AngouriMath
SolveABinomialInTheThirdCaseWithASymbolicExponent(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
A binomial differential x^m (a + b x^n)^p with a symbol in its exponents, in
Chebyshev's third case at a whole(m + 1)/n + p + 1 = -k of at most 0: Rubi's
1/(a + b x^n)^((1 + 2 n)/n) .
Chebyshev's third case at a whole
Remarks
At k = 0 the integral is x^(m + 1) (a + b x^n)^(p + 1)/(a (m + 1)) , whose
derivative is the integrand there. Below it Rubi's reduction raisesp by one,
andk steps of it reach the first. SolveABinomialDifferential(AngouriMath.Entity,AngouriMath.Entity.Variable) reads
the three cases with numbers for the exponents; with a symbol, a power of x beside the
binomial atk = 0 is SolveAsTheDerivativeOfAProductOfPowers(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)'s, which
leaves a power on its own and everyk past 0, and those were declined: Rubi's
1.1.3.2:3301-3304. The answer is differentiated back at sampled points before it is
returned.
https://github.com/asc-community/AngouriMath/issues/718
derivative is the integrand there. Below it Rubi's reduction raises
J(p) = -x^(m + 1) (a + b x^n)^(p + 1)/(a n (p + 1)) + (m + 1 + n (p + 1))/(a n (p + 1)) J(p + 1)and
the three cases with numbers for the exponents; with a symbol, a power of x beside the
binomial at
leaves a power on its own and every
1.1.3.2:3301-3304. The answer is differentiated back at sampled points before it is
returned.
https://github.com/asc-community/AngouriMath/issues/718
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