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, $$G(f)=\int_0^\infty g^{-1}(x)\cos(2\pi f x)=K_0(2\pi f),$$
Now I would need to take the convolution of two Bessel functions to obtain the desired $H_p(f)$ for $p=3/2$. From the plot below I 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 $f>1$. For small $f$ there is a plateau at $|K_0(2\pi i)|=0.4992$, not exactly $1/2$ but close.
So at least qualitatively, we can understand what happens: upon convolution, the function $G$ smoothes out the peak near $f=1$ in $H_{1/2}$, while leaving the flat portions for smaller and larger $f$ unaffected.
Plot of $|H_{1/2}(f)|$ (blue) and $|H_{3/2}(f)|$ (gold).
Plot of $|H_{3/2}(f)|$ for $f<1$, to show that it is almost but not quite flat, and almost but not quite $1/2$ for $f\rightarrow 0$. The sharp peak at $f=1$ that was present in $|H_{1/2}(f)|$ has been greatly suppressed by the convolution with $G$.