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,
$$G(f)=\int_0^\infty g^{-1}(x)\cos(2\pi f x)=K_0(2\pi f).$$

The key thing to note at this point is that the Bessel function $K_0$ for real argument is basically a broadened delta function. So $H_{1/2}$ decays rapidly for $f>1$. For small $f$ there is a plateau at $|K_0(2\pi i)|=0.4992$, not exactly $1/2$ but close.

Now I need to take the convolution of $H_{1/2}$ with $G$ to obtain the desired $H_{3/2}(f)$. Since $G(f)$ is a broadened delta function, this convolution smears out the sharp peak in $H_{1/2}$ at $f=1$, while leaving the plateaus for smaller and larger $f$ unaffected. 

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

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

<IMG SRC="https://ilorentz.org/beenakker/MO/H32_detail.png"/>

*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 $G$.*