2 ax+a^2)}{x^2+b^2} \ d x\\
=& \int_{-\infty}^{\infty}\bigg( \int_0^{\pi/4}\frac{2ax\cos t}{x^2+2ax \sin t+a^2} dt+ {\ln (x^2+a^2)}\bigg) \frac{dx}{x^2+b^2}\\
 =& \int_0^{\pi/4}\frac{-2\pi a\sin t}{a^2+2ab\cos t+b^2} dt\ + \frac{2\pi}b \ln(a+b)\\
= &\ \overset{}{\frac{\pi}{ b} }\ln \left(a^2+b^2+\sqrt{2} a b\right) 
\end{align}
A: 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.
