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SolveByScalingTheVariable​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

 Method (no overloads)

Summary

An integrand carrying one symbolic parameter, scaled by it — x = c t — so that
what is left to integrate has the variable alone in it.

Remarks

1/(a^3 + x^3) had no antiderivative, and neither did 1/(a^4 - x^4),
1/(a^5 + x^5) or any of the family 1/(x^k (a^n ± x^n)) — while
1/(8 + x^3) and 1/(16 - x^4) are answered at once. The parameter is the
whole difference: the rational rules read a denominator as a polynomial over the
rationals
, and a^3 is not a rational coefficient, so the factoring that
answers x^3 + 8 has nothing to work with on x^3 + a^3.
Scaling puts the parameter where it does no harm. With x = c t and
dx = c dt, a homogeneous integrand becomes c^k times a function of
t alone: 1/(a^3 + x^3) becomes a^(-2) / (1 + t^3), whose
denominator has integer coefficients again. The parameter comes back out as a constant
factor and the answer is read at t = x/c.
Homogeneity is checked rather than assumed, and it is what makes this
terminate. The scaled integrand has to be exactly c^k h(t) for a whole
k, which is verified by dividing it out; only h is handed on, so the
sub-problem has one variable and cannot be scaled again. Without that check the rule
would hand on something still carrying c and scale it once more at every level.
What it is not defined at.c = 0 is not a scaling, and the answer this
produces is undefined there rather than wrong — a^(-2) G(x/a) has no value at
a = 0, which is the honest report for a substitution that does not exist. The
integrand itself is a different function there (1/(a^3 + x^3) is 1/x^3),
and is answered on its own if asked that way.
https://github.com/asc-community/AngouriMath/issues/718

























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