AngouriMath
SolveAHalfOddPowerOfALinearPlusTheRootOfAQuadraticByEulersSubstitution(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A rational function of x and R = sqrt(Q) beside a half-odd power of
L + R , for a quadratic Q and a linear L with Q - L^2 = c (x - r)^2 ,
in Euler'su = (R - L)/(x - r) : x - r = 2 l u/D , R = L + u (x - r) and
L + R = 2 c l/D , for l = L(r) , D = c - 2 b u - u^2 and b the slope
ofL , so that the integrand is rational in u beside a half-odd power of the
quadraticD . Bondarenko's sqrt(1 + sqrt(x) + sqrt(1 + 2x + 2 sqrt(x))) is
8 sqrt(2) u (1 + u^2)/D^(7/2) under x = t^2 .
in Euler's
of
quadratic
Remarks
With s = x - r , R^2 = L^2 + c s^2 and R = L + u s give
s (c - 2 b u - u^2) = 2 l u , (L + R)(R - L) = c s^2 gives L + R = c s/u ,
anddx = 2 l (c + u^2)/D^2 du . The power (L + R)^(k/2) is asked as
(2 c l)^(k/2)/D^(k/2) , which it is up to a sign, and the answer in u is
multiplied by the quotient of the two, whose square is one: it holds on every branch.
Declined before, since no substitution any rule tries makes the nested root rational.
The answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233
and
multiplied by the quotient of the two, whose square is one: it holds on every branch.
Declined before, since no substitution any rule tries makes the nested root rational.
The answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233
Angouri © 2019-2023 · Project's repo · Site's repo · Octicons · Transparency · 4862 pages online