Ask a about hard integral of $\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$ I want to evaluate the integral:
$$I(a,b)=\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$$
Attempt:$$\frac{\partial ^2I}{\partial a\partial b}=4ab\int_{0}^{\infty}\frac{\log x}{(a^2+x^2)(b^2+x^2)}dx=\frac{4ab}{b^2-a^2}\int_{0}^{\infty}\log x\left(\frac{1}{a^2+x^2}-\frac{1}{b^2+x^2}\right)dx$$
$$=\frac{4ab}{b^2-a^2}\frac{\pi}{2}\left(\frac{\log a}{a}-\frac{\log b}{b}\right)=\frac{2\pi(b\log a-a\log b)}{b^2-a^2}$$
Then $$I(a,b)=2\pi\int_{0}^{b}\int_{0}^{a}\frac{(y\log x-x\log y)}{y^2-x^2}dxdy$$
But this integral very hard to solve,can anyone help me,thank you! 

One way is to replace $\ln x$ with $x^p$, then the integrand becomes a product of two linear Meijer G-functions after the change of variables $t = 1/x^2$. We obtain
$$I(p) = \int_0^\infty x^p
 \ln \left(1 + \frac {a^2} {x^2} \right)
 \ln \left(1 + \frac {b^2} {x^2} \right) dx = \\
\frac 1 2 \int_0^\infty t^{(-3-p)/2}
 G_{2, 2}^{1, 2} \left( a^2 t \middle| {1, 1 \atop 1, 0} \right)
 G_{2, 2}^{1, 2} \left( b^2 t \middle| {1, 1 \atop 1, 0} \right) dt = \\
\frac {a^{1+p}} 2 G_{4, 4}^{3, 3} \left( \frac {b^2} {a^2} \middle|
  {1, 1, \frac {1+p} 2, \frac {3+p} 2 \atop
   1, \frac {1+p} 2, \frac {1+p} 2, 0} \right),$$
which is expressible in terms of the Lerch transcendent. Then
$$\int_0^\infty \ln x
 \ln \left(1 + \frac {a^2} {x^2} \right)
 \ln \left(1 + \frac {b^2} {x^2} \right) dx =
I'(0) = \\
-\pi \left( \frac {a \omega (1-\omega)} 2
  \Phi \!\left( \omega^2, 2, \frac 1 2 \right) + \frac {\pi^2 b} 2 - \\
 (2b(1 - \ln b) - (a-b) \ln(1-\omega)) \ln \omega -
 (a+b) (\ln(a b) - 2) \ln(1+\omega) \right), \\
0 < b < a, \quad\omega = \frac b a.$$
