A root of a quadratic in the tangent is rotated until the quadratic has no linear term - #1475
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…as no linear term `1/sqrt(1 + 2 tan(x) + 3 tan(x)^2)` was left as written, though `1/((1 + t^2) sqrt(A + C t^2))` -- what the tangent substitution leaves of it once the linear term is gone -- was already answered. Under `x = y + arctan(m)` the tangent becomes `(tan(y) + m)/(1 - m tan(y))`, the Möbius map that preserves `1 + tan^2`, and the quadratic's linear coefficient becomes `b(1 - m^2) + 2(c - a)m`: zero for a root of `b m^2 + 2(a - c)m - b`, whose discriminant `4((a - c)^2 + b^2)` is never negative, so the rotation is always real. The two roots are the quadratic form's perpendicular directions and either will do. The radicand is assembled -- `A = a + bm + cm^2`, `C = am^2 - bm + c` -- and not substituted into: writing the quotient inside the root leaves a nesting that nothing downstream reduces, which is how the first version of this rule turned a decline into a ninety-second search. The modulus that `sqrt(N/(1 - mS)^2)` leaves comes out in front as `sgn(1 - m tan(y))`, constant between its zeros. A numeric rotation only. With symbolic coefficients `m` is a nested surd, the rotated quadratic is written in it, and what the rational integrator is handed is a quotient over that field: two minutes and no answer, where the integrand is declined in under a second unrotated. That bound is on the work rather than on the mathematics, and a test pins that the symbolic shape is still declined quickly. Family 4 of the Rubi suite: 290/422 either way, 0 wrong -- its rows of this family are symbolic; family 6 377/417 and the 1774-problem suite 1707 unchanged. Suite 12655 passed; allocation gate passed on all 19 gated benchmarks. Each new row differentiates back at five points. 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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1/sqrt(1 + 2 tan(x) + 3 tan(x)^2)was left as written — though1/((1 + t^2) sqrt(A + C t^2)), which is what the tangent substitution leaves of it once the linear term is gone, was already answered.Under
x = y + arctan(m)the tangent becomes(tan(y) + m)/(1 - m tan(y)): the Möbius map that preserves1 + tan^2, the factor the tangent substitution'sdxbrings. The quadratic's linear coefficient becomesb(1 - m^2) + 2(c - a)m, zero for a root ofb m^2 + 2(a - c)m - b, whose discriminant4((a - c)^2 + b^2)is never negative — the two roots are the quadratic form's perpendicular directions, and either will do.Two things the measurements decided:
A = a + bm + cm^2,C = am^2 - bm + c. Writing the Möbius quotient inside the root leaves a nesting that neitherSimplifynor the chain reduces, so the substitution never sees the rotated quadratic: the first version of this rule turned a 787 ms decline into a 90 s search. The modulus thatsqrt(N/(1 - mS)^2)leaves comes out in front assgn(1 - m tan(y)), constant between its zeros as every such sign in these rules is.a, b, cthe rotationmis a nested surd, the rotated quadratic is written in it, and the rational integrator is handed a quotient over that field — two minutes and no answer, against a 0.8 s decline unrotated. The bound is on the work, not on the mathematics; a test pins that the symbolic shape is still declined quickly.Measured (final build):
int dx/sqrt(1 + 2 tan x + 3 tan^2 x)and its kin.bench.sh: PASSED on all 19 gated benchmarks.tan(x)^2/(2 + tan(x) + tan(x)^2)^(3/2).The symbolic case wants a rational integrator that stays in a quadratic field rather than expanding into it; that is the next piece of this, and it is not in here.
Part of #718.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura