A whole power of a perfect square in a power of the variable is written as a power of its root

[Rafael-SOWNet]

1/(a^2 + 2 a b x^2 + b^2 x^4)^2 was left unevaluated. The quartic is b^2 (x^2 + a/b)^2, so a whole power of it is (b^2)^n (x^2 + a/b)^(2n), exactly and with no sign. The rational integrator reads a power of x^2 + a/b where it declined the quartic. This extends #1491, which read half-odd powers of the same squares, to whole ones.

It applies at the question asked, to a square with a symbol in it, where x is below the bar. Two more restrictions each come from a measurement:

• Not in a polynomial. x (a^2 + 2 a b x^2 + b^2 x^4)^2 is answered as a polynomial already, and written as a power of x^2 + a/b it only changed form.
• Only a square written out as a sum. The hyperbolic-tangent substitution asks its integrand as the same question, and for Rubi 6.5.7's coth(c + d x)^2/(a + b sech(c + d x)^2)^2 that integrand holds ((a + b - b u^2)^2)^(-1), a square already written as one. Expanded and rewritten, it went from 842 ms to over 60 s.

Two answers that came out in the expanded square before now come out in x^2 + a/b; they're in the changelog with their 2.5.0 values.

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	388	409, no row lost, 9 timeouts either way
Rubi families 1, 4–7, the 7.1 files and the 1774-problem suite	—	unchanged, row for row
unit suite	12,724	12,727 (three cases added), 0 failed, run to completion
allocation gate	—	PASSED on all 19 gated benchmarks

Part of #718.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura