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A power of a constant multiple of the radicand a logarithm holds is written over it - #1487
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…ritten over it `acosh(a x)^2/sqrt(1 - a^2 x^2)` and `x (a + b acosh(c x))/(d - c^2 d x^2)^3` ran out of time. `acosh(c x)` is `ln(c x + sqrt(c^2 x^2 - 1))`, and the substitution `u = acosh(c x)` wants the root its derivative carries, `sqrt(c^2 x^2 - 1)`; beside it the integrand holds a constant multiple of that radicand, `d - c^2 d x^2 = -d (c^2 x^2 - 1)`, which nothing read as one. A whole power of the multiple is now the product of the powers, and a root `(lambda M)^(k/2)` is `K^k M^(k/2)` with `K = sqrt(lambda M)/sqrt(M)`: plus or minus `i`, or `sqrt(lambda)`, constant on every interval where it is defined, so it is carried through the integration as a symbol and written out in front of the answer, the way Rubi writes it. `arcosh(2 x)^2/sqrt(1 - 4 x^2)` is real on `(-1/2, 1/2)`, where `arcosh(2x)` is `i arccos(2x)`, and the answer is checked there too. Three things the measurement found. The polynomial reader does not take the monic spelling the normalisation gives a power with a symbolic leading coefficient -- `d^0 (-1)^(-1) c^(-2) + x^2` -- nor its simplification, which is `x^2 - c^(-2) provided not d = 0`; it is read bare. The ratio of leading coefficients is simplified where it holds symbols, `-c^2 d/c^2` being `-d`. And the rule answers the question asked only: below it the substitutions write logarithms of their own, and there it found a genuine multiple in a by-parts remainder, rewrote it and asked a whole search again that was going to decline -- `atan(c x/sqrt(a - c^2 x^2))^2/sqrt(d - c^2 d x^2/a)` took thirty-one seconds to decline where master takes seven. The inverse cosine is in the integrand as asked, and a round of parts differentiates it away, so nothing below needs it. Rubi's 7.2.4 and 7.2.5, all 277 problems that count: 127 to 255, no row lost, 121 timeouts to 18. Family 7 223 to 235 of 270; families 1, 4, 5 and 6 and the 1774-problem suite unchanged, row for row. 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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… holds are written as that radicand (#1488) `(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
Rafael-SOWNet
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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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acosh(a x)^2/sqrt(1 - a^2 x^2)andx (a + b acosh(c x))/(d - c^2 d x^2)^3ran out of time.acosh(c x)isln(c x + sqrt(c^2 x^2 - 1)), and the substitutionu = acosh(c x)wants the root its derivative carries,sqrt(c^2 x^2 - 1); beside it the integrand holds a constant multiple of that radicand,d - c^2 d x^2 = -d (c^2 x^2 - 1), which nothing read as one.A whole power of the multiple is now the product of the powers, and a root
(lambda M)^(k/2)isK^k M^(k/2)withK = sqrt(lambda M)/sqrt(M)— plus or minusi, orsqrt(lambda), constant on every interval where it is defined — so it is carried through the integration as a symbol and written out in front of the answer, the way Rubi writes it (sqrt(-1 + c x) sqrt(1 + c x)/sqrt(d - c^2 d x^2)and kin).arcosh(2 x)^2/sqrt(1 - 4 x^2)is real on(-1/2, 1/2), wherearcosh(2x)isi arccos(2x); the existingDeclinedOrRighttest checks it there, and a new theory requires it answered.Three things the measurement found:
d^0 (-1)^(-1) c^(-2) + x^2— nor its simplification,x^2 - c^(-2) provided not d = 0. It is read bare.-c^2 d/c^2written into a whole power cost the integrand its time.atan(c x/sqrt(a - c^2 x^2))^2/sqrt(d - c^2 d x^2/a)declining in 31 s where master takes 7. Two calls of the rule took 12.5 s each: below the top the substitutions write logarithms of their own, and in a by-parts remainder the rule found a genuine multiple, rewrote it and asked a whole search again that was going to decline. A depth-two scope and a numeric screen for the multiples were each measured and neither was the cause; answering the question asked only is, and it costs nothing here — the inverse cosine is in the integrand as asked, and a round of parts differentiates it away. That decline is 6.7 s now.Measured
Of the 7.2 answers, 105 are for integrands complex-valued along the whole real line and are checked there as complex numbers; seven of the shapes were also checked independently at five points each, exact, including a negative
d.Part of #718.
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