Here is an approach using contour integration in case anyone is
interested. An effort has been made to use pen-and-paper type
manipulations only. These are simple yet demand a certain care with
the algebra. Suppose we seek to verify that
$$\int_0^\infty \frac{4x^2}{(x^4+2x^2+2)^2} dx =
\frac{\pi}{4} \sqrt{5\sqrt{2}-7}$$
or alternatively
$$\int_0^\infty \frac{x^2}{(x^4+2x^2+2)^2} dx =
\frac{\pi}{16} \sqrt{5\sqrt{2}-7}.$$
We use a semicircular contour in the upper half plane with two
straight components $\Gamma_0$ and $\Gamma_1$ on the positive and
negative real axis and having radius $R$ ($\Gamma_2.$)
The denominator here is
$$((x^2+1)^2+1)^2$$
so the poles are double and located at
$$\rho_{0,1,2,3} = \pm\sqrt{-1\pm i}.$$
We convert this to polar form in order to determine which poles are in
the upper half plane, getting
$$\pm\sqrt{\sqrt{2} \exp(\pi i \pm \pi i/4)}
= \sqrt[4]{2} \exp(\pi i/2 \pm \pi i/8 + \pi i/2 \pm \pi i/2)
\\ = \sqrt[4]{2} \exp(\pi i \pm \pi i/8 \pm \pi i/2).$$
Fortunately we can see by inspection that only the
two poles
$$\rho_0 = \sqrt[4]{2} \exp(\pi i - \pi i/8 - \pi i/2)
= \sqrt[4]{2} \exp(3\pi i/8)
\\ \quad\text{and}\quad
\rho_1 = \sqrt[4]{2} \exp(\pi i + \pi i/8 - \pi i/2)
= \sqrt[4]{2} \exp(5\pi i/8)$$
are inside the contour (arguments are $3\pi/8$ and $5\pi/8$, the other
two are at $-3\pi /8$ and $-5\pi /8.$)
For the residue we get
$$\frac{1}{2\pi i}
\int_{|z-\rho_0|=\epsilon} \frac{z^2}{(z^4+2z^2+2)^2} \; dz.$$
In order to get a pole that is amenable to easy algebra we introduce
$w = z\exp(-3\pi i/8)/\sqrt[4]{2}$ and
$z = w\exp(3\pi i/8)\sqrt[4]{2}$
which maps $\rho_0$ to $1$ so we obtain
$$\exp(3\pi i/4+3\pi i/8)\sqrt{2}\sqrt[4]{2}
\\ \times \frac{1}{2\pi i}
\int_{|w\exp(3\pi i/8)\sqrt[4]{2}-1|=\epsilon}
\frac{w^2}{(-2iw^4+2w^2(-1+i)+2)^2} \; dw
\\ = - \exp(9\pi i/8) \frac{2^{3/4}}{4} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{w^2}{(w-1)^2\times (w^2-i)^2(w+1)^2} \; dw.$$
The residue is thus given by
$$- \exp(9\pi i/8) \frac{2^{3/4}}{4} \lim_{w\rightarrow 1}
\left(\frac{w^2}{(w^2-i)^2(w+1)^2}\right)'
\\ = - \exp(9\pi i/8) \frac{2^{3/4}}{4} \lim_{w\rightarrow 1}
\left(\frac{2w}{(w^2-i)^2(w+1)^2}
\\ - \frac{w^2}{(w^2-i)^4(w+1)^4}
(2(w^2-i) 2w (w+1)^2 + (w^2-i)^2 2(w+1))\right).$$
This works out to
$$-\exp(9\pi i/8) \frac{2^{3/4}}{4} \times
\frac{1}{8} (2-i)
= i\exp(-3\pi i/8) \frac{2^{3/4}}{4} \times
\frac{1}{8} \sqrt{5} \exp(-i\beta)$$
where $2-i = \sqrt{5}\exp(-i\beta).$
Continuing with the second pole we we introduce
$w = z\exp(-5\pi i/8)/\sqrt[4]{2}$ and
$z = w\exp(5\pi i/8)\sqrt[4]{2}$
and obtain
$$\exp(5\pi i/4+5\pi i/8)\sqrt{2}\sqrt[4]{2}
\\ \times \frac{1}{2\pi i}
\int_{|w\exp(5\pi i/8)\sqrt[4]{2}-1|=\epsilon}
\frac{w^2}{(2iw^4+2w^2(-1-i)+2)^2} \; dw
\\ = - \exp(15\pi i/8) \frac{2^{3/4}}{4} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{w^2}{(w-1)^2\times (w^2+i)^2(w+1)^2} \; dw.$$
This is the same as in the previous pole except the sign in the
$w^2-i$ term has been flipped. Re-using the derivative thus yields
$$-\exp(15\pi i/8) \frac{2^{3/4}}{4} \times
\frac{1}{8} (2+i)
= i\exp(3\pi i/8) \frac{2^{3/4}}{4} \times
\frac{1}{8} \sqrt{5} \exp(i\beta).$$
Adding the two residues we thus obtain
$$\frac{2^{3/4}}{32}\times\sqrt{5}\times 2i\cos(\beta+3\pi /8).$$
Returning to the main computation, on the part of the contour that is
on the negative real axis which is $\Gamma_1$ we trivially obtain
$$\int_{-R}^0 \frac{x^2}{(x^4+2x^2+2)^2} \; dx$$
which yields with $z=-x$
$$- \int_R^0
\frac{z^2}{(z^4+2z^2+2)^2} \; dz
= \int_{\Gamma_0} \frac{z^2}{(z^4+2z^2+2)^2} \; dz.$$
Finally we have by the ML bound for the semicircular component
$$\lim_{R\rightarrow\infty}
\left|\int_{\Gamma_2} \frac{z^2}{(z^4+2z^2+2)^2} \; dz\right|
\le \lim_{R\rightarrow\infty} 2\pi R/2 \times
\frac{R^2}{(R^4-2R^2+2)^2} = 0.$$
It follows that
$$\int_0^\infty \frac{x^2}{(x^4+2x^2+2)^2} \; dx
= \frac{1}{2}\times 2\pi i \times
\frac{2^{3/4}}{32}\times\sqrt{5}\times 2i\cos(\beta+3\pi /8)
\\ = -\frac{\pi}{16} 2^{3/4} \sqrt{5} \cos(\beta+3\pi /8).$$
To manipulate this to match the form in the introduction we use
angle sum and half-angle formulae as in
$$\sqrt{5}\cos(\beta+3\pi /8) =
\sqrt{5}\cos\beta\cos(3\pi /8) - \sqrt{5}\sin\beta\sin(3\pi /8)
\\ = 2\cos(3\pi /8) - \sin(3\pi /8).$$
As we are integrating a function that is never negative on the
integration interval we see that the sign on this last term must be
negative. Observe that
$$\cos(3\pi/8) = \sqrt{\frac{1+\cos(3\pi/4)}{2}}
= \sqrt{\frac{1-\sqrt{2}/2}{2}}$$
and
$$\sin(3\pi/8) = \sqrt{\frac{1-\cos(3\pi/4)}{2}}
= \sqrt{\frac{1+\sqrt{2}/2}{2}}.$$
Squaring we obtain
$$4 \frac{1-\sqrt{2}/2}{2} + \frac{1+\sqrt{2}/2}{2}
- 4 \sqrt{\frac{1-2/4}{4}}
= \frac{5}{2} - \frac{3}{4}\sqrt{2}-\sqrt{2}
\\ = \frac{5}{2} - \frac{7}{4}\sqrt{2}.$$
We thus have for the end result
$$-\frac{\pi}{16} 2^{3/4} \times
- \sqrt{\frac{5}{2} - \frac{7}{4}\sqrt{2}}
= \frac{\pi}{16} \times
\sqrt{\frac{5}{2}2^{3/2} - \frac{7}{4}2^2}
\\ = \frac{\pi}{16} \times \sqrt{5\sqrt{2}-7}.$$
This is the claim.
