AngouriMath
SolveAnEvenQuotientOverAPowerOfALinearInTheSquareAndABiquadratic(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A polynomial in x^2 over a power of a linear in x^2 and a biquadratic, with a
symbol in it:N(x^2)/(L(x^2)^k Q(x^2)) , L(w) = l (w - r) , split in
s = x^2 - r . Written in powers of s , the quotient is the one
SolveAnEvenPolynomialOverASymbolicBiquadratic(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) splits overw^k Q , and the
terms ins^(-j) go by the reduction
int dx/(x^2 + p)^j = x/(2 p (j - 1) (x^2 + p)^(j - 1)) + (2 j - 3)/(2 p (j - 1)) int dx/(x^2 + p)^(j - 1) ,
p = -r , down to atan(x/sqrt(p))/sqrt(p) , which holds for every complex
p but zero, as the biquadratic's own terms do.
symbol in it:
SolveAnEvenPolynomialOverASymbolicBiquadratic(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) splits over
terms in
Remarks
Under t = sqrt(c + d u) Rubi's sqrt(c + d u)/((a + b u)^n (1 + u^2)) , which the
tangent substitution makes ofsqrt(c + d tan(x))/(a + b tan(x))^n , is
t^2/((a d + b (t^2 - c))^n ((t^2 - c)^2 + d^2)) up to a constant. Split in t , the
sum of two squares went over its conjugates, and the answer to
sqrt(c + d tan(x)) (A + B tan(x) + C tan(x)^2)/(a + b tan(x))^3 ran to 3.5 million
characters. Only one linear inx^2 , to a power, beside one biquadratic: with
r zero it is SolveAnEvenPolynomialOverASymbolicBiquadratic(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)'s, and a root
r shared with the biquadratic declines.
https://github.com/asc-community/AngouriMath/issues/718
tangent substitution makes of
sum of two squares went over its conjugates, and the answer to
characters. Only one linear in
https://github.com/asc-community/AngouriMath/issues/718
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