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The other four trigonometric functions are written as sine and cosine before the complement of a substitution is, and never where that leaves an odd power of the complement under a root - #1312
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… before the complement of a substitution is, and never where that leaves an odd power of the complement under a root Charlwood's `arctan(sqrt(sec(x) - 1)) sin(x)` had no antiderivative. By parts it leaves `sin(x)/sqrt(sec(x) - 1)`, and under `u = cos(x)` -- the complement of the written sine, a candidate since #1307 -- that is `-1/sqrt(1/u - 1)`, a root of a quotient of two linears, which is the linear-radical substitution's since #1308. With the secant left standing it kept its `x` and the candidate was refused. The pass that writes the complement's even powers in `u` writes the secant, cosecant, tangent and cotangent of the argument as quotients of the sine and cosine first now, and substitutes `u` again where that has written the function anew. And refuses the rewriting where it leaves an odd power of the complement under a root. The first draft did not, and Timofeev's `(sec(x)^2 - 3 sqrt(4 sec(x)^2 + 5 tan(x)^2) tan(x))/(sin(x)^2 (...)^(3/2))` under `u = sin(x)` became a root of a quotient by `cos(x)^2`, which the simplification the substitution runs takes as `.../cos(x)` -- its value where the cosine is positive and its negative elsewhere -- and the integrand was answered wrongly on half the line, measured at two of six sampled points and pinned as declined-or-right on both sides. An even power of the complement under a root is `1 - u^2` by then and is no such hazard. One answer changes shape: `sin(x) sin(2x)` over `cos(x)` is `2 sin(x)^2` once the simplification has written the double angle out, and the second pass reads the sine that wrote as `u` again, so it is `2 sin(x)^3/3` now rather than product-to-sum's cosines of `x + 2x`; the test that pinned the old shape pins the new one. ## Measured Every answer differentiated back on both sides of zero. Rubi corpus, 463-problem sample, on the same evening: master at #1309 401/463, 0 wrong, 0 timeouts, 100 s with this 402/463, 0 wrong, 0 timeouts, 99 s One more, nothing lost: Charlwood 32. The five test suites are green. Part of #718. Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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* Breaking-changes entries for #1828 to #1832 and #1374 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura * And for #1368, #1611 and #1517, each measured on 2.5.0 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura * And for #1289, #1312, #1315, #1377, #1380 and #1371, each measured on 2.5.0 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura --------- Co-authored-by: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
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Charlwood's
arctan(sqrt(sec(x) - 1)) sin(x)had no antiderivative. By parts it leavessin(x)/sqrt(sec(x) - 1), and underu = cos(x)-- the complement of the writtensine, a candidate since #1307 -- that is
-1/sqrt(1/u - 1), a root of a quotient oftwo linears, which is the linear-radical substitution's since #1308. With the
secant left standing it kept its
xand the candidate was refused. The passthat writes the complement's even powers in
uwrites the secant, cosecant,tangent and cotangent of the argument as quotients of the sine and cosine
first now, and substitutes
uagain where that has written the function anew.And refuses the rewriting where it leaves an odd power of the complement under
a root. The first draft did not, and Timofeev's
(sec(x)^2 - 3 sqrt(4 sec(x)^2 + 5 tan(x)^2) tan(x))/(sin(x)^2 (...)^(3/2))underu = sin(x)became a root of a quotient bycos(x)^2, which the simplification thesubstitution runs takes as
.../cos(x)-- its value where the cosine is positiveand its negative elsewhere -- and the integrand was answered wrongly on half the
line, measured at two of six sampled points and pinned as declined-or-right on
both sides. An even power of the complement under a root is
1 - u^2by then andis no such hazard.
One answer changes shape:
sin(x) sin(2x)overcos(x)is2 sin(x)^2once thesimplification has written the double angle out, and the second pass reads the
sine that wrote as
uagain, so it is2 sin(x)^3/3now rather thanproduct-to-sum's cosines of
x + 2x; the test that pinned the old shape pinsthe new one.
Measured
Every answer differentiated back on both sides of zero.
Rubi corpus, 463-problem sample, on the same evening:
One more, nothing lost: Charlwood 32.
The five test suites are green.
Part of #718.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura