Verdict: falsified. I ran the numerical experiment and it fails at two independent walls. Full writeup: Could AAA-fit rationals replace Slater orbitals?.

Experiment

AAA fits $e^{-r}$ on $[0, 20]$ to machine precision (max error $5 \times 10^{-14}$ with 11 support points). But the fitted rational has no constraint to decay correctly outside the sample interval — it decays algebraically at best, and on this fit it grows past $r \approx 20$. The consequence: the most basic integral in quantum chemistry, $\int_0^\infty r^2,R(r)^2,dr$, blows up to $2.4 \times 10^{12}$ instead of the closed-form $1/4$. The tail dominates everything.

Classical $[0/n]$ Padé recovers integrability (guaranteed $1/r^n$ decay), and the pole-residue summation of the radial integral then does work — it matches scipy.quad to $10^{-11}$. So the one positive piece survives: given an integrable rational, residue summation computes the 1-center integral.

Why it fails structurally

Even granting an integrable rational, the multi-center integral does not reduce. Gaussians work because of Boys’ product theorem — a product of two Gaussians on different centers is a single Gaussian on a third center (it is the parallel axis theorem in disguise). Rationals have no analog: a product of rationals on different centers stays a multi-center rational, and partial fractions in one variable leaves the other untouched. There is no rational product theorem, so the 4-center 2-electron integral cost never drops.

What would un-falsify it

Exhibit a closed-form reduction identity for the product of two rational-in-distance functions on different centers — the rational analog of $e^{-\alpha|\vec r - \vec A|^2} e^{-\beta|\vec r - \vec B|^2} = K,e^{-(\alpha+\beta)|\vec r - \vec P|^2}$. No such identity is known, and the linear/non-quadratic dependence on distance is the obstruction (the completing-the-square trick is specific to quadratic exponents). Find one and this claim flips back to open.