AngouriMath
SolveAsTheArctangentOfAPowerOfXOverABinomial(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
The derivative of w = x^k/L , for a binomial L = x^p + l , over a multiple of
L^2 plus a power of x, c L^2 + d x^(2k) : the integrand is
(m/c) w'/(1 + (d/c) w^2) , an arctangent of w or a hyperbolic one. Rubi's
1.3.2(e - 4 f x^3)/(e^2 + 4 d f x^2 + 4 e f x^3 + 4 f^2 x^6) is
arctan(2 sqrt(d f) x/(e + 2 f x^3))/(2 sqrt(d f)) , and its rows in symbolic
powers of x are the same inw = x^(m + 1)/(e + 2 f x^n) .
1.3.2
powers of x are the same in
Remarks
The denominator is read as four terms, each a constant times a power of x: one is
d x^(2k) , and the other three are a + b x^p + c x^(2p) with
b^2 = 4 a c , which is c L^2 for L = x^p + b/(2c) . The numerator is
a constant multiple ofk x^(k - 1) L - p x^(k + p - 1) , which is w' L^2 ;
the constant is read atx = 1 , and the answer is differentiated back. Past a
quartic these were declined: partial fractions over a sextic in x, or over a sum of
symbolic powers of it, are out of reach, and nothing read the quotient.
https://github.com/asc-community/AngouriMath/issues/718
a constant multiple of
the constant is read at
quartic these were declined: partial fractions over a sextic in x, or over a sum of
symbolic powers of it, are out of reach, and nothing read the quotient.
https://github.com/asc-community/AngouriMath/issues/718
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