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Read hyperbolic integrands through cosh^2 = (1 + i sinh)(1 - i sinh), a symbolic power's own variable, and a phase - #2018

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Part of #233.

Five shapes of Rubi's 6.1.1 and 6.7.1, each answered through an identity of the hyperbolic functions that the substitutions did not use. One test class per shape.

  1. (g + f x)^m cosh(c + d x)^k/(a + i a sinh(c + d x)) and its sinh^k rows. Since cosh^2 = (1 + i sinh)(1 - i sinh), the numerator divides by the denominator, leaving a polynomial and a constant over it.
  2. 1/(sech(x) ± i tanh(x))^4 is cosh(x)^4/(1 ± i sinh(x))^4, which the same identity cancels to (1 ∓ i sinh)^2/(1 ± i sinh)^2. It applies only to cosh^(2m) over (1 ± i sinh)^(2m) and the other way up; cancelling where something else is left made other answers longer.
  3. A symbolic power of sinh, cosh, sech or csch beside an odd power of the other, under u = that function. A power of csch stays a power of u, since it is not sinh^(-n) where the sine is negative.
  4. (cosh(x) coth(x))^(3/2) and (csch(x) + sinh(x))^(3/2) are both (cosh(x)^2/sinh(x))^(3/2), which is cosh^3 sinh^(-3/2) times a constant on each side of zero.
  5. Half-odd powers of S = b cosh(x) + c sinh(x) ± sqrt(b^2 - c^2). With T = b sinh + c cosh, d/dx (T S^(n-1)) = n S^n - a (2n - 1) S^(n-1) steps the power to ±1/2. The branch first also stepped whole powers up to T/(a S). On today's master the exponential substitution writes 1/S^2, 1/S^3 and 1/S^4 in 182, 272 and 416 characters, where the recurrence gave 350, 539 and 728, so whole powers are now left to it.

Measured.

  • Chain against master 4877ed2, the branch's base, with the harness at its current revision on both arms:
    • pocket of 1,756 rows: 1,686 → 1,723 solved, 0 wrong;
    • samples: unchanged or better;
    • the 43 rows that moved, re-run alone: 4 → 42 solved.
  • Re-measured with the probe on master f7f2a7a, after the rebase. 14 of those 43 rows are new there:
    • twelve rows of (e + f x)^k cosh^m/(a + i a sinh) go from past 20 s to 0.1–0.9 s;
    • (csch(x) + sinh(x))^(3/2) and ^(5/2) go from declined to answered.
  • The other 29 answer on both builds, most of them much cheaper here:
input master f7f2a7a this
sech(a+b*x)^n*tanh(a+b*x)^3 11.2 s, 9.12 GB, 180 characters 71 ms, 0.02 GB, 183
cosh(a+b*x)^3*sinh(a+b*x)^n 1.16 GB, 326 0.02 GB, 149
1/(sech(x)+i*tanh(x))^4 13.5 s, 10.5 GB, 551 0.08 GB, 241
(sech(x)+i*tanh(x))^4 2,916 251
(g+f*x)*cosh(c+d*x)^2/(a+i*a*sinh(c+d*x)) 26.9 s, 27.5 GB, 740 0.07 GB, 366
1/sqrt(b*cosh(x)+c*sinh(x)+sqrt(b^2-c^2)) 0.42 GB, 467 0.004 GB, 409

Every answer was checked by differentiating back at the probe's points; the test classes pin each shape the same way. 426 tests pass locally across the five classes, the hyperbolic tests and the corpus gate.

The BREAKING-CHANGES.md entry has 2.5.0's column measured. Its "Now" column and its comparison with the unreleased master are re-measured on this build and on f7f2a7a.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

… a symbolic power's own variable, and a phase

Five shapes of Rubi's 6.1.1 and 6.7.1, each through an identity the
substitutions did not use:

- (g + f x)^m cosh(c + d x)^k/(a + i a sinh(c + d x)) and its sinh^k rows:
  cosh^2 = (1 + i sinh)(1 - i sinh), so the numerator divides by the
  denominator, leaving a polynomial and a constant over it.
- 1/(sech(x) ± i tanh(x))^4 is cosh(x)^4/(1 ± i sinh(x))^4, which the same
  identity cancels to (1 ∓ i sinh)^2/(1 ± i sinh)^2; cosh^(2m) over
  (1 ± i sinh)^(2m) and the other way up only.
- A symbolic power of sinh, cosh, sech or csch beside an odd power of the
  other, under u = sinh, cosh, sech or csch; a power of csch stays a power of
  u, since it is not sinh^(-n) where the sine is negative.
- (cosh(x) coth(x))^(3/2) and (csch(x) + sinh(x))^(3/2) are
  (cosh(x)^2/sinh(x))^(3/2), cosh^3 sinh^(-3/2) times a constant on each side
  of zero.
- Half-odd powers of S = b cosh(x) + c sinh(x) ± sqrt(b^2 - c^2): with
  T = b sinh + c cosh, d/dx (T S^(n-1)) = n S^n - a (2n - 1) S^(n-1) steps
  the power to ±1/2. The whole powers are left to the exponential
  substitution, which writes 1/S^2, 1/S^3 and 1/S^4 in 182, 272 and 416
  characters, where the recurrence gave 350, 539 and 728.

Chain against master 4877ed2: pocket of 1,756 rows, 1,686 -> 1,723 solved,
0 wrong; samples unchanged or better; the 43 rows that moved, alone, 4 -> 42
solved. Re-measured on master f7f2a7a after the rebase: 14 of the 43 are
new there (12 of (e + f x)^k cosh^m/(a + i a sinh) past 20 s -> 0.1-0.9 s,
(csch(x) + sinh(x))^(3/2) and ^(5/2) declined -> answered), and the other
29 answer on both, most much cheaper here: sech(a + b x)^n tanh^3 9.12 ->
0.02 GB, 1/(sech(x) + i tanh(x))^4 10.5 -> 0.07 GB and 551 -> 241
characters.

Part of #233.

Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
@Rafael-SOWNet Rafael-SOWNet added this to the 2.6.0 milestone Oct 21, 2026
@Rafael-SOWNet
Rafael-SOWNet merged commit 6e16aae into master Oct 21, 2026
38 of 40 checks passed
@github-project-automation github-project-automation Bot moved this from Todo to Done in AngouriMath Oct 21, 2026
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