Generalization of Laplace's integral

Question

A generalization of the Gaussian integral $$ \int_0^{\infty} e^{-\pi at^2} \, dt = \frac{1}{2\sqrt{a}}, \hspace{0.5cm} a>0 $$ is Laplace's integral: $$ \int_0^{\infty} e^{-\pi at^2-\pi b/t^2} \, dt = \frac{1}{2\sqrt{a}}e^{-2\pi \sqrt{ab}}, \hspace{0.5cm} a,b>0. $$ Given that the Gaussian integral may be generalized to $$ \int_0^{\infty} \cos(2\pi xt) e^{-\pi at^2} \, dt = \frac{1}{2\sqrt{a}}e^{-\pi x^2/a}, \hspace{0.5cm} a>0, \, x \in \mathbb{R}, $$ I am wondering if a similar 'cosine generalization' of Laplace's integral exists. That is, can we evaluate $$ \int_0^{\infty} \cos(2\pi xt)e^{-\pi at^2-\pi b/t^2} \, dt, \hspace{0.5cm} a,b>0, \, x \in \mathbb{R} $$ in closed form? I have tried to do this by Taylor expanding the cosine and evaluating $$ \int_0^{\infty} t^{2n} e^{-\pi at^2-\pi b/t^2} \, dt $$ for all $n \geq 0$ by differentiating Laplace's integral repeatedly with respect to $a$, but the formulas get complicated very quickly. Is there a nicer way to attempt this?