An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$. 
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$

What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because of parity of the integrand (nothing else worked for me): $$2\int_0^\infty \frac{x^4 + 1}{(x^4 - \sqrt3x^2 + 1)(x^4 + \sqrt3x^2 + 1)}dx = $$ $$ = \int_0^\infty \frac1{x^4 - \sqrt3x^2 + 1}dx + \int_0^\infty\frac1{x^4 + \sqrt3x^2 + 1}dx = $$ $$ = \int_0^\infty \frac1{(x^2 - \frac{\sqrt3}2)^2 + \frac14}dx + \int_0^\infty \frac1{(x^2 + \frac{\sqrt3}2)^2 + \frac14}dx.$$
I don't see what would be continuation of this. Any help is appreciated.
Thank you for any help. Appreciate it.

More a comment than an answer:
According to Wolfy,
the indefinite integrals
involve arctan and
complex values.
But the definite integrals
come out like this:
$\int_0^∞ \dfrac{dx}{x^4 - \sqrt{3} x^2 + 1}  
= \frac12 \sqrt{2 + \sqrt{3}} π≈3.0345
$
and
$\int_0^∞ \dfrac{dx}{x^4 + \sqrt{3} x^2 + 1} dx = \dfrac{π}{2 \sqrt{2 + \sqrt{3}}}
≈0.8131
$
