Does $\int { y\cosh \left(\beta y^2\right)}J_0\left(\gamma y^2 \right) dy$ have a closed form I am trying to solve the following indefinite integral
$$F_Y(y) = \int {y\cosh \left(\beta y^2\right)}J_0\left(\gamma y^2 \right) dy$$
Where $J_0$ is the Bessel function of the first kind.
I tried to expand the $J_0$ using it's Taylor Series expansion like J.M showed here but the expression became much too unwieldy afterwards. 
Is it possible to express this integral in a closed form (preferably, using elementary functions, Bessel functions, integers and basic constants) maybe with hyper geometric function ? 
 A: Some notes to consider:


*

*Let $\beta \rightarrow ia$ to obtain the form
\begin{align}
F(a, \gamma) = \int y \cos(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy
\end{align}
or, more generally, 
\begin{align}\tag{1}
F(a, \gamma) = \int y e^{ia y^{2}} \ J_{0}(\gamma y^{2}) \ dy.
\end{align}

*Let $t = y^{2}$ for which (1) becomes
\begin{align}\tag{2}
F(a, \gamma) = \frac{1}{2} \int e^{i at} \ J_{0}(\gamma t) \ dt.
\end{align}


After the calculation of (2) take the real part of the value obtained and then let $a \rightarrow - i \beta$ to obtain the desired result.  
Example: If the limits are $(0, \infty)$ then
\begin{align}
\int_{0}^{\infty} e^{-iat} \ J_{0}(\gamma t) \ dt = \frac{-i}{\sqrt{a^{2} - \gamma^{2}}}. 
\end{align}
This leads to 
\begin{align}
\int_{0}^{\infty} y \ \cos(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy &= 0 \\
\int_{0}^{\infty} y \ \sin(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy &= \frac{1}{\sqrt{a^{2} - \gamma^{2}}}.
\end{align}
In the desired form of what the question is seeking the results are
\begin{align}
\int_{0}^{\infty} y \ \cosh(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy &= 0 \\
\int_{0}^{\infty} y \ \sinh(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy &= \frac{-1}{\sqrt{a^{2} + \gamma^{2}}}.
\end{align}
