How might I go about solving this definite integral? I am having trouble solving this integral, I have tried using Mathematica and Sympy but had no luck.
$$ \int_{0}^{\infty} \dfrac{1}{(\omega_0^2-\omega^2)^2 + (\Gamma_0\omega)^2} d\omega $$
If it helps, for the case of interest $\Gamma_0 < \omega_0$ and $\Gamma_0 > 0$ and $\omega_0 > 0$.
 A: Hint:
The quartic polynomial can be factored as the product of two quadratic ones. Then the integrand will decompose in simple quadratic fractions with numerators of the first degree. The numerators can be adjusted to be the derivatives of the denominators, and constant numerators will remain. The inverse of a quadratic is turned, by a linear change of variable, to the derivative of $\arctan$ or $\text{artanh}$, depending on the number of real roots.
Once you have factored the initial polynomial, the rest is quite tractable.

After reduction to simple fractions, you can use
https://www.wolframalpha.com/input/?i=integrate+(px%2Bq)%2F(ax%5E2%2Bbx%2Bc)
(or the definite integral).

The roots of the biquadratic equation
$$(a^2-\omega^2)^2+2b^2\omega^2=\omega^4+2(b^2-a^2)\omega^2+a^4=0$$ are
$$\omega=\pm\sqrt{b^2-a^2\pm\sqrt{(b^2-a^2)^2-a^4}}.$$ 
They can be real or complex. In the case of complex roots, the factorization is
$$(\omega+c+id)(\omega+c-id)(\omega-c+id)(\omega-c-id)=((\omega+c)^2-d^2)((\omega-c)^2-d^2).$$
A: Mathematica 11.1 gives me 
$$
\frac{i \pi  \left(\sqrt{-i \Gamma_0 \sqrt{4 {\omega_0}^2-\Gamma_0^2}+\Gamma_0^2-2 {\omega_0}^2}-\sqrt{i \Gamma_0 \sqrt{4 {\omega_0}^2-\Gamma_0^2}+\Gamma_0^2-2 {\omega_0}^2}\right)}{\sqrt{2} \Gamma_0 \sqrt{i \Gamma_0 \sqrt{4 {\omega_0}^2-\Gamma_0^2}+\Gamma_0^2-2
   {\omega_0}^2} \sqrt{\left(\Gamma_0^2-4 {\omega_0}^2\right) \left(i \Gamma_0 \sqrt{4 {\omega_0}^2-\Gamma_0^2}-\Gamma_0^2+2 {\omega_0}^2\right)}}
$$
but the imaginart part may be spurious comparing to numerical trials if the inputs are real.
