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A function comes out of a fractional power of its even power with its sign - #1399
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… sign (a sin(x)^2)^(5/2) is a^(5/2) |sin(x)|^5, which is a^(5/2) sgn(sin(x)) sin(x)^5, and the sign is a constant between the sine's zeros: it comes out in front of the integral, and what is left is a whole power any rule reads. An even power of a real function is not negative, so (c q)^p = c^p q^p holds for the principal powers whatever c is, and the test differentiates every answer back at a = -3 as well as a = 2, on both sides of a zero of the function. Whole products of the exponents only, at the top of the integration only, after the rule that does the same for a polynomial under a square root, and declining a base that is itself a fractional power. Rubi's x sqrt(sin(x)^2), (csc(x)^2)^(3/2) and 1/(csc(x)^2)^(7/2), unevaluated since #1387 made the root of an even power the modulus, are answered, and (a sin(x)^2)^(5/2), 1/sqrt(a cot(x)^2) and (b tan(x)^2)^(5/2) no longer come back in powers of abs(sin(x)) over sgn(cos(x)). The 1774-problem suite is unchanged at 1706 with 0 wrong; the family 4 sample answers eight more. Part of #718. Part of #1387. Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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#1387 made
(sin(x)^2)^(3/2)the modulus|sin(x)|^3and left three of Rubi's family-4 integrands unevaluated for it, with two more answered in powers of the modulus through the substitutions. The modulus is the function times its sign, and the sign is a constant between the function's zeros — the same observationSolveByTakingASquareFactorOutOfARootmakes for a polynomial under a square root, made here for a function:(a sin(x)^2)^(5/2)isa^(5/2) sgn(sin(x)) sin(x)^5, the sign comes out in front of the integral, and what is left is a whole power any rule reads.Why it needs no assumption on the constant. An even power of a real function is not negative, so
(c q)^p = c^p q^pholds for the principal powers whatevercis (arg(c q) = arg(c)forq ≥ 0), and(f^(2k))^p = |f|^(2kp). The test differentiates every answer back ata = -3as well asa = 2, on both sides of a zero of the function. Whole products of the exponents only, so that what is handed on is a whole power; at the top only, as the polynomial rule: a substitution's variable is one it knows the sign of, and a sign written for it is one nothing differentiates. A base that is itself a fractional power is declined, since it is not real where its own base is negative.x*sqrt(sin(x)^2)sgn(sin(x)) (sin(x) - x cos(x))(csc(x)^2)^(3/2),1/(csc(x)^2)^(7/2)sgn(csc(x))times the integral ofcsc^3, ofsin^7(sin(x)^2)^(3/2)(2/3 tan(x/2)^5 + 2 tan(x/2)^3) ((2 tan(x/2))^2)^(1/2)/(tan(x/2)^2 + 1)^3sgn(sin(x)) (cos(x)^3/3 - cos(x))(a*sin(x)^2)^(5/2)sgn(sin(x)) a^(5/2) (…)in(1 - abs(sin(x))^2)^(1/2)oversgn(cos(x))sgn(sin(x)) a^(5/2) (2 cos(x)^3/3 - cos(x) - cos(x)^5/5)1/sqrt(a*cot(x)^2)-sgn(tan(x)) ln(1/sqrt(1 + abs(tan(x))^2))/sqrt(a)sgn(cot(x)) ln(1 + tan(x)^2)/(2 sqrt(a))Measured. The 1,774-problem independent suite: 1706/1774, 0 wrong, 0 timeouts, per-problem verdicts identical to master
374d6c1a, 90.4 s → 88.2 s in total. Rubi's family 4 sample (422, six per file): 243/422 with 0 wrong; eight rows are this rule's shapes —(a*sec(x)^2)^(1/2),1/(a*csc(x)^2)^(1/2),x*sqrt(sin(x)^2),(csc(x)^2)^(3/2)and1/(csc(x)^2)^(7/2)were unevaluated on master374d6c1a(probed), and(a*sin(x)^2)^(5/2),(b*tan(x)^2)^(5/2)and1/sqrt(a*cot(x)^2)were answered in powers ofabs(sin(x))— and every other row that differs from the previous family-4 run is identical between the two builds, so it is the merged PRs' since. All four test suites green; newEvenPowerSignTest(10 rows) with its.editorconfigsection; BREAKING-CHANGES entry under the #1387 one.Part of #718. Part of #1387.
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