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A root of a multiple of the radicand beside Euler's sum is the same root - #1871
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Rafael-SOWNet merged 3 commits intoOct 13, 2026
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… and a radicand over a constant is read Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…e-of-the-radicand-beside-the-euler-sum
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Part of #718.
Five of Rubi's 1.3.2 problems are
(d + e x + f sqrt(Q))^nin spellings thatSolveByEulerSubstitutiondid not read:sqrt(g Q)sits above or below the bar beside the sum. The rule took exactly one radicand.sqrt(g Q)/sqrt(Q)is constant wherever both roots are defined, whatever the sign ofg.WithTheRootsOfMultiplesAsOnenow writes each root of a constant multiple of the sum's own radicand as a named constant timessqrt(Q), asks the question, and substitutes the ratio back.(a f^2 + e x (2 d + e x))/f^2.TryGetPolyQuadraticdoes not read a quotient by a constant, soTryReadAQuadraticOverAConstantreads the numerator and divides.Sizes, measured on master 683ab15, on 2.5.0 and on this branch:
(d+e*x+f*sqrt((a*f^2+e*x*(2*d+e*x))/f^2))^n(d+e*x+f*sqrt((a*f^2+e*x*(2*d+e*x))/f^2))^n/((a*f^2+e*x*(2*d+e*x))/f^2)^(1/2)(a*g+2*d*e*g*x/f^2+e^2*g*x^2/f^2)^(1/2)*(d+e*x+f*sqrt(a+2*d*e*x/f^2+e^2*x^2/f^2))^n(d+e*x+f*sqrt(a+2*d*e*x/f^2+e^2*x^2/f^2))^n/(a*g+2*d*e*g*x/f^2+e^2*g*x^2/f^2)^(1/2)(d+e*x+f*sqrt((a*f^2+e*x*(2*d+e*x))/f^2))^n/((a*f^2*g+e*g*x*(2*d+e*x))/f^2)^(1/2)Each was checked exactly, derivative against integrand at 50, 300 and 2,000 digits, with
g = 5/2and withg = -5/2. The tests differentiate them back atg = 2.5andg = -2.5.Rubi corpus, compared with master 915f742:
The allocation gate passed. The suite passed: 15,315 passed, 0 failed. Master 683ab15 is merged in, and the build passes for every target framework.
🤖 Generated with Claude Code
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