AngouriMath
SolveANegativePowerOfXOverAWholePowerOfABinomialByRaisingThePowerOfX(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
The reduction of SolveANegativePowerOfXBesideARootOfABinomialByRaisingThePowerOfX(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) for x^m B^p alone with a whole p <= -2 and a symbol in B : Rubi's
1.1.2.21/(x^6 (a + b x^2)^10) is three steps from 1/(a + b x^2)^10 .
1.1.2.2
Remarks
The partial fractions take the binomial apart over its symbolic roots, and past the
twenty-fourth degree below the bar they ran out the budget:1/(x^4 (a + b x^2)^10) took a second, 1/(x^6 (a + b x^2)^10) and 1/(x^7 (a + b x^2)^10) more than
twenty, and1/(x^5 (a + b x^2)^10) about two, for 67,808 characters where this gives
1,860. Asked before the reduction over a power of a symbolic quadratic, which reads x as
a linear beside it, and past the twenty-fourth degree only, so that what the partial
fractions answered in time is answered as before.
https://github.com/asc-community/AngouriMath/issues/233
twenty-fourth degree below the bar they ran out the budget:
twenty, and
1,860. Asked before the reduction over a power of a symbolic quadratic, which reads x as
a linear beside it, and past the twenty-fourth degree only, so that what the partial
fractions answered in time is answered as before.
https://github.com/asc-community/AngouriMath/issues/233
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