Proving $\int_{0}^{\infty}\frac{4x^2}{(x^4+2x^2+2)^2}dx\stackrel?=\frac{\pi}{4}\sqrt{5\sqrt2-7}$

Hint. One may write $$ \begin{align} \int_{0}^{\infty}\frac{4x^2}{(x^4+2x^2+2)^2}dx&=4\int_0^{\infty}\frac1{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2}\:\frac{dx}{x^2} \\\\&=2\sqrt{2}\int_0^{\infty}\frac1{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2}\:dx \quad (x \to \sqrt{2}/x) \\\\&=\frac1{\sqrt{2}}\int_{-\infty}^{\infty}\frac{\left(1+\frac{\sqrt{2}}{x^2}\right)}{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2}\:dx \\\\&=\frac1{\sqrt{2}}\int_{-\infty}^{\infty}\frac{d\left(x-\frac{\sqrt{2}}x\right)}{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2} \\\\&=\frac1{\sqrt{2}}\int_{-\infty}^{\infty}\frac{du}{(u^2+2+2\sqrt{2})^2} \\\\&=\color{red}{\frac14 \sqrt{5\sqrt{2}-7} \:\pi} \\\\ \end{align} $$

where we have made $u:=\sqrt{2+2\sqrt{2}}\:\sinh v$ to get the last step.

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.