Computation of Definite Integral of Rational Function I am dealing with a definite integral of a rational function that seems quite hard to get a nice closed form/explicit expression for. Let $ -1 < z < 1 $, then my aim is to determine an expression for the integral $ I $ in terms of $ z $:
$$ I(z) = \int_{0}^{\infty} \frac{(1+z)t^4 + (1-z)}{(1+z)^2 t^6 + 3(1+z)(5+z)t^4 + 3(1-z)(5-z)t^2 + (1-z)^2} \ dt $$
Any help would be appreciated.

Current Attempts
Thank you @Claude Leibovici for your answer. Following the process of the comment, one can arrive at the following.
If we let $ (1+z)^2Q(t) := (1+z)^2 t^6 + 3(1+z)(5+z)t^4 + 3(1-z)(5-z)t^2 + (1-z)^2 $ for $ t \in (0,\infty) $, then $ Q $ has 6 roots $ \pm \omega_i \in \mathbb{C} $ for $ i = 1,2,3 $, all of which are dependent on $ z \in (-1,1) $. In particular, it can be shown that
$$ \omega_k(z)^2 = \frac{4\sqrt{4z + 5}\cos\left(\frac{1}{3}\left(\arccos\left(-\frac{2z^2 + 14z + 11}{(4z + 5)^{3/2}}\right) - 2\pi(k-1)\right)\right) - (5+z)}{(1+z)} < 0, $$
for $ z \in (-1,1) $.
Then using a partial fractions approach yields
$$ I(z) = \int_{0}^{\infty} \frac{(1+z)t^4 + (1-z)}{(1+z)^2Q(t)} \ dt \\ = \frac{i\pi}{2(1+z)^2}\frac{\omega_1\omega_2\omega_3(\omega_1\omega_2 + \omega_1\omega_3 + \omega_2\omega_3)(1 + z) + (\omega_1 + \omega_2 + \omega_3)(1 - z)}{\omega_1\omega_2\omega_3(\omega_1 + \omega_2)(\omega_1+\omega_3)(\omega_2 + \omega_3)}, $$
where it is known, by Vieta's formula that
$$ \omega_1^2 + \omega_2^2 + \omega_3^2 = -\frac{3(5+z)}{(1+z)} $$
$$ \omega_1^2\omega_2^2 + \omega_1^2\omega_3^2 + \omega_2^2\omega_3^2 = \frac{3(1-z)(5-z)}{(1+z)^2} $$
$$ \omega_1^2\omega_2^2\omega_3^2 = -\frac{(1-z)^2}{(1+z)^2} $$

A clearer question would be: is there a way to simplify the above evaluation of $ I(z) $ into a nicer expression?


Consider the integrand $$\frac{(1+z)t^4 + (1-z)}{(1+z)^2 t^6 + 3(1+z)(5+z)t^4 + 3(1-z)(5-z)t^2 + (1-z)^2} $$ and rewrite it as
$$\frac 1 {z+1} \frac{t^4+a}{(t^2-b)(t^2-c)(t^2-d)}$$ where $a=\frac {1-z}{1+z}$ and $(b,c,d)$ are the roots of the cubic equation in $t^2$. Now, using partial fraction decomposition
$$\frac{t^4+a}{(t^2-b)(t^2-c)(t^2-d)}=$$
$$-\frac{a+b^2}{(b-c) (b-d) \left(t^2-b\right)}+\frac{a+c^2}{(b-c) (c-d) \left(t^2-c\right)}-\frac{a+d^2}{(b-d)
   (c-d) \left(t^2-d\right)}$$ and the antiderivative will not make any problem.
A numerical analysis for $-1 < z < 1$ shows that $(b,c,d)$ are all negative. So, we face three integrals looking like
$$I_k=\int \frac {dt}{t^2+k}=\frac{\tan ^{-1}\left(\frac{t}{\sqrt{k}}\right)}{\sqrt{k}}\implies J_k=\int_0^\infty \frac {dt}{t^2+k}=\frac{\pi }{2 \sqrt{k}}$$
What is left is just the computation of $(b,c,d)$. Since they are all real, I suggest you use the trigonometric method for cubic equations.
