 Solution

For the uniform-gravity Hamiltonian, take the three integrals $p_x,p_y,H$. Since the potential does not depend on $x,y$,For $g\ne 0$, the $mgdz$ term in $dH$ makes these three differentials independent. Thus the system is Liouville integrable, as the explicit separated solution already demonstrates.

Replacing $mgz$ by a sufficiently smooth arbitrary function $f(z)$ preserves the same commuting integrals. They are independent wherever $p_z$ and $f^{\prime }(z)$ do not vanish simultaneously. The vertical motion reduces toThe horizontal motion remains uniform. Thus an elementary antiderivative is unnecessary: the system is still integrable by a one-dimensional quadrature on regular branches, joined at turning points. Critical equilibria may be singular common level sets without destroying generic integrability. If $f$ is constant, the three momenta themselves give an everywhere independent commuting set, avoiding the rank loss of $p_x,p_y,H$ at $p_z=0$.