Skip to content

Integrate a power of x beside a power of a logarithm of a binomial in that binomial - #2093

Merged
Rafael-SOWNet merged 1 commit into
masterfrom
a-power-of-x-beside-a-function-of-a-binomial
Oct 23, 2026
Merged

Rafael-SOWNet merged 1 commit into
masterfrom
a-power-of-x-beside-a-function-of-a-binomial

Conversation

@Rafael-SOWNet

Copy link
Copy Markdown
Member

Integrates x^k beside a power of a logarithm of one binomial a + b x^q by writing it in t = a + b x^q. Part of #2075.

The rule. Where N = (k + 1)/q is a whole number above 0, x^k dx is (t - a)^(N - 1) dt/(q b^N). That holds on every branch, since x^(k + 1) = (x^q)^N and x^(k + 1 - q) = (x^q)^(N - 1) are whole powers of x^q. The integrand in t is then a polynomial beside the logarithm's power, which the rules answer. For example, (a + b ln(c (d + e/sqrt(x))))^p/x^2 is -2/e^2 times the integral of (t - d) (a + b ln(c t))^p.

Only a power of a logarithm is taken. The first version took an exponential of a binomial too, and answered f^(a + b/x)/x^4 as e^(ln f (a + b/x)), where the exponential rules write it in its own base, f^(a + b/x). A test that pins that form failed.

Measured on Rubi's suite, with the harness that reads GAMMA, against 68211405, the whole corpus each side. This branch sits on 2be40bad, which adds #2088 to that base.

68211405 this
solved 48,907 of 49,052 48,926
wrong 0 0
rows lost 0

The gains are 3.4's 14 rows, 3.4:696-750. The other 5 are #2088's: 6.1.1:205-207 and 6.1.3:98/100. Families 2 and 3 were run again after the narrowing, 2,274 rows: the same 14 gains, 0 wrong, 0 lost. Allocation per stream is 1.0002, 1.0001 and 0.9946 of the base's, and time 1.015, 1.026 and 0.922. One row's allocation grew in the run, 6.1.7:196, which is the same alone on both builds (1.079 GB), as it was for #2088 and #2090.

Three integrals differentiate back in the tests. The suite passes (16,955), and so do the MCP tests (329).

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

… that binomial

x^k beside a power that is not whole of a logarithm of one binomial a + b x^q is written in
t = a + b x^q where N = (k + 1)/q is a whole number above 0. Then x^k dx is (t - a)^(N - 1) dt/(q b^N)
on every branch, since x^(k + 1) = (x^q)^N and x^(k + 1 - q) = (x^q)^(N - 1) are whole powers of x^q,
and the integrand in t is a polynomial beside the logarithm's power, which the rules answer:
(a + b ln(c (d + e/sqrt(x))))^p/x^2 is -2/e^2 times the integral of (t - d) (a + b ln(c t))^p. An
exponential of a binomial is left to the exponential rules, which write it in its own base.

Part of #2075.

Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
@Rafael-SOWNet
Rafael-SOWNet force-pushed the a-power-of-x-beside-a-function-of-a-binomial branch from a0dd149 to 0a3790e Compare October 23, 2026 14:49
@Rafael-SOWNet
Rafael-SOWNet merged commit e2cf38a into master Oct 23, 2026
34 checks passed
@github-project-automation github-project-automation Bot moved this from Todo to Done in AngouriMath Oct 23, 2026
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment

Labels

None yet

Projects

Status: Done

Development

Successfully merging this pull request may close these issues.

1 participant