Two linear factors whose product is a multiple of the radicand arsinh holds are written as that radicand - #1488
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… holds are written as that radicand `(a + b arsinh(c x)) sqrt(d + i c d x) sqrt(f - i c f x)` ran out of time. The two linear factors multiply to `d f (1 + c^2 x^2)`, a constant multiple of the radicand `arsinh(c x) = ln(c x + sqrt(c^2 x^2 + 1))` holds -- the case #1487 answers, spelled as two factors. Rubi's 7.1.4 writes seventy-two of its 663 rows this way. `L1^p L2^q` with `p - q` whole is now `L1^(p - q) L1^q L2^q`, and `L1^q L2^q` is `lambda^q M^q` for a whole `q` and `K^k M^(k/2)` for `q = k/2`, with `K = sqrt(L1) sqrt(L2)/sqrt(M)` carried as a symbol and written in front of the answer: constant wherever it is defined, as #1487's is. Tried only where no single base is a multiple of the radicand, on the factors of both sides of the bar, with a nested power read through and a factor below the bar read with its exponent negated. Rubi's 7.1.4 and 7.1.5, all 376 problems that count: 308 to 345, no row lost, 56 timeouts to 27. Family 7 235 to 237 of 270; the 7.2 files, families 1, 4, 5 and 6 and the 1774-problem suite unchanged, row for row. The answers are complex-valued along the whole line and are exact at five points each compared as complex numbers, with a negative `d` among the pins. Part of #718. Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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arcsin(u) holds the radicand 1 - u^2 without writing it: its derivative is u'/sqrt(1 - u^2). Beside arcsin(c x), the factors d + c d x and f - c f x multiply to d f (1 - c^2 x^2), a multiple of that radicand. The rule that writes such a multiple over the radicand (#1487, #1488) took its radicands only from logarithms, so the pair ran out of time. The radicand of an inverse sine or cosine of a linear is now one of the radicands it reads. Two factors whose product is the radicand itself, sqrt(1 + c x) sqrt(1 - c x), are written as its one root too. Rubi's 5.1.4, 5.1.5, 5.2.4 and 5.2.5, all 504 problems that count: 405 to 463, no row lost, 77 timeouts to 19. Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura Co-authored-by: Claude Opus 5.5 <noreply@anthropic.com>
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(a + b arsinh(c x)) sqrt(d + i c d x) sqrt(f - i c f x)ran out of time. The two linear factors multiply tod f (1 + c^2 x^2), a constant multiple of the radicand thatarsinh(c x) = ln(c x + sqrt(c^2 x^2 + 1))holds. That is the case #1487 answers, spelled as two factors, and Rubi's 7.1.4 writes 72 of its 663 rows this way.L1^p L2^qwithp - qwhole is nowL1^(p - q) L1^q L2^q, andL1^q L2^qislambda^q M^qfor a wholeqandK^k M^(k/2)forq = k/2, withK = sqrt(L1) sqrt(L2)/sqrt(M)carried as a symbol and written in front of the answer, constant wherever it is defined, as #1487's is. It is tried only where no single base is a multiple of the radicand, over the factors of both sides of the bar.Measured
(d + i c d x)^(5/2)(a + b arsinh(c x)) sqrt(f - i c f x)The integrands are complex-valued along the whole line. Four of the shapes are exact at five points each compared as complex numbers, with a negative
damong the pins, and the tests use numeric coefficients as well as symbolic ones. Still declined: the pair wholly below the bar,(a + b arsinh(c x))/((d + i c d x)^(5/2)(f - i c f x)^(3/2)), where the rewritten integral is the slow part.Part of #718.
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