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A power of a multiple of a quadratic's derivative beside a power of the quadratic is a binomial (#1691)
Rubi's 1.2.1.2 (b d + 2 c d x)^m (a + b x + c x^2)^p, with a power that
is not whole, was declined or ran out of time: no substitution the
search tried saw that the linear is a multiple of the quadratic's
derivative. Under t = b d + 2 c d x the quadratic is
(t^2/d^2 - (b^2 - 4 a c))/(4 c), exactly, since the substitution is
affine, and the integrand is a binomial in t.
The quadratic is written in t from its coefficients rather than by
substituting, which would leave a linear term zero as a value and not
as written; and each constant of the binomial that is a compound of
the coefficients is named by a symbol of its own and written back into
the answer. The rules answer t^(-7) sqrt(c t^2 + k) in 63 ms, and
with a - b^2/(4 c) for k they ran past twenty seconds. A quadratic with
no linear term is a binomial already and is left alone, else the rule
would ask its own question with t scaled.
Part of #718.
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
Co-authored-by: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
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