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 "Tables of Integral Transforms" (Volume I).

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 would need to take the convolution of two Bessel functions to obtain the desired $H_p(f)$ for $p=3/2$. I do note that the sharp drop of the Fourier transform for $f>1$ is already present in $H_{1/2}$, presumably because the integrand in the Fourier transform decays in the upper half of the complex plane provided $f>1$.

<IMG SRC="https://ilorentz.org/beenakker/MO/Hp_plot.png"/>

*Plot of $|H_{1/2}(f)|$ (blue) and $|H_{3/2}(f)|$ (gold).*