$\int_{-\infty}^{\infty} {\rm exp}\left(\frac{i b t - a}{t^2 + 1}\right) - {\rm exp}\left(\frac{i b}{t} - \frac{1}{t^2}\right) {\rm d}t$
Because of the symmetry, your integral becomes
$$\begin{aligned}
&2\int_{0}^{\infty}
\left ( e^{\displaystyle{-\frac{\alpha}{1+t^2} } }
\cos\left ( \frac{\beta t}{1+t^2} \right )
-e^{\displaystyle{-\frac{1}{t^2} } }
\cos\left ( \frac{\beta}{t} \right )
\right ) \text{d}t\\
=&2\int_{0}^{\infty}
\frac{1}{t^2} \left ( e^{-\alpha} e^{\displaystyle{\frac{\alpha}{1+t^2} }}
\cos\left ( \frac{\beta t}{1+t^2} \right )-e^{-t^2}\cos(\beta t)\right ) \text{d}t.
\end{aligned}$$
The second integral is trivial,
$$
\int_{0}^{\infty}t^{\rho-1}e^{-t^2}\cos(\beta t)\text{d}t
=\frac{1}{2}\Gamma\left ( \frac{\rho}{2} \right )
{}_1F_1\left ( \frac{\rho}{2};\frac12;-\frac{\beta^2}{4} \right ).
$$
To evaluate the first one, let me introduce two identity,
$$
\int_{0}^{\infty} t^{\mu-1}e^{\displaystyle{\frac{\alpha}{1+t^2} }}\text{d}t
=\frac{\pi}{\displaystyle{2\sin\left(\frac{\pi \mu}{2} \right)}}
\cdot\alpha\cdot{}_1F_1\left (1-\frac{\mu}{2};2;\alpha \right ).
$$
$$\int_{0}^{\infty}x^{\mu-1}\cos\left ( \frac{2\beta x}{1+x^2} \right ) \text{d}x
=-\beta^2\cdot\frac{\pi}{\displaystyle{2\sin \left ( \frac{\pi\mu}{2} \right ) }}
\cdot\mu\cdot{}_2F_3\left ( 1+\frac{\mu}{2},1+\frac{\mu}{2};\frac{3}{2},\frac{3}{2},2;-\frac{\beta^2}{4} \right ).$$
(The proof is omitted.)
Therefore, we might write
$$
\int_{0}^{\infty} t^{\rho-1}e^{\displaystyle{\frac{\alpha}{1+t^2} }}
\cos\left ( \frac{2\beta x}{1+x^2} \right )\text{d}t
=-\frac{\pi^2\alpha\beta^2}{4}\cdot\frac{1}{2\pi i}
\int_{c-i\infty}^{c+i\infty} \frac{{}_2F_3\left ( 1+\frac{\mu}{2},1+\frac{\mu}{2};\frac{3}{2},\frac{3}{2},2;-
\frac{\beta^2}{4}\right)}{\displaystyle{\sin \left ( \frac{\pi\mu}{2} \right ) }}
\frac{{}_1F_1\left (1-\frac{\rho-\mu}{2};2;\alpha \right )}{\displaystyle{\sin\left(\frac{\pi (\rho-\mu)}{2} \right)}}
\mu\text{d}\mu
$$
By using Mellin convolution theorem. Finally, combine two integrals and take the limit as $\rho$ goes to $-1$, the final expression will be found.