AngouriMath
SolveByDividingAPolynomialByAPowerOfALinearOfTheLogarithm(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A polynomial P of degree m times a function G of one logarithm of a
quotient of linears, over the powerL^k of a linear proportional to one of them,
withm <= k - 2 : P written in powers of L at its root,
P = p_0 + p_1 L + ... + p_m L^m , makes the integrand sum p_j G/L^(k - j) exactly, and each term is a power of one of the logarithm's linears beside a function of
it, which SolveByTheQuotientOfTheLogarithmsLinears(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) answers under the
quotient. Rubi's 7.3.4x (a + b atanh(c x))^2/(d + c d x)^3 is
((a + b atanh(c x))^2/(d + c d x)^2 - d (a + b atanh(c x))^2/(d + c d x)^3)/(c d) .
https://github.com/asc-community/AngouriMath/issues/233
quotient of linears, over the power
with
it, which SolveByTheQuotientOfTheLogarithmsLinears(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) answers under the
quotient. Rubi's 7.3.4
https://github.com/asc-community/AngouriMath/issues/233
Remarks
The rule for a rational function times a power of a logarithm took it by parts, and the
remainder ran past the budget. Not where a term would be overL alone, a
dilogarithm nothing here answers, and not whereG is the logarithm's first power,
which that rule answers in one step and keeps the answer it gives.
remainder ran past the budget. Not where a term would be over
dilogarithm nothing here answers, and not where
which that rule answers in one step and keeps the answer it gives.
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