We first notice that
\begin{align*}
\int_{0}^{\infty}\frac{\operatorname{gd}(x)}{e^x-1}\,dx
&= \int_{0}^{\infty} \frac{1}{e^x-1} \left( \int_{0}^{x} \frac{dy}{\cosh y} \right) \, dx \\
&= \int_{0}^{\infty} \frac{1}{\cosh y} \left( \int_{y}^{\infty} \frac{dx}{e^x - 1} \right) \,d y \\
&= -2 \int_{0}^{\infty} \frac{\log(1 - e^{-y})}{e^y + e^{-y}} \, dy \\
&=-2\int_{0}^{\frac{\pi}{4}}\log(1-\tan\theta)\,d\theta \tag{$e^{-x}=\tan\theta$}.
\end{align*}
The last integral is our starting point. We introduce two tricks to evaluate this.
Step 1. Notice that $\tan(\frac{\pi}{4}-\theta)=\frac{1-\tan\theta}{1+\tan\theta}$. So by the substitution $\theta \mapsto \frac{\pi}{4}-\theta$, it follows that
$$ \int_{0}^{\frac{\pi}{4}}\log(1+\tan\theta)\,d\theta
= \int_{0}^{\frac{\pi}{4}}\log\left(\frac{2}{1+\tan\theta}\right)\,d\theta $$
and hence both integrals have the common value $\frac{\pi}{8}\log 2$. Applying the same idea to our integral, it then follows that
\begin{align*}
-2\int_{0}^{\frac{\pi}{4}}\log(1-\tan\theta)\,d\theta
&= -2\int_{0}^{\frac{\pi}{4}}\log\left(\frac{2\tan\theta}{1+\tan\theta}\right)\,d\theta \\
&= -2\int_{0}^{\frac{\pi}{4}}\log\tan\theta \, d\theta - \frac{\pi}{4}\log 2.
\end{align*}
Step 2. In order to compute the last integral, we notice that for $\theta\in\mathbb{R}$ with $\cos\theta\neq0$, we have
\begin{align*}
-\log\left|\tan\theta\right|
&= \log\left|\frac{1+e^{2i\theta}}{1-e^{2i\theta}}\right|
= \operatorname{Re} \log\left(\frac{1+e^{2i\theta}}{1-e^{2i\theta}}\right) \\
&= \operatorname{Re}\left( \sum_{n=1}^{\infty} \frac{1+(-1)^n}{n} e^{2in\theta} \right) \\
&= \sum_{k=0}^{\infty} \frac{2}{2k+1}\cos(4k+2)\theta.
\end{align*}
So by term-wise integration, we obtain
\begin{align*}
-2\int_{0}^{\frac{\pi}{4}}\log\tan\theta \, d\theta
&= \sum_{k=0}^{\infty} \frac{4}{2k+1} \int_{0}^{\frac{\pi}{4}} \cos(4k+2)\theta \, d\theta \\
&= 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}
= 2K,
\end{align*}
where $K$ is the Catalan's constant.
As said, at the price of the "monster" given by @Raffaele in comments, you can compute it.
What you could also do (but at the price of another infinite summation, is to use
$$\sqrt{\cos(x)}=t \implies x=\cos ^{-1}\left(t^2\right)\implies dx=-\frac{2 t}{\sqrt{1-t^4}}\,dt$$ to make
$$I=-2\int \frac{t^2}{\sqrt{1-t^4}}\, \cos ^{-1}\left(t^2\right)\,dt$$
Now, using
$$\frac{t^2}{\sqrt{1-t^4}}=\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\, t^{4 n+2}$$ we face the problem of
$$J_n=\int t^{4 n+2}\, \cos ^{-1}\left(t^2\right)\,dt$$ which are
$$J_n=\frac{t^{4 n+3} \left(2 t^2 \,
_2F_1\left(\frac{1}{2},\frac{4n+5}{4};\frac{4n+9}{4};t^4\right)+(4 n+5) \cos
^{-1}\left(t^2\right)\right)}{(4 n+3) (4 n+5)}$$
But, these integrals express also in terms of elliptic integrals of the first kind. Their general form is
$$J_n= t \sqrt{1-t^4}P_{n}(t^4)+\frac {t^{4n+3}}{4n+3} \, \cos ^{-1}\left(t^2\right)+a_n\, F\left(\left.\sin ^{-1}(t)\right|-1\right)$$
Concerning the $a_n$'s, they form the sequence
$$\left\{\frac{2}{9},\frac{10}{147},\frac{30}{847},\frac{26}{1155},\frac{442}{27797},
\frac{1326}{110561},\frac{11050}{1168101},\frac{320450}{41575743},\cdots\right\}$$ which do not seem to be known in $OEIS$.
However, in a private discussion, @Raymond Manzoni did identify the sequence
$$a_n=\frac{2}{3 (4 n+3)}\prod_{k=1}^n \frac{4 k+1}{4 k+3}=\frac{2}{3 (4 n+3)}\frac{\Gamma \left(\frac{7}{4}\right) \Gamma \left(n+\frac{5}{4}\right)}{\Gamma
\left(\frac{5}{4}\right) \Gamma \left(n+\frac{7}{4}\right)}$$
I think that the above would be interesting from a computing point of view.
Edit
If we are concerned by
$$I=\int_0^{\frac \pi 2} x\sqrt{\cos (x)}\,dx=\frac{4}{15} \,
_3F_2\left(1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};1\right)$$
Now, using
$$I=-2\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\, J_n$$ Integrated between $0$ and $1$, the first two terms of the expressions for $J_n$ are $0$ and we are left with
$$J_n=a_n\, K(-1)=\frac{2\, \sqrt{\pi } \,\Gamma \left(n+\frac{5}{4}\right)}{(4 n+3)^2 \,\Gamma \left(n+\frac{3}{4}\right)}$$
$$I=-4 \sqrt{\pi }\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\,\frac{ \Gamma
\left(n+\frac{5}{4}\right)}{(4 n+3)^2 \,\Gamma \left(n+\frac{3}{4}\right)}$$
$$I=-\frac{\sqrt{\pi }\, \Gamma \left(\frac{1}{4}\right)}{9 \,\Gamma
\left(\frac{3}{4}\right)}\,
_3F_2\left(\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{7}{4},\frac{7}{4};1\right
)$$
Update
All of the above is so complex that, may be, approximations could be used.
For example, using the $1,400$ years od approximation
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for}\qquad -\frac \pi 2 \leq x\leq\frac \pi 2 $$ in this range
$$\int x\sqrt{\cos (x)}\,dx \sim \frac{1}{2} \sqrt{\left(\pi ^2-4 x^2\right) \left(x^2+\pi ^2\right)}+\frac{5}{4}
\pi ^2 \sin ^{-1}\left(\frac{2 \sqrt{x^2+\pi ^2}}{\sqrt{5} \pi }\right)$$ which, integrated between $0$ and $\frac \pi 2$ would give
$$\frac{5 \pi ^3}{8}-\frac{\pi ^2}{4} \left(2+5 \sin
^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)\approx 0.785221$$ while the exact value is $\approx 0.784608$.
Best Answer
Since$$x^4-x^3+x^2-x+1=\frac{x^5+1}{x+1},$$the roots of $x^4-x^3+x^2-x+1$ are the fifth roots of $-1$ other than $-1$. That is, they are $e^{\pi i/5}$, $e^{3\pi i/5}$, $e^{7\pi i/5}$, and $e^{9\pi i/5}$; besides, $e^{9\pi i/5}=\overline{e^{\pi i/5}}$ and $e^{7\pi i/5}=\overline{e^{3\pi i/5}}$. So\begin{align}x^4-x^3+x^2-x+1&=\left(x-e^{\pi i/5}\right)\left(x-e^{9\pi i/5}\right)\left(x-e^{3\pi i/5}\right)\left(x-e^{7\pi i/5}\right)\\&=\left(x^2-2\cos\left(\frac\pi5\right)x+1\right)\left(x^2-2\cos\left(\frac{3\pi}5\right)x+1\right)\\&=\left(x^2-\frac{1+\sqrt5}2x+1\right)\left(x^2-\frac{1-\sqrt5}2x+1\right).\end{align}So, you can write$$\frac{x^2-2x+3}{x^4-x^3+x^2-x+1}\quad\text{as}\quad\frac{Ax+B}{x^2-\frac{1+\sqrt5}2x+1}+\frac{Cx+D}{x^2-\frac{1-\sqrt5}2x+1}.$$In order to get the coefficients $A$, $B$, $C$, and $D$, you sum these two fractions on the right and the asserting that this sum is equal to the expression that you want to integrate, you solve the system$$\left\{\begin{array}{l}B+D=3\\A+\frac{\sqrt{5}-1}2B+C-\frac{1+\sqrt5}2D=-2\\\frac{\sqrt{5}-1}2A+B-\frac{\sqrt{5}+1}2C+D=1\\A+C=0.\end{array}\right.$$You will get that$$A=-\frac2{\sqrt5},\ B=\frac{15-\sqrt5}{10},\ C=\frac2{\sqrt5},\text{ and }D=\frac{15+\sqrt5}{10}.$$It follows from this that\begin{multline}\int\frac{x^2-2x+3}{x^4-x^3+x^2-x+1}\,\mathrm dx=\\=-\frac{\log\left(2x^2-\left(1+\sqrt5\right)x+2\right)}{\sqrt5}+\frac15\left(\sqrt5-1\right)\sqrt{2\left(5+\sqrt5\right)}\arctan\left(\frac{4x-\sqrt5-1}{\sqrt{10-2 \sqrt5}}\right)+\\+\frac{\log \left(2x^2-\left(1-\sqrt5\right)x+2\right)}{\sqrt5}+\frac25\sqrt{2\left(5+\sqrt5\right)} \arctan\left(\frac{4x+\sqrt{5}-1}{\sqrt{2\left(5+\sqrt5\right)}}\right).\end{multline}