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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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Rafael-SOWNet merged 3 commits intoOct 4, 2026
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…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
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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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 ofa ± i a sinh, six of them after searches past the budget. 2.5.0 declined them too:287c69a7x^3 sqrt(a + i a sinh(c + f x))x sqrt(a + i a sinh(c + f x))sqrt(a + i a sinh(c + f x))/xx^3 (a + i a sinh(c + f x))^(3/2)1/sqrt(a - i a sinh(c + f x))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 ofa (1 ± i sinh(y))isa^ptimes an odd power ofcosh(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 fora ± 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 whateverais.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 master8f3757cd, the branch's base: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 = 181.8witha = 2.1,c = 1.9,d = 1.7, where the integrand is about2·10^-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 atx = -181.8and more than 390 atx = ±0.5,±3and±40.Measured then on the Rubi corpus against master
8f3757cd: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 sinhbeside a power ofxor besideA + 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 master287c69a7,7667efdf, 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