Taking a constant factor out of a denominator made an unanswerable integral four times slower to decline

[Rafael-SOWNet]

1/(b^2*cos(x)^2 + a^2*sin(x)^2) has no elementary antiderivative this library can find, and it did
not find one before either. What changed is how long it takes to say so: 107 s to over 400 s,
measured on one machine, both arms.

Where it comes from

TakeConstantFactorOutOfDenominator, added in #1249. It reads a quotient whose denominator mixes a
factor free of the variable with one holding it, takes the constant out, and hands
numerator / variablePart back to the integrator:

return Integration.ComputeIndefiniteIntegral(numerator / variablePart, x, integrateByParts)
    ?.Pipe(i => i / constantPart);

That is a fresh integral for the whole chain to work through, and on this integrand the search
underneath produces a great many denominators of that shape — every one of them a new problem.

Attributed by ablation, not by reading

Two hypotheses about which rule was responsible were wrong before this one was right, so the method
is worth recording. Returning null at the top of the method under an environment variable, one
rule at a time:

2dbeedf7, before #1249	136 s
master	> 400 s
master, SolveByScalingTheVariable ablated	> 400 s
master, TakeConstantFactorOutOfDenominator ablated	107 s

The scaling rule — the other half of #1249, and the one I first suspected because it is the visible
new rule — is not involved.

Two narrowings that do not help

Both were plausible and both measured no better than unguarded, which is the useful part of the
report:

• Only fire where the variable part is a rational function of x. The reasoning was that the
  rules wanting a polynomial denominator are the rational ones, so a trigonometric denominator gains
  nothing from losing its constant factor. Still > 400 s: the expensive firings are on denominators
  that are rational — after the half-angle rewrite everything under this integrand is rational in
  the new variable with a and b as parameters. The rule is expensive inside its intended
  domain, not outside it.
• Only fire where the constant part is symbolic, since a numeric factor never needed this rule.
  Also > 400 s, for the same reason: a and b are symbolic.

So the cost is not a matter of firing on the wrong shapes. It is that the rule multiplies the number
of distinct sub-problems on integrands that have no answer, and no structural condition separates
"will pay off" from "will not".

What it is not

Not a wrong answer, and not a hang: the integral is declined either way, and every suite is green.
It is a slower refusal, and the same kind of cost #1245 records for the half-angle rewrite.

It is, however, why a full intbench run over Rubi's suites no longer completes on this machine —
the harness sits on this problem long past its budget.

What would fix it

A cost bound rather than another structural condition. WorkBudget from #896 is the instrument:
a rule that hands on a speculative sub-problem should be able to say how much of the remaining
budget it is prepared to spend, and stop. Inventing another predicate here would be a third guess of
the same kind as the two above.

Filed against #718.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

[Rafael-SOWNet]

Re-measured, and this no longer reproduces. 1/(b^2*cos(x)^2 + a^2*sin(x)^2) is now answered in 0.3 s: it goes through the tangent substitution and comes back as a piecewise on the sign of a^2 b^2, on master 1293b56 and on 1a7f58d (today's master with #1602 on top). The answer differentiates back to the integrand at seven points for two sets of values of a and b. Full intbench runs over Rubi also complete again: last night's covered all 2253 rows with %i in the integrand, the families sample and family 0.

That leaves no measured case behind the cost bound proposed here, so I'm closing this. The next slow decline that shows the constant factor multiplying sub-problems gets its own issue, with its integrand.