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Half-odd powers of a + i a sinh are integrated by the half angle at which they are squares - #1753

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a-root-of-one-plus-an-imaginary-hyperbolic-sine
Oct 4, 2026
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Rafael-SOWNet merged 3 commits into
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a-root-of-one-plus-an-imaginary-hyperbolic-sine

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Part of #718.

x^3 sqrt(a + i a sinh(c + f x)) was declined, with the rest of Rubi's 6.1.1 and 6.1.5 with a half-odd power of a ± i a sinh, six of them after searches past the budget. 2.5.0 declined them too:

integrand 2.5.0 master 287c69a7 this
x^3 sqrt(a + i a sinh(c + f x)) declined declined after 4.8 s in 0.8 s
x sqrt(a + i a sinh(c + f x)) declined declined after 2.5 s in 0.2 s
sqrt(a + i a sinh(c + f x))/x declined declined in 0.2 s
x^3 (a + i a sinh(c + f x))^(3/2) declined declined after 2.3 s in 1.8 s
1/sqrt(a - i a sinh(c + f x)) declined searched past 8 s in 0.4 s

Each answer is differentiated back with the symbols pinned and compared as a complex number at x = -2.7, -1.3, -0.4, 0.4, 1.3, 2.7; the times include that.

What changes. 1 + i sinh(y) is (cosh(y/2) + i sinh(y/2))^2, so a half-odd power of a (1 ± i sinh(y)) is a^p times an odd power of cosh(y/2) ± i sinh(y/2) -- a sum of exponentials of the half angle -- up to a constant on every interval where both are continuous. The rest of the integrand is written in the half angle, as the rule for a ± a cosh(y) beside it does, and the question asked again; the answer is the antiderivative times each power as written over the form it was rewritten to, a quotient whose logarithmic derivative is zero, so it holds whatever a is.

Tests: OnePlusAnImaginaryHyperbolicSineIntegralTest, five rows compared as complex numbers on both sides of zero; none is answered on master.

Measured first on the 47 problems of 6.1.1 and 6.1.5 with a half-odd power of a ± i a sinh, 32 run, at the corpus's 5-second budget, against master 8f3757cd, the branch's base:

master this
solved 0 29
wrong 0 1
past the budget 6 2

The one the harness counts wrong here, 1/sqrt(a + i a sinh(c + d x)), is right. It disagrees at one point of the harness's scan, x = 181.8 with a = 2.1, c = 1.9, d = 1.7, where the integrand is about 2·10^-68 (1 - i): the harness's test for a real value is absolute below |z| = 1, and takes that for one. Evaluated with 400 digits there, the derivative agrees with the integrand to 341 digits, and to 343 at x = -181.8 and more than 390 at x = ±0.5, ±3 and ±40.

Measured then on the Rubi corpus against master 8f3757cd:

master this
family 0, independent suites (1814) 1766 1766
family 1, 40 a file (1381) 1296 1294
families 2 to 8, sampled (2410) 2215 2220

The harness counts no other answer wrong. Of the 39 problems the two builds disagreed on, pocket and sample together, run again one build at a time, master answers 2 and this 30. Twenty-nine are the pocket's, half-odd powers of a ± i a sinh beside a power of x or beside A + B sinh(x), answered here in 0.2 to 17 seconds; one, sqrt(x)/((a + b x^2) (c + d x^2)^3), is answered here at twenty seconds and ran past the harness's patience on master. The two master answers, at 23 and 24 seconds, ran past it here; run again, twice on each build, both run past it on both. 1/(a + i a sinh(c + d x))^(5/2), declined on master, runs past the budget here, and five more run past it on both.

The suite passes on the commit measured, 322b50ec, 14,596 tests with 13 skipped, and the allocation gate with it. On the merge with master 287c69a7, 7667efdf, the 4,142 calculus and corpus tests that run pass, with 2 skipped. Every row of the first table is as it says on the merge.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

Rafael-SOWNet and others added 2 commits October 4, 2026 08:25
…hich they are squares

x^3 sqrt(a + i a sinh(e + f x)) was declined, with the rest of Rubi's
6.1.1 and 6.1.5 that put a half-odd power of a ± i a sinh beside a power
of x. 1 + i sinh(y) is (cosh(y/2) + i sinh(y/2))^2, so such a power is
a^p times an odd power of cosh(y/2) ± i sinh(y/2) up to a constant on
every interval where both are continuous. The rest of the integrand is
written in the half angle and asked again, and the answer is the
antiderivative times the power as written over that form.

Part of #718.

Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
@Rafael-SOWNet Rafael-SOWNet added this to the 2.6.0 milestone Oct 4, 2026
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
@Rafael-SOWNet
Rafael-SOWNet merged commit d276b34 into master Oct 4, 2026
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