Call $I(u,v,w)$ the integral to compute and note that this can be defined only when $4vw>u^2$. Using the definition of $\sinh$ and the parity of the function to be integrated one sees that $ 4I(u,v,w)=\int_{-\infty}^\infty\int_{-\infty}^\infty{\mathrm dp\mathrm dq}\,\frac{e^{upq}}{q^2 - p^2}pq\, e^{-vq^2 - wp^2}, $ that is, $4I(u,v,w)=\partial_uJ(u,v,w)$ with $ J(u,v,w)=\iint{\mathrm dp\mathrm dq}\,\frac{e^{upq}}{q^2 - p^2}\, e^{-vq^2 - wp^2}. $ The function $J(u,\cdot,\cdot)$ is symmetric and $ \partial_wJ(u,v,w)-\partial_vJ(u,v,w)=\iint{\mathrm dp\mathrm dq}\,e^{upq}\,e^{-vq^2-wp^2}. $ The exponent in the exponential is a quadratic form in $(p,q)$ and one knows that $ \iint e^{-\frac12\xi^*C\xi}\,\text{d}\xi=2\pi\det(C)^{-1/2}, $ hence $ \partial_wJ(u,v,w)-\partial_vJ(u,v,w)=\frac{2\pi}{\sqrt{4vw-u^2}}. $ This is enough to recover $J(u,v,w)$, hence $I(u,v,w)$. Since $J(u,\frac12(v+w),\frac12(v+w))=0$ by symmetry, one gets $J(u,v,w)$ as an integral of $\partial_tJ(u,\frac12(v+w)-t,\frac12(v+w)+t)$, that is, $ J(u,v,w)=\int\limits_{0}^{(w-v)/2}\frac{2\pi \text{d}t}{\sqrt{4\left(\frac12(v+w)+t\right)\left(\frac12(v+w)-t\right)-u^2}}, $ which is $ J(u,v,w)=\int\limits_{0}^{w-v}\frac{\pi \text{d}t}{\sqrt{s^2-t^2}},\quad s^2=(v+w)^2-u^2. $ Hence, $ J(u,v,w)=\pi\text{Arcsin}\left(\frac{w-v}{s}\right). $ Differentiating this with respect to $u$ yields finally $ 4I(u,v,w)=\frac{\pi(w-v)u}{((v+w)^2-u^2)\sqrt{4vw-u^2}}. $