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SolveTwoLinearPowersWhoseExponentsSumToAWholeNumber​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

 Method (no overloads)

Summary

Powers of two linears whose exponents sum to a whole number -k, k >= 2,
beside a polynomial of degree at most k - 2: (a + b x)^m (c + d x)^(-3 - m),
x^(n - 4)/(a + b x)^n, (e + f x)^2 (a + b x)^m (c + d x)^(-4 - m). Under
t = L1/L2 each is t^A beside a polynomial in t, so the answer is
the two powers as written times a polynomial in x.

Remarks

With L1 = a1 + b1 x, L2 = a2 + b2 x and D = a1 b2 - a2 b1, the
substitution makes x = (a1 - a2 t)/(b2 t - b1), L2 = D/(b2 t - b1) and
dx = -D dt/(b2 t - b1)^2, so L1^A L2^B P(x) dx is, up to a factor constant
on each interval, -D^(1 - k) t^A S(t) dt for the polynomial
S(t) = sum_j p_j (a1 - a2 t)^j (b2 t - b1)^(k - 2 - j). Its antiderivative is
sum_i s_i t^(A + i + 1)/(A + i + 1), and since t^(A + i + 1) is
t^A t^(i + 1) for a whole i + 1, the constant factor and t^A come
back as the two powers as written: the answer is
L1^A L2^B sum_i -D^(1 - k) s_i L1^(i + 1) L2^(k - 1 - i)/(A + i + 1), exact
wherever the integrand is defined. The generic case, as the integrator answers
everywhere: A + i + 1 is taken to be nonzero.
 Only the sum -2 was answered, as the derivative of the product of the two powers
raised by one; Rubi's (a + b x)^m (c + d x)^n files hold the rest, and under
u = sin(y) so do the powers of 1 ± sin(y) beside one another. A closed
rule, so at any depth. https://github.com/asc-community/AngouriMath/issues/718

























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