How to show that $\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx=3\pi$ 
Consider
  $$I=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx
\qquad J=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx$$
  I want to show that $I=3\pi$ and that $I=J$.

First, we noticed that $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$
So it gives us an idea to try and factorise $x^4-x^2+1$ but cannot find any factors.
Integrate $I$ (We try some substitutions to see where it will get us to)
$x=\sqrt{u}$ then $dx={2\over \sqrt{u}}du$
$$I=16\cdot{1\over 2}\int_{0}^{\infty}{u^{3/2}\over (1-u+u^2)^4}\mathrm du$$
$u=\tan(y)$ then $du=\sec^2(y)dy$
$$I=8\int_{0}^{\pi/2}{\tan^{3/2}(y)\over (1-\tan(y)+\tan^2(y))^4}{\mathrm dy\over \cos^2(y)}$$
then simplified down to
$$I=128\int_{0}^{\pi/2}{\cos^6(y)\tan^{3/2}(y)\over (2-\sin(2y))^4}\mathrm dy$$
we further simplified down to
$$I={128\over 2^{3/2}}\int_{0}^{\pi/2}{\cos^3(y)\sin^{3/2}(2y)\over (2-\sin(2y))^4}\mathrm dy$$
Not so sure what is the next step.
 A: Consider the transformation $x=1/y$. Then
$$
J=16\int_0^{\infty} dy\frac{1}{y^2} \frac{1/y^4}{(1-1/y^2+1/y^4)^3}=8\int_{-\infty}^{\infty} dy\frac{1}{(y^2-1+1/y^2)^3}=\\8\int_{-\infty}^{\infty} dy\frac{1}{((y-1/y)^2+1)^3}\underbrace{=}_{(\star)}8\int_{-\infty}^{\infty}dz\frac{1}{(z^2+1)^3}=3\pi
$$
and your proof is complete (for the next to last equality $(\star)$ we applied Glasser's Master theorem which is not difficult to proof for this special case)
Furthermore applying the same substitution $x=1/y$ again, it is a matter of straightforward algebra that
$$
\Delta=J-I=-\Delta
$$
so 

$$
\Delta=0 \,\,\text{or}\,\,I=J=3\pi
$$

QED
A: $\begin{align}I&=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx
\qquad\\
&=\int_{0}^{\infty} \dfrac{16x^4(1+x^2)^4}{(1+x^6)^4} dx\\
&=\int_{0}^{\infty} \dfrac{16\cdot {{x}^{12}}+64\cdot {{x}^{10}}+96\cdot {{x}^{8}}+64\cdot {{x}^{6}}+16\cdot {{x}^{4}}}{(1+x^6)^4} dx\\
\end{align}$
Perform the change of variable $y=x^6$,
$\begin{align}
I&=\int_{0}^{\infty} \dfrac{\tfrac{8}{3}x^{\tfrac{7}{6}}+\tfrac{32}{3}x^{\tfrac{5}{6}}+16x^{\tfrac{1}{2}}+\tfrac{32}{3}x^{\tfrac{1}{6}}+\tfrac{8}{3}x^{-\tfrac{1}{6}}}{(1+x)^4} dx\\
&=\dfrac{8}{3}\text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)+\dfrac{32}{3}\text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)+16\text{B}\left(\dfrac{3}{2},\dfrac{5}{2}\right)+\dfrac{32}{3}\text{B}\left(\dfrac{7}{6},\dfrac{17}{6}\right)+\dfrac{8}{3}\text{B}\left(\dfrac{5}{6},\dfrac{19}{6}\right)\\
&=\dfrac{8}{3}\times \dfrac{35\pi}{648}+\dfrac{32}{3}\times \dfrac{35\pi}{648}+16\times \dfrac{\pi}{16}+ \dfrac{32}{3}\times\dfrac{55\pi}{648}+\dfrac{8}{3}\times \dfrac{91\pi}{648}\\
&=\boxed{3\pi}
\end{align}$
Addendum:
$B$ is the Euler beta function.
$\begin{align}
\text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)&=\dfrac{\Gamma\left(\dfrac{13}{6}\right)\Gamma\left(\dfrac{11}{6}\right)}{\Gamma\left(4\right)}\\
&=\dfrac{1}{6} \left(\dfrac{7}{6}\Gamma\left(\dfrac{7}{6}\right)\right)\times \left(\dfrac{5}{6}\Gamma\left(\dfrac{5}{6}\right)\right)\\
&=\dfrac{1}{6} \left(\dfrac{7}{6}\times\dfrac{1}{6}\Gamma\left(\dfrac{1}{6}\right)\right)\times \left(\dfrac{5}{6}\Gamma\left(\dfrac{5}{6}\right)\right)\\
&=\dfrac{35}{1296}\Gamma\left(\dfrac{1}{6}\right)\Gamma\left(\dfrac{5}{6}\right)\\
&=\dfrac{35}{1296}\times \dfrac{\pi}{\sin\left(\pi\times \tfrac{1}{6}\right)}\\
&=\boxed{\dfrac{35\pi}{648}}
\end{align}$
A: Note that the integrands are even, so $$I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}dx
 $$ and $$J=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}dx.
 $$ Now let us consider the semicircular contour of radius $R$ centred in the origin on the upper plane of the complex plane. It is not difficult to note that the integral over the semicircumference vanish as $R\rightarrow\infty
 $ so $$\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}dx=2\pi i\left(\underset{x=\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}\right)+\underset{x=-\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}\right)\right)
 $$ $$=2\pi i\left(\frac{247+77i\sqrt{3}}{54\left(-\sqrt{3}+3i\right)}+\frac{247-77i\sqrt{3}}{54\left(\sqrt{3}+3i\right)}\right)=6\pi
 $$ hence $$I=\color{red}{3\pi}.
 $$ In a similar manner $$\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}dx=2\pi i\left(\underset{x=\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}\right)+\underset{x=-\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}\right)\right)
 $$ $$=2\pi i\left(\frac{13+i5\sqrt{3}}{3\left(-\sqrt{3}+3i\right)}+\frac{13-i5\sqrt{3}}{3\left(\sqrt{3}+3i\right)}\right)=6\pi
 $$ hence $$\color{blue}{I=J}.$$
A: The second integral  $$J=I-\int_{0}^{\infty}{(2x)^4(x^2-x^4)\over (1-x^2+x^4)^4}\, dx$$
The integral on the  right is zero by $x \to 1/x$.
