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We note that $x\mapsto \mathrm{\Im}\left(f\left(x\right)\right)$ is the integrand above. We also note that the poles of $f$ are ${z}^{\pm}=-1\pm 2i$. By choosing the right contour, one can show that $$ and thus $${\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}\frac{16\mathrm{sin}\left(2x+\pi \right)}{5{\left({x}^{2}+2x+5\right)}^{2}}dx=\frac{\pi \mathrm{sin}\left(2\right)}{{e}^{4}}$$