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.

Reduce the rational degrees with successive  substitutions as follows
$$\begin{align}
\int_{-\infty}^{\infty}\frac{(x^{4}+1)^{2}}{x^{12}+1}dx=&\ 2\int_{0}^{\infty}\frac{x^{4}+1}{x^{8}-x^{4}+1}\overset{x\to\frac1x}{dx}=\int_{0}^{\infty}\frac{(x^{4}+1)(x^2+1)}{x^{8}-x^{4}+1}
\overset{ x-\frac1x\to x}{ dx} \\
=& \ 2\int_0^\infty \frac{x^2+2}{x^4+4x^2+1} \overset{x\to\frac1x}{dx}= 3\int_{0}^{\infty}\frac{x^2+1}{x^4+4x^2+1}
\overset{x-\frac1x\to x}{ dx} \\
=&\ 6\int_0^\infty \frac1{x^2+6}dx
=\sqrt{\frac32}\pi
\end{align}$$
