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Read the derivative of a sum of exponentials over another off the integrand over the other's square (#2040)
Rubi's 6.7.1 x (b - a sinh(x))/(a + b sinh(x))^2 ran past the budget: the split
took the bracket's terms apart, the substitution in e^x answered each over the
square of a quadratic with a piecewise on its discriminant, and parts beside x
integrated that piecewise again. For a sum D of exponentials of x, the
integrand over D^2 is now read as (N/D)', N a constant plus a constant times
each exponential in D. The exponentials' span is closed under the derivative,
so N' D - N D' is linear in N's constants, and with each exponential a whole
power of the first, w, the identity read at six values of w is a linear
system. N and N + c D share a derivative over D; the solution with no
constant is tried first, which is -cosh(x)/(a + b sinh(x)). Beside x, by
parts, the integral of N/D is written as a constant times ln(D) where N is a
constant times D', and asked otherwise. The trigonometric twin is
SolveAsTheDerivativeOfASineRaisedFromNothingBesidePowersOfTheCosine's; there
are no hyperbolic nodes, since sinh is parsed into exponentials.
Part of #233.
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
Co-authored-by: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
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@@ -570,6 +570,24 @@ tanh keep the answers the exponential substitution gives them ([#233](https://gi
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| `"sin(tanh(a+b*x))".ToEntity().Integrate("x")` | `integral(...)` | `Si` and `Ci` of `1 - tanh(a + b x)` and `1 + tanh(a + b x)` |
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| `"tanh(x)/sqrt(a+b*tanh(x)^2+c*tanh(x)^4)".ToEntity().Integrate("x")` | `integral(...)` | a logarithm of the root, piecewise on the sign of `a + b + c` |
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### The derivative of a sum of exponentials over another is read off the integrand over the other's square
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**Answers where there were none.** Rubi's 6.7.1 `x (b - a sinh(x))/(a + b sinh(x))^2` and `x (b + a cosh(x))/(a + b cosh(x))^2`
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were declined. The bracket over the square is the derivative of `-cosh(x)/(a + b sinh(x))`, and of
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`sinh(x)/(a + b cosh(x))`. So by parts the integral is `x` times that quotient, plus `ln(a + b sinh(x))/b` or minus
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`ln(a + b cosh(x))/b`. The integrals without x were answered with a piecewise on a discriminant, from the
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substitution in `e^x`; they are now the quotient itself. For a sum `D` of exponentials, the rule reads the
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integrand over `D^2` as `(N/D)'`, with `N` a constant times each exponential in `D`
| `"(b-a*sinh(x))/(a+b*sinh(x))^2".ToEntity().Integrate("x")` | `integral(...)` | 71 characters, where before this change master gave 7,861 |
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| `"(b+a*cosh(x))/(a+b*cosh(x))^2".ToEntity().Integrate("x")` | `integral(...)` | 70 characters, where before this change master gave 7,631 |
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| `"sech(x)^2*tanh(x)^2/(2+tanh(x)^3)^2".ToEntity().Integrate("x")` | no answer in thirty seconds | 60 characters, `-1/3/(2 + tanh(x)^3)`, in 55 ms; master gave 117 in 16 s and 9.7 GB |
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### A hyperbolic function of `a + 2 ln(x)` is integrated with `e^a` as a symbol too
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**Answers where there were none.** `tanh(a + 2 ln(x))^2/x^3` was declined, and so were nineteen more of Rubi's 6.3.2,
else if (underneath && sum is null && factor is Powf(Sumf or Minusf, Number.Integer { EInteger: var two } ) squared && two.Equals(EInteger.FromInt32(2)))
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