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Integrate a negative power of an inverse function of c + d x^2 in sine and cosine integrals - #1906
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Rafael-SOWNet merged 4 commits intoOct 16, 2026
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For u the inverse sine or cosine of w = c + d x^2 with c^2 = 1, its hyperbolic cosine with c^2 = 1, or its hyperbolic sine with c^2 = -1, integral (a + b u)^n is x (a + b u)^n + 2 e n b H (a + b u)^(n - 1) - 4 e b^2 n (n - 1) integral (a + b u)^(n - 2), with H = 1/(d x u'(w)), and it closes at the first two powers. Rubi's 5.1.5, 5.2.5 and 7.2.5 (a + b acos(1 + d x^2))^4 and the like were declined. Part of #718 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…e and cosine integrals w - c is a constant times g(u/2 + beta)^2 for a sine or a cosine g, so x/g is constant on each side of 0 and dx = (x/g)(g'/2) du. The powers -1 and -2 are then sine and cosine integrals of (a + b u)/(2 b), hyperbolic ones for the inverse hyperbolic cosine, and the recurrence run backwards gives the rest. Rubi's 5.1.5, 5.2.5 and 7.2.5 1/(a + b acos(1 + d x^2)) and the like, to the first three negative powers, were declined. Part of #718 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…-an-inverse-function-of-a-quadratic # Conflicts: # Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs # Sources/AngouriMath/Functions/Continuous/Integration/Integration.Definition.cs # Sources/Tests/UnitTests/Calculus/APowerOfAnInverseFunctionOfAQuadraticIntegralTest.cs
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. Follows #1904.
#1904 integrates
(a + b u)^nfor a wholen >= 1, withuthe inverse sine or cosine ofc + d x^2(c^2 = 1) or its inverse hyperbolic cosine. This adds the negative powers.w - cis a constant timesg(theta)^2withtheta = u/2 + beta:uc = 1c = -1acos(w)cos u - 1 = -2 sin(u/2)^2cos u + 1 = 2 cos(u/2)^2asin(w)sin u - 1 = -2 sin(u/2 - pi/4)^2sin u + 1 = 2 cos(u/2 - pi/4)^2acosh(w)cosh u - 1 = 2 sinh(u/2)^2cosh u + 1 = 2 cosh(u/2)^2So
x = kappa g(theta), withkappa = x/g(theta)constant on each side of 0, anddx = (kappa/2) g'(theta) du. Withv = (a + b u)/(2 b),I(-1) = (kappa/(2 b)) integral g'(v - gamma)/v dvisCi/Si(Chi/Shiforacosh).I(-2)follows by parts, and the recurrence of #1904 run backwards gives the lower powers.The answers are exact on the integrand's real domain. Off it, where the inverse function is complex,
x/gcan change sign across a cut. The exact checks found 114 such points, every one with the integrand non-real there. The rule refuses an answer only where it differs from the integrand where the integrand is real (DerivativeDiffersWhereReal).Checks
APowerOfAnInverseFunctionOfAQuadraticIntegralTest: 6 more cases, pinned on the real domain.Corpus, measured against master 55a8044, before #1904 (this branch then merged 2d17cd1):
No losses and no slower rows. Gate passed; suite 15,427 passed, 0 failed.
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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura