Let $\varepsilon$ be distributed according to the following $t$-distribution:
$$p(\varepsilon) = \frac{\Gamma\left(\frac{\nu+1} 2\right)}{\Gamma\left(\frac \nu 2\right)[(\nu-2) \pi\sigma^2]^{1/2}}\left[1+\frac 1 {\nu-2}\left(\frac \varepsilon \sigma \right)^2\right]^{-\frac{\nu+1} 2}$$
so that $\operatorname{E}[\varepsilon]=0$ and $\operatorname{Var}[\varepsilon] = \sigma^2$. Consider the r.v. $\zeta = \frac \varepsilon \sigma$ which has unit variance.
Is it possible to compute in closed form $\operatorname{E}\left[\frac{\zeta^2}{(a+\zeta^2)^2}\right],\quad a\in\mathbb{R}$?