Evaluating $\int_{-\infty}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^n \ \mathrm{d}x$ I am trying to find a general form of
$$I_n = \int_{-\infty}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^n \ \mathrm{d}x$$
as well as methods of solving it (preferably elementary). This is easy enough to solve for individual $n$, which leads me to believe that there is likely a $\frac{\pi}{\sqrt{3}}$ term in the general expression, but I am having trouble finding a general solution. I have tried constructing a recurrence via IBP, but I always have an extra $\frac{(1 - x^2)^m}{(1 + x^2 + x^4)^n}$ term with $m \neq n$. I also tried creating a $2D$ recurrence, but that doesn't seem solvable, nor can I figure out how to incorporate the requirement of $m \leq 2n - 1$ (for the integral to be finite).
Additionally, I tried using a semicircular contour in the upper half-plane, but computing the residues at $z = e^{\frac{i\pi}{3}}, e^{\frac{2i\pi}{3}}$ seems very bashy.
I have tried using Mathematica, and that just outputs the input.
Any help would be great.
 A: A systemic procedure is to utilize the reduction formula
$$\int_0^\infty \frac{x^{2m}}{(1+x^2+x^4)^n}dx =I_{m,n}=-2I_{m-1,n}+ \frac{4n-2m-3}{2(n-1)}I_{m-1, n-1}$$
Then, given the initial values
$$I_{0,1}=I_{1,1}=\frac\pi{2\sqrt3}, \>\>\>I_{0,2}=\frac\pi{3\sqrt3}, \>\>\> I_{0,3}=\frac{13\pi}{48\sqrt3}, \>\>\>\cdots$$
we evaluate recursively
\begin{align}
&I_{1,2}=-2 I_{0,2}+\frac32 I_{0,1}=\frac\pi{12\sqrt3}\\
 &I_{2,2}=-2 I_{1,2}+\frac12 I_{1,1}=\frac\pi{12\sqrt3}\\
 &I_{1,3}=-2 I_{0,3}+\frac74 I_{0,2}=\frac\pi{24\sqrt3}\\
 &I_{2,3}=-2 I_{1,3}+\frac54 I_{1,2}=\frac\pi{48\sqrt3}\\
 &I_{3,3}=-2 I_{2,3}+\frac34 I_{2,2}=\frac\pi{48\sqrt3}\\
&\ \cdots
\end{align}
As a result
\begin{align}
&\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )dx=I_{0,1}- I_{1,1}=0\\
&\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^2dx=I_{0,2}- 2I_{1,2}+ I_{2,2} =\frac\pi{4\sqrt3}\\
 &\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^3 dx= I_{0,3}- 3I_{1,3}+ 3I_{2,3} -I_{3,3}=\frac{3\pi}{16\sqrt3}\\
 &\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^4 dx=\cdots=\frac{\pi}{6\sqrt3}\\
 &\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^5 dx=\cdots\\
\end{align}
