A definite integral $\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx$ I need to find a value of this definite integral:
$$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx.$$
Its numeric value is approximately $0.7875720991394284$, and lookups in Inverse Symbolic Calculator Plus and WolframAlpha did not return a plausible closed-form candidate. 
Do you have any ideas how I can approach this problem?
 A: There is an alternate way to compute this integral w/o a summation over $n$ in the middle steps.
Notice
$$\frac{2-\cos z}{5 - 4\cos z} = \frac12 \left[\frac{(2-e^{iz})+(2-e^{-iz})}{(2-e^{iz})(2-e^{-iz})}\right]
= \frac12\left[\frac{1}{2-e^{iz}} + \frac{1}{2-e^{-iz}}\right]
$$
and $\displaystyle\;\frac{1}{1+z^4}$ is an even function, we have
$$\mathcal{I} \stackrel{def}{=} \int_0^\infty \frac{2-\cos x}{(1+x^4)(5-4\cos x)} dx =
\frac12\int_{-\infty}^\infty \frac{1}{(1+z^4)(2 - e^{iz})}dz$$
Since $\displaystyle\;\frac{1}{2-e^{iz}}$ is entire on the upper half plane $\Im z \ge 0$ and $\displaystyle\;\left|\frac{1}{2-e^{iz}}\right| \le \frac{1}{2-1} = 1$ there, we can complete the contour in the upper half-plane and
$$\mathcal{I} = \lim_{R\to\infty}\frac12 \oint_{C_R} \frac{1}{(1+z^4)(2 - e^{iz})}dz
\quad\text{ where }\quad
C_R = [-R, R ] \cup \big\{\; Re^{i\theta} : \theta \in [0,\pi]\big\}
$$
$\displaystyle\;\frac{1}{1+z^4}$ has $4$ poles $\omega_k = e^{\frac{(2k+1)i}{4\pi}}, k = 0..3$ over $\mathbb{C}$. Two of them $\omega_0 = \frac{1+i}{\sqrt{2}}$ and $\omega_1 = \frac{-1+i}{\sqrt{2}}$ belongs to the upper half-plane. Since
$$\frac{1}{1+z^4} = \sum_{k=0}^3 \frac{1}{(z-\omega_k) 4\omega_k^3}
= -\frac14 \sum_{k=0}^3\frac{\omega_k}{z-\omega_k}$$
The residue of the integrand at $\omega_k$ are $\displaystyle\;-\frac14 \frac{\omega_k}{2 - e^{i\omega_k}}$ for $k = 0, 1$. This leads to
$$\begin{align}
\mathcal{I} 
&= \frac12\left[ -\frac{2\pi i}{4}\left(\frac{\omega_0}{2 - e^{i\omega_0}} + \frac{\omega_1}{2 - e^{i\omega_1}} \right)\right]\\
&= \frac{-\pi i}{4\sqrt{2}}\left(
\frac{1+i}{2 - e^{-1/\sqrt{2}} e^{i/\sqrt{2}}} +
\frac{-1+i}{2 - e^{-1/\sqrt{2}} e^{-i/\sqrt{2}}}\right)\\
&= \frac{-\pi i}{4\sqrt{2}}
\left[
\frac{
  (1+i)(2 - e^{-1/\sqrt{2}} e^{-i/\sqrt{2}})
+ (-1+i)(2 -e^{-1/\sqrt{2}} e^{ i/\sqrt{2}})
}{
4 - 4 e^{-1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + e^{-\sqrt{2}}
}
\right]\\
&=
\frac{\pi}{2\sqrt{2}}
\left[
\frac{2 - e^{-1/\sqrt{2}}\left(
\cos\left(\frac{1}{\sqrt{2}}\right)
- \sin\left(\frac{1}{\sqrt{2}}\right)
\right)
}{
4 - 4 e^{-1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + e^{-\sqrt{2}}
}
\right]\\
&=
\frac{\pi e^{1/\sqrt{2}} }{2\sqrt{2}} 
\left[
\frac{2 e^{1/\sqrt{2}} - \left(
\cos\left(\frac{1}{\sqrt{2}}\right)
- \sin\left(\frac{1}{\sqrt{2}}\right)
\right)
}{
4 e^{\sqrt{2}} - 4 e^{1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + 1
}
\right]
\end{align}
$$
Reproducing what Vladimir get in his answer.
A: Yes, there is an elementary closed form for this integral:
$$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx=\frac{\pi}{2\,\sqrt2}\cdot\exp\left(\frac1{\sqrt2}\right)\cdot\frac{\sin\left(\frac1{\sqrt2}\right)-\cos\left(\frac1{\sqrt2}\right)+2\,\exp\left(\frac1{\sqrt2}\right)}{1-4\,\exp\left(\frac1{\sqrt2}\right)\cos\left(\frac1{\sqrt2}\right)+4\,\exp\left(\sqrt2\right)}\tag1$$

Proof:
Let us denote the integral in question as
$$\mathcal{I}=\int_0^\infty\frac{2-\cos x}{\left(x^4+1\right)\,\left(5-4\cos x\right)}dx\tag2$$
Note that the trigonometric part of the integrand is a periodic function and can be expanded to a Fourier series with particularly simple coefficients:
$$\frac{2-\cos x}{5-4\cos x}=\sum_{n=0}^\infty\frac{\cos(n\,x)}{2^{n+1}}\tag3$$
(this can be easily checked by expressing cosines via exponents of an imaginary argument).
Now we can integrate it term-wise:
$$\mathcal{I}=\sum_{n=0}^\infty\left(\frac1{2^{n+1}}\int_0^\infty\frac{\cos(n\,x)}{x^4+1}dx\right)=\sum_{n=0}^\infty\left(\frac1{2^{n+1}}\cdot\frac{\pi}{2\,\sqrt2}\cdot\exp\left(-\frac{n}{\sqrt2}\right)\cdot\left(\sin\left(\frac{n}{\sqrt2}\right)+\cos\left(\frac{n}{\sqrt2}\right)\right)\right)\tag4$$
(for the integral, see DLMF 1.14, vii, Table 1.14.2, $4^{th}$ row).
Trig functions in the last sum can again be expressed via exponents of an imaginary argument, and then the sum is easily evaluated. Converting exponents back to trig functions and getting rid of complex numbers, we get the final result $(1)$.
