AngouriMath
SolveByTheQuotientOfTheLogarithmsLinearsOntoTheExponentialIntegral(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
SolveByTheQuotientOfTheLogarithmsLinears(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) where
the logarithm's factor is a negative whole power and the integrand int is a power
oft and a polynomial beside it, t^q (b - d t)^j (A + B ln(K t^n))^(-k) : the
powers of the linears sum to-2 - j , and SolveAPowerOverAPowerOfALogarithm(AngouriMath.Entity,AngouriMath.Entity.Variable) answers it onto the exponential integral, a power of t at a time. Rubi's
1/((a g + b g x)^2 (A + B ln(e (a + b x)/(c + d x)))^2) is
t^(-2) (A + B ln(e t))^(-2)/(g^2 D) .
the logarithm's factor is a negative whole power and the integrand in
of
powers of the linears sum to
Remarks
Before the polynomial term, which takes b^2 g^2 out of (a g + b g x)^2 and asks
1/((a/b + x)^2 (A + B ln(...))^2) again, and a level below that the substitution
search ran past the budget: the rule above, far down the chain, was not reached. Of Rubi's
3.2.1 those rows were from 1.3 s to 23 s, and four past the budget.
https://github.com/asc-community/AngouriMath/issues/233
search ran past the budget: the rule above, far down the chain, was not reached. Of Rubi's
3.2.1 those rows were from 1.3 s to 23 s, and four past the budget.
https://github.com/asc-community/AngouriMath/issues/233
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