clicliclic@freenet.de wrote:
>
> Wow! In Legendre's pseudo-elliptic integral:
>
> INT(x/((4 - x^3)*SQRT(1 - x^3)), x)
> = 2^(1/3)/18*(ATANH(SQRT(1 - x^3))
> - 3*ATANH((1 + 2^(1/3)*x)/SQRT(1 - x^3))
> - SQRT(3)*ATAN((2^(1/3) - 2^(2/3)*x - x^2)
> /(SQRT(3)*2^(1/3)*SQRT(1 - x^3))))
>
This does not verify. AFAICS the third atan should be
atan((1 - 2^(1/3)*x - 2^(-1/3)*x^2)/(sqrt(3)*sqrt(1 - x^3)))
> as evaluated by Fricas 1.3.5:
>
> 4+1 arc tangents occur of which the first four appear to be sextupled.
> So undoubling alone won't be enough, untripling is called for as well.
> Note in passing that my model ATAN can be split in two simpler ones:
>
> ATAN((2^(1/3) - 2^(2/3)*x - x^2)/(SQRT(3)*2^(1/3)*SQRT(1 - x^3)))
> = ATAN(SQRT(3)/SQRT(1 - x^3))
> + ATAN(SQRT(3)*(1 - 2^(1/3)*x)/SQRT(1 - x^3))
>
> For the trivially related:
>
> INT(x/((1 - x^3)*SQRT(1 - 4*x^3)), x)
> = 1/9*(ATANH(SQRT(1 - 4*x^3)) - 3*ATANH((1 + 2*x)/SQRT(1 - 4*x^3))
> - SQRT(3)*ATAN((1 - 2*x - 2*x^2)/(SQRT(3)*SQRT(1 - 4*x^3))))
>
> FriCAS 1.3.5 returns the much more compact:
>
> (3*log(((6*x^4 + (-18)*x^3 + (-9)*x^2 + (-6)*x)*((-4)*x^3 + 1)^(1/2)
> + (2*x^6 + (-30)*x^5 + 12*x^4 + 14*x^3 + 21*x^2 + 6*x + 2))
> /(x^6 + 3*x^5 + 6*x^4 + 7*x^3 + 6*x^2 + 3*x + 1))
> + ((-2)*3^(1/2)*atan((2*x^3 + 6*x^2 + 3*x + 1)*3^(1/2)
> *((-4)*x^3 + 1)^(1/2)/(12*x^4 + 12*x^3 + (-3)*x + -3))
> + (-2)*3^(1/2)*atan((2*x^3 + 6*x^2 + (-6)*x + 1)*3^(1/2)
> *((-4)*x^3 + 1)^(1/2)/(24*x^4 + (-12)*x^3 + (-6)*x + 3))))/54
>
> which still needs untripling. Let's risk a closer look. Although the
> second ATAN could be undoubled right away:
>
> ATAN((1 - 6*x + 6*x^2 + 2*x^3)/(SQRT(3)*(1 - 2*x)*SQRT(1 - 4*x^3)))
> = 2*ATAN(SQRT(3)*(1 - 2*x)/SQRT(1 - 4*x^3))
>
> the full antiderivative contains this latter ATAN element thrice, not
> twice, and the untripling can succeed only if the original two ATANs
> are merged:
>
> ATAN((1 - 6*x + 6*x^2 + 2*x^3)/(SQRT(3)*(1 - 2*x)*SQRT(1 - 4*x^3)))
> - ATAN((1 + 3*x + 6*x^2 + 2*x^3)/(SQRT(3)*(1 + x)*SQRT(1 - 4*x^3)))
> = ATAN((2 + x - 11*x^3 - x^4)*(1 - 2*x - 2*x^2)
> /(3*SQRT(3)*x*(1 - 3*x^2 - x^3)*SQRT(1 - 4*x^3)))
> = 3*ATAN((1 - 2*x - 2*x^2)/(SQRT(3)*SQRT(1 - 4*x^3)))
>
> Oof! Why are compound arc tangents as arbitrary as these two produced
> in the first place? How could their reduction be handled automatically?
Mathematically, if we have
f = c1*f1 + c2*f2 + c3*f3
and if A is invertible integer matrix, (d1, d2, d3) = A^(-1)(c1, c2, c3)
(h1, h2, h3) = A(f1, f2, f3), then
f = d1*h1 + d2*h2 + d3*h3
If fi-s are logarithms (or atans), then some combinations are
multiples (can be undoubled, untripled, etc.), some are not.
For rational functions there are resonably natural "canonical"
choices. For algebraic functions this is more tricky, there
are equivalent expressions without clear indication which
one is "better". In particular, two bases above are not
equivalent: space spanned by logs and atan from the two expressions
has dimension 4. Expression with minimal number of terms
has two terms (one log and one atan).
Now, expression produced by FriCAS is not arbitrary, it directly
follows from the integral. More precisely, FriCAS looks at
residues. In the second case residure are root of (two) polynomials
of degree 2, which allows expression as four log terms.
FriCAS does not notice that roots can be expressed in terms
of single square root which would reduce number of terms to two.
Later in attempt to simplify answer FriCAS manages to tranform
this into 3 terms. In first integral FriCAS gets residues as
roots of polynomial of degree 6. FriCAS does not see that 2^(1/3)
is a trivial scale factor, which leads to 6 terms. Later
conversion to atans adds a more terms.
--
Waldek Hebisch
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