Integrate $\log_2(1+b(ax^2+c)^{-2})$ I am wondering how to calculate $\int \log_2(1+b(ax^2+c)^{-2})$, where $a,b,c>0$.
I have used wolframalpha to calculate it, and the answer is like

While it gives me a nice result, I am wondering how to get it. (Wolfram|Alpha does not provide Step-by-Step Solution for this problem)
Answer:
Most results are from the answer of @Claude Leibovici, and part is from the suggestion of @Robert Z. Thanks for their help.
$$
 \int\log_2 \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\,dx = \frac{1}{\log\left(2\right)}\int\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\,dx
$$
We denote $I \triangleq \int\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\,dx$. Use integration by parts to get rid of the logarithm,
$$
 u\triangleq\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\implies du=-\frac{4 a b x}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}
$$
$$I=x\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)+\int\frac{4 a b x^2}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}\,dx$$
Now, partial fraction decomposition
$$\frac{4 a b x^2}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}=\frac{4 \left(a c x^2+b+c^2\right)}{a^2 x^4+2 a c x^2+b+c^2}-\frac{4 c}{a x^2+c}$$
Now, factor the denominator
$$a^2 x^4+2 a c x^2+b+c^2=a^2\left(x^2+\frac{c+ \sqrt{-b}}{a}\right) \left(x^2+\frac{c- \sqrt{-b}}{a}\right)$$
$$\frac{4 \left(a c x^2+b+c^2\right)}{a^2 x^4+2 a c x^2+b+c^2} = \frac{2 c+2 \sqrt{-b}}{a x^2+c+ \sqrt{-b}} + \frac{2c-2 \sqrt{-b}}{a x^2+ c- \sqrt{-b}} = \frac{2}{\frac{a}{c+ \sqrt{-b}} x^2+1} + \frac{2}{\frac{a}{c- \sqrt{-b}} x^2+1}$$
Thus
\begin{equation}
 \begin{aligned}
  I = x\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right) 
  &+ 2  \sqrt{ \frac{ c+\sqrt{-b} }{a} } \tan^{-1} \left(\sqrt{\frac{a}{c+\sqrt{-b}}} x \right)\\
  &+ 2  \sqrt{ \frac{ c-\sqrt{-b} }{a} } \tan^{-1} \left(\sqrt{\frac{a}{c-\sqrt{-b}}} x \right)\\
  &-  4 \sqrt{ \frac{c}{a} } \tan^{-1} \left(\sqrt{\frac{a}{c}} x \right)
 \end{aligned}
\end{equation}
 A: $$I=\int\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\,dx$$ Use integration by parts to get rid of the logarithm
$$u=\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\implies du=-\frac{4 a b x}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}$$
$$I=x\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)+\int\frac{4 a b x^2}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}\,dx$$ Now, partial fraction decomposition
$$\frac{4 a b x^2}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}=\frac{4 \left(a c x^2+b+c^2\right)}{a^2 x^4+2 a c x^2+b+c^2}-\frac{4 c}{a x^2+c}$$
Now, factor the denominator
$$a^2 x^4+2 a c x^2+b+c^2=a^2\left(x^2+\frac{c+i \sqrt{b}}{a}\right) \left(x^2+\frac{c-i \sqrt{b}}{a}\right)$$ Partial fractions again .....
Edit
Considering the more general case of
$$I_d=\int\log \left(1+\frac{b}{\left(a x^2+c\right)^d}\right)\,dx$$ where $d$ is an integer (if not, no hope for any result). As you wrote in comment
$$I_d=x\log \left(1+\frac{b}{\left(a x^2+c\right)^d}\right)+2abd \int\frac{x^2}{\left(b + \left(ax^2+c\right)^d\right)\left(ax^2+c\right)}\,dx$$ The denominator is a polynomial of degree $(d+1)$ in $x^2$ which can be written as
$$\left(b + \left(ax^2+c\right)^d\right)\left(ax^2+c\right)=a^{d+1} \prod_{i=1}^{d+1} (x^2-r_i)$$ where the $r_i$'s are the roots of the polynomial. So, partial fraction decomposition will give
$$2abd \int\frac{x^2}{\left(b + \left(ax^2+c\right)^d\right)\left(ax^2+c\right)}\,dx=\frac {2bd}{a^d}\,\sum_{i=1}^{d+1}\alpha_i \int \frac {dx}{x^2-r_i}$$ which does not make much problem (as long as you found the $r_i$'s and the $\alpha_i$'s (the latest depending on the former).
