The exponential substitution reads a fractional slope, and takes the sign of its base from the radicals - #1290
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…sign of its base from the radicals `e^(x/2)/sqrt(e^x - 1)` had no antiderivative. It is `2/sqrt(u^2 - 1)` under `u = e^(x/2)`, and the exponential substitution refused the half: a slope had to be a whole number. The base is now `e^(k x)` with `k` the greatest common divisor of the slopes as rationals -- numerators' gcd over denominators' lcm -- so every exponential in the integrand is a whole power of it, and `e^(x/2)` beside `e^x` is `u` beside `u^2`. The sign of `k` is chosen for the radicals. `sqrt(1 + e^(-x))/sinh(x)` under `u = e^x` is a root of `1 + 1/u`, a quotient, which nothing rationalises; under `u = e^(-x)` it is `sqrt(1 + u)`, a root of something linear, which the linear-radical substitution answers. So where every exponential under a root has a negative slope the base is `e^(-k x)`; `u` is positive either way, being an exponential, and nothing about the substitution changes but the spelling of what comes out. ## Measured Every answer differentiated back. Rubi corpus, 463-problem sample, against the master it was cut from (#1288), timed on the same evening: master 325/463, 0 wrong, 0 timeouts, 80 s with this 328/463, 0 wrong, 0 timeouts, 81 s Three more, nothing lost: Timofeev's `e^(x/2)/sqrt(e^x - 1)`, and Bondarenko's `sqrt(1 + e^(-x))/(e^x - e^(-x))` and `sqrt(1 + e^(-x))/sinh(x)`. `sqrt(1 + tanh(4x))`, the fourth of that family, becomes `sqrt(2u^2/(u^2 + 1))` and wants `sqrt(u^2) = u` for a `u` the rule knows to be positive; that is a rewrite this rule could make and the simplifier rightly will not, and it is not here yet. The five test suites are green. Part of #718. Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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…them, and the closed exponential-trigonometric rule reads a power (#1291) `e^(arcsin(x)) x^3/sqrt(1 - x^2)` had no antiderivative. Under `x = sin(u)` it is `e^u sin(u)^3`; `x arcsec(x)/sqrt(x^2 - 1)` is `u sec(u)^2` under `x = sec(u)`. The general substitution does not find either, because it substitutes for a subtree and asks what is left, and what is left beside `arcsin(x)` is `x` itself -- which only the inverse substitution removes, the same reason the logarithm has one. The radical goes by construction: `sqrt(1 - x^2)` is `cos(u)` on the branch the arcsine is defined on, where substituting and simplifying would leave `sqrt(1 - sin(u)^2)`, which the simplifier is right not to call `cos(u)`. Admitted only where there is a radical to remove or an exponential of the inverse to turn into `e^u`; `x arcsin(x)` stays by parts', and its answer stays in `x`. Three things the substituted integrand needed, each a gap in its own right: - The closed rule for `P(x) e^(ax) trig(bx)` read one bare sine or cosine, so `e^u sin(u)^3` and `e^u/sec(u)` fell out. A power of the sine or the cosine, or a product of the two, is a sum of sines and cosines of multiples of the argument -- `sin^3 u = (3 sin u - sin 3u)/4` -- and each term is the shape the rule closes, so it runs once per term and the answers add. A secant or cosecant under the bar is read as a cosine or sine above it. - By parts read a product and not a quotient with a polynomial over the bar: `x/cos(x)^2` reached nothing where `x sec(x)^2` is one step. The polynomial factors of the numerator are gathered against the rest of the quotient, at the top only, and not over an algebraic function, which is the rational integrator's or Euler's. And the polynomial step hands its `dv` to the closed `sin^p cos^q` rule directly, which is scoped to the top and would otherwise not see it: `u sin(u)^3/cos(u)^2` was nine seconds of declining through the half-angle substitution and is half a second. - `ComputeAsAQuestionOfItsOwn` went back through the entry check, where a depth of zero means a fresh question and empties the set of integrals in progress -- so the integrand a substitution was rewriting lost its place in the cycle guard. It now enters past that check. And the truncation flag that gates caching a decline is scoped to the computation it is about, rather than to everything asked after some unrelated branch above ran out of depth. ## Measured Every answer differentiated back at points on the branch the inverse function is defined on. Rubi corpus, 463-problem sample, against the master it was rebased onto (#1290), both measured on the same evening: master 337/463, 0 wrong, 0 timeouts, 89 s with this 343/463, 0 wrong, 0 timeouts, 88 s Six more, nothing lost: Charlwood's `e^asin(x) x^3/sqrt(1-x^2)`, `x asec(x)/sqrt(x^2-1)` and `asin(x)/(1+sqrt(1-x^2))`, Timofeev's `(x^2-1)^(3/2) asec(x)^2/x^5`, Bronstein's `(x - tan(x))/tan(x)^2` and Welz's `x/(a e^(px) + b e^(-px))^2`. Two inputs were measured slower in the corpus run and turned out to cost the same on master alone -- warm-cache ordering, not this change. The five test suites are green. Part of #718. Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura Co-authored-by: Claude Opus 5 (1M context) <noreply@anthropic.com>
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e^(x/2)/sqrt(e^x - 1)had no antiderivative. It is2/sqrt(u^2 - 1)underu = e^(x/2), and the exponential substitution refused the half: a slope had to be awhole number. The base is now
e^(k x)withkthe greatest common divisor of theslopes as rationals -- numerators' gcd over denominators' lcm -- so every exponential
in the integrand is a whole power of it, and
e^(x/2)besidee^xisubesideu^2.The sign of
kis chosen for the radicals.sqrt(1 + e^(-x))/sinh(x)underu = e^xis a root of
1 + 1/u, a quotient, which nothing rationalises; underu = e^(-x)it issqrt(1 + u), a root of something linear, which the linear-radical substitutionanswers. So where every exponential under a root has a negative slope the base is
e^(-k x);uis positive either way, being an exponential, and nothing about thesubstitution changes but the spelling of what comes out.
Measured
Every answer differentiated back.
Rubi corpus, 463-problem sample, against the master it was cut from (#1288), timed on
the same evening:
Three more, nothing lost: Timofeev's
e^(x/2)/sqrt(e^x - 1), and Bondarenko'ssqrt(1 + e^(-x))/(e^x - e^(-x))andsqrt(1 + e^(-x))/sinh(x).sqrt(1 + tanh(4x)),the fourth of that family, becomes
sqrt(2u^2/(u^2 + 1))and wantssqrt(u^2) = ufor a
uthe rule knows to be positive; that is a rewrite this rule could make andthe simplifier rightly will not, and it is not here yet.
The five test suites are green.
Part of #718.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura