Evaluating $ \int_0^{\infty}\frac{v}{\sqrt{v + c}}e^{-\frac{y^2}{2(v + c)} - \frac{(u-v)^2}{u^2v}}dv$ While working on mixture (variance) of normal distribution and keep running into these two integrals
$$ \int_0^{\infty}\dfrac{v}{\sqrt{v + c}}e^{-\dfrac{y^2}{2(v + c)} - \dfrac{(u-v)^2}{u^2v}}dv,$$
$$\int_0^{\infty}\dfrac{v^{-1}}{\sqrt{v + c}}e^{-\dfrac{y^2}{2(v + c)} - \dfrac{(u-v)^2}{u^2v} }dv,$$
where $c>0, u>0 ,y\in \mathbb R$.
I was wondering are they solvable? Can they be expressed as some known function or in elementary terms?
Any help would be appreciated.

For this integral, the best result seems to be a Taylor series in $c$ with coefficients in closed form.
In the typical case $c=1$, even specializing to the easiest case of $u=1$, $y=0$, we get the integrals $\int_0^\infty v^{\pm1}(v+1)^{-1/2}e^{-v-1/v}dv$,
for which neither Mathematica nor Gradshteyn and Ryzhik has any answer.
However, there is an explicit expression in the limit case $c=0$, which includes Gradshteyn and Ryzhik 3.471.15.  Setting $z=\sqrt{4+2y^2}/u$, Mathematica gives the two integrals as:
$$\frac{u^3(1+z)}{2}\sqrt{π}\,e^{\,z(\sqrt{2}−1)} \text{ and }
\frac{2}{zu}\sqrt{π}\,e^{\,z(\sqrt{2}−1)}.$$
When we expand the integrands above as power series in $c$, the integrals of each term have similar closed-form expressions.
