I think that the Hamiltonian describing such a system would be highly complex because the expression to describe interactions between electrons and nucleus (potential energy term) doesn’t simplify at all if the multiple expansion is not applied. Also, in that regard, the complexity of the Hamiltonian doesnt allow us to discern which effect is more important as the multipole expansion does (dipole, quadrople and higher effects are just too small to be considered), which would insanely increase the calculation effort (if possible).
Thinking on PT basics, the fact that the multipole expansion allows for quantification of the “size” of the perturbing term, if there is no multipole expansion at all, I would imagine the PT is simply nonapplicable cause, again, the perturbation Hamiltonian looks like very complex and is not necessarily “small” compared with the unperturbed case.

I would appreciate any comment 😀