A perfect square in a power of the variable is read as one

[Rafael-SOWNet]

sqrt(a^2 + 2 a b x^2 + b^2 x^4) sqrt(c + e x + d x^2) was left unevaluated. The radicand is (a + b x^2)^2, so its root is |a + b x^2|, but the rule that reads a root of a perfect square (#1477) read only a quadratic in x. It now reads the same quadratic in a power of x, a x^(2k) + b x^k + c. It puts sgn(x^k + h) in front of the integral, as it does for k = 1, and leaves the sign off where k is even and h is a positive number. The shift h is simplified where it holds a symbol, so 2 a b/(2 b^2) becomes a/b, for k = 1 as well.

Measured

	before	after
Rubi 1.2.2.2, 1.2.2.4, 1.2.2.7 and 1.2.3.2: the 574 problems that count with a perfect square written in symbols	182	388, no row lost, 19 timeouts → 9, 732 s → 478 s
Rubi family 1	160/228	166/228
Rubi families 4–7, the 7.1 files and the 1774-problem suite	—	unchanged, row for row
unit suite	12,709	12,712 (three cases added), 0 failed, run to completion
allocation gate	—	PASSED on all 19 gated benchmarks

Two rows went from a decline to a timeout: (d x)^(23/2)/(a^2 + 2 a b x^2 + b^2 x^4)^(5/2) and its 21/2 neighbour. Both become a high power of x over the fifth power of x^2 + a/b.

Still declined: the same square in a symbolic power x^n, and whole powers of the square, such as 1/(a^2 + 2 a b x^2 + b^2 x^4)^2. The whole powers are 84 of the remaining rows, and I'm taking them next.

Part of #718.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura