I am trying to evaluate the integral

$$
\int_0^1 \frac{e^{-y^2(1+v^2)}}{(1+v^2)^n}dv = e^{-y^2}\int_0^1 \frac{e^{-y^2v^2}}{(1+v^2)^n}dv
$$
for $n\in \mathbb{N}$.For n=1 one finds Owen's T function, i.e.
\begin{align}
\int_0^1 \frac{e^{-y^2(1+v^2)}}{(1+v^2)}dv=2\pi \operatorname{T}\left(\sqrt{2} y,1\right) = \frac{\pi}{2} \operatorname{erfc}(y) \left(1 - \frac{1}{2} \operatorname{erfc}(y)\right)
\end{align}
A nice source on the Owen's T function is [2]. In [3] they state that
\begin{align}
\int \frac{e^{-v^2}}{v^2 + 1} dv ,
\end{align}
has no anti-derivative. Hence, I do not suspect one can find one for our integral.

This integral occurs in a series I am integrating over for a approximation I am performing. Hence, it would already be nice if I could find the second (n=2) and third (n=3) term. Has someone an idea how to evaluate the integral. Many thank in advance!