AngouriMath
TryWriteAPairOfLinearsOverARadicand(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Collections.Generic.List{AngouriMath.Entity},AngouriMath.Entity.Variable,AngouriMath.Entity@,AngouriMath.Entity@)
Method (no overloads)
Summary
Two linear factors whose product is a constant multiple of a radicand a logarithm holds,
written as that radicand: `d + i c d x` and `f - i c f x` beside `arsinh(c x)`, whose
product is `d f (1 + c^2 x^2)`. `L1^p L2^q` is `L1^(p - q) L1^q L2^q` where `p - q` is
whole, and `L1^q L2^q` is `lambda^q M^q` for a whole `q` and `K^k M^(k/2)` for
`q = k/2`, with `K = sqrt(L1) sqrt(L2)/sqrt(M)` constant wherever it is defined -- the
same factor the single base is written with. Rubi's 7.1.4 states seventy-two of its
rows this way; `(d + i c d x)^(5/2) sqrt(f - i c f x) (a + b arsinh(c x))` ran past
seventy seconds where the same thing over `1 + c^2 x^2` takes two thirds of one.
written as that radicand: `d + i c d x` and `f - i c f x` beside `arsinh(c x)`, whose
product is `d f (1 + c^2 x^2)`. `L1^p L2^q` is `L1^(p - q) L1^q L2^q` where `p - q` is
whole, and `L1^q L2^q` is `lambda^q M^q` for a whole `q` and `K^k M^(k/2)` for
`q = k/2`, with `K = sqrt(L1) sqrt(L2)/sqrt(M)` constant wherever it is defined -- the
same factor the single base is written with. Rubi's 7.1.4 states seventy-two of its
rows this way; `(d + i c d x)^(5/2) sqrt(f - i c f x) (a + b arsinh(c x))` ran past
seventy seconds where the same thing over `1 + c^2 x^2` takes two thirds of one.
Remarks
above the bar too: `(d + i c d x)^(3/2) (f - i c f x)^(5/2)` is `K^3 (f - i c f x) M^(3/2)`,
a polynomial beside the root. Paired the other way it was `K^5 M^(5/2)/(d + i c d x)`,
which ran past the budget, as did the same with `(a + b arsinh(c x))^2`; the polynomial
takes half a second.
less, the whole power `L1^n` is written out as a sum of monomials, each over the root:
`sqrt(f - i c f x)/(d + i c d x)^(3/2)` is `K^(-3) (f - i c f x)^2/M^(3/2)`, and as that
product the rules for a power of a quadratic do not read it -- ten of Rubi's 7.1.4
rows, from `(f - i c f x)^2` to `(f - i c f x)^5` over `M^(3/2)` and `M^(5/2)`, ran past
the budget -- while each `x^k (a + b arsinh(c x))/M^(3/2)` is answered in a tenth of a
second. Over `sqrt(M)` the product is answered as it is, and is left so; and both
factors below the bar are left as they are paired.
https://github.com/asc-community/AngouriMath/issues/233
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