This is not quite an answer, but it may be helpful to note that the Fourier transform $H_p(f)$ of $h_p(x)=g^{-p}(x)\exp[-2\pi ig(x)]$, with $g(x)=\sqrt{1+x^2}$ has a closed form expression for $p=1/2$:
$$H_{1/2}(f)=\int_0^\infty h_{1/2}(x)\cos(2\pi f x)=K_0\left[2\pi\sqrt{f^2-1}\right],$$
see page 17 of Erdelyi's table of Fourier transforms.

The Fourier transform of $1/g(x)$ is also a Bessel function,
$$G(f)=\int_0^\infty g^{-1}(x)\cos(2\pi f x)=K_0(2\pi f),$$

So now I just need to take the convolution of two Bessel functions to obtain the desired $H_p(f)$ for $p=3/2$