You can actually see directly that there is no elementary expression for the integral, using contour integration. For your integral is the same as
$$\mathrm{Im}\int_0^\infty {e^{iax} \over x^2 + b^2}\mathrm dx$$
Note that
$${1 \over x^2 + b^2} = {1 \over 2ib}\left({1 \over x - ib} - {1 \over x + ib}\right)$$
So it suffices to evaluate the integrals
$$\int_0^\infty {e^{iax} \over x \pm bi}\,\mathrm dx$$
These integrals converge despite the denominator having only degree one because the exponential factor modulates it, similar to for $\dfrac{\sin(x)}{x}$. You can use your quarter circle idea on each of these two terms (without even having to worry about indentations)... the upper quarter circle for $+bi$ and the bottom quarter circle for $-bi$. For example you get that the first one is equal to
$$\int_0^\infty {e^{-ax} \over x + b}\,\mathrm dx$$
Letting $x = z - b$ this becomes
$$e^{ab}\int_b^\infty {e^{-az} \over z}\,\mathrm dz$$
Then letting $z = \dfrac{w}{a}$ this turns into
$$e^{ab}\int_{ab}^\infty {e^{-w} \over w}\,\mathrm dw$$
$$= e^{ab}\mathrm{Ei}(-ab)$$
Here $\mathrm{Ei}$ is the exponential function which is defined in terms of the above integral. So unless the exponential functions from the two quarter circles somehow cancel out (which I seriously doubt in view of Sasha's answer), you won't be able to get an elementary expression for the integral.
Integrals of the form
$$\int_{-\infty}^\infty \frac{p(x)}{\cosh x}\,dx,$$
where $p$ is a polynomial can be evaluated by shifting the contour of integration to a line $\operatorname{Im} z \equiv c$. We first check that the integrals over the vertical segments connecting the two lines tend to $0$ as the real part tends to $\pm\infty$:
$$\lvert \cosh (x+iy)\rvert^2 = \lvert \cosh x\cos y + i \sinh x\sin y\rvert^2 = \sinh^2 x + \cos^2 y,$$
so the integrand decays exponentially and
$$\left\lvert \int_{R}^{R + ic} \frac{p(z)}{\cosh z}\,dz\right\rvert
\leqslant \frac{K\,c}{\sinh R}\left(R^2+c^2\right)^{\deg p/2} \xrightarrow{R\to \pm\infty} 0.$$
Since $\cosh \left(z+\pi i\right) = -\cosh z$, and the only singularity of the integrand between $\mathbb{R}$ and $\mathbb{R}+\pi i$ is a simple pole at $\frac{\pi i}{2}$ (unless $p$ has a zero there, but then we can regard it as a simple pole with residue $0$) with the residue
$$\operatorname{Res}\left(\frac{p(z)}{\cosh z};\, \frac{\pi i}{2}\right) = \frac{p\left(\frac{\pi i}{2}\right)}{\cosh' \frac{\pi i}{2}} = \frac{p\left(\frac{\pi i}{2}\right)}{\sinh \frac{\pi i}{2}} = \frac{p\left(\frac{\pi i}{2}\right)}{i},$$
the residue theorem yields
$$\begin{align}
\int_{-\infty}^\infty \frac{p(x)}{\cosh x}\,dx
&= 2\pi\, p\left(\frac{\pi i}{2}\right) + \int_{\pi i-\infty}^{\pi i+\infty} \frac{p(z)}{\cosh z}\,dz\\
&= 2\pi\, p\left(\frac{\pi i}{2}\right) - \int_{-\infty}^\infty \frac{p(x+\pi i)}{\cosh x}\,dx\\
&= 2\pi\, p\left(\frac{\pi i}{2}\right) - \sum_{k=0}^{\deg p} \frac{(\pi i)^k}{k!}\int_{-\infty}^\infty \frac{p^{(k)}(x)}{\cosh x}\,dx.\tag{1}
\end{align}$$
Since $\cosh$ is even, only even powers of $x$ contribute to the integrals, hence we can from the beginning assume that $p$ is an even polynomial, and need only consider the derivatives of even order.
For a constant polynomial, $(1)$ yields
$$\int_{-\infty}^\infty \frac{dx}{\cosh x} = 2\pi - \int_{-\infty}^\infty \frac{dx}{\cosh x}\Rightarrow \int_{-\infty}^\infty \frac{dx}{\cosh x} = \pi.$$
For $p(z) = z^2$, we obtain
$$\begin{align}
\int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx &= 2\pi \left(\frac{\pi i}{2}\right)^2 - \int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx - (\pi i)^2\int_{-\infty}^\infty \frac{dx}{\cosh x}\\
&= - \frac{\pi^3}{2} - \int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx + \pi^3,
\end{align}$$
which becomes
$$\int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx = \frac{\pi^3}{4}.$$
Best Answer
We can use some instances of Schroder's Integral, which evaluates Gregory coefficients, namely:
$$(-1)^{n-1}G_n=\int_0^\infty\frac{1}{(\pi^2+\ln^2 t)(1+t)^n}dt;\quad G_1=\frac12, \ G_2=-\frac{1}{12}$$
But let's adjust things first.
$$I=\int_0^\infty \frac{(\pi x - 2\log{x})^3}{\left(\log^2{x} + \frac{\pi^2}{4}\right)(1+x^2)^2} dx \overset{x^2=t}=2\int_0^\infty \frac{(\pi\sqrt t-\ln t)^3}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$=2\pi^3\int_0^\infty \frac{t}{(\pi^2+\ln^2 t)(1+t)^2}dt-6\pi^2 \int_0^\infty \frac{\sqrt t\ln t}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$+6\pi \int_0^\infty \frac{\ln^2 t}{(\pi^2+\ln^2 t)(1+t)^2}dt-2\int_0^\infty \frac{\ln^3 t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$=2\pi^3 I_1-6\pi^2I_2+6\pi I_3-2 I_4$$
$$I_1=\int_0^\infty \frac{t}{(\pi^2+\ln^2 t)(1+t)^2}dt=\int_0^\infty \frac{(1+t)-1}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$=\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)}dt-\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$=G_1+G_2=\frac12-\frac{1}{12}=\frac{5}{12}$$
The $I_2$ integral can be found here.
$$I_2=\int_0^\infty \frac{\sqrt t\ln t}{(\pi^2+\ln^2 t)(1+t)^2}dt\overset{t=\frac{1}{x}}=-\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2}\frac{dx}{\sqrt x}=\frac{\pi}{24}$$
$$I_3=\int_0^\infty \frac{\ln^2 t}{(\pi^2+\ln^2 t)(1+t)^2}dt = \int_0^\infty \frac{(\pi^2 +\ln^2 t)-\pi^2}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$=\int_0^\infty \frac{1}{(1+t)^2}dt-\pi^2\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$=1+\pi^2 G_2=1-\frac{\pi^2}{12}$$
$$I_4=\int_0^\infty \frac{\ln^3 t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}=\int_0^\infty \frac{\ln t((\pi^2+\ln^2 t ) -\pi^2 )}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$=\int_0^\infty \frac{\ln t}{(1+t)^2}\frac{dt}{\sqrt t}-\pi^2 \int_0^\infty \frac{\ln t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$=-\pi+\pi^2 I_2=-\pi +\frac{\pi^3}{24}$$
$$\require{cancel} \Rightarrow I=\cancel{2\pi^3{\frac{5}{12}}} -\cancel{6\pi^2{\frac{\pi}{24}}}+6\pi \left({1-\cancel{\frac{\pi^2}{12}}}\right)-2\left({-\pi +\cancel{\frac{\pi^3}{24}}}\right)={8\pi}$$