Richard Fateman wrote:
>
> Unfortunately it doesn't. The "steps" indicated are not explained
> algorithmically, and in fact there may be
> extraordinary difficulty in making them algorithmic. That is why the
> programs to do them have probably not
> been written in the last several decades.
I don't know if a "general" program has been written or not to do elliptic
integrals, (i thought that Trager had something like that in Axiom, but i
have just checked Axiom doesn't do elliptic integrals) but i have a
tendency to think that special cases are much more useful. Anyways Maple
has no problem to do:
rose% maple
|\^/| Maple 10 (X86 64 LINUX)
> lprint(int(1/sqrt(x^4+1),x));
1/(1/2*2^(1/2)+1/2*I*2^(1/2))*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*
EllipticF(x*(1/2*2^(1/2)+1/2*I*2^(1/2)),I)
> quit
bytes used=2397208, alloc=2227816, time=0.06
It does also the case x^3+1 but the result is obviously more complicated
since exp(2i pi/3) appears.
By the way if we multiply the abelian differential by a reasonable rational
function, maple is able to express the result on elliptic integrals of
first and third kind:
> lprint(int((x^3+5*x+7)/((x^2+6*x+8)*sqrt(x^4+1)),x));
1/2*arcsinh(x^2)-6/(1/2*2^(1/2)+1/2*I*2^(1/2))*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(
x^4+1)^(1/2)*EllipticF(x*(1/2*2^(1/2)+1/2*I*2^(1/2)),I)-77/1028*257^(1/2)*
arctanh(1/514*(2+32*x^2)*257^(1/2)/(x^4+1)^(1/2))-77/8*(-1)^(3/4)*(1-I*x^2)^(1/2
)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticPi((-1)^(1/4)*x,-1/16*I
(-I)^(1/2)/(-1)^
(1/4))+11/68*17^(1/2)*arctanh(1/34*(2+8*x^2)*17^(1/2)/(x^4+1)^(1/2))+11/4*(-1)^(
3/4)*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticPi((-1)^(1/4)*x,-1/4*
I,(-I)^(1/2)/(-1)^(1/4))
The prefactors are quite complicated, i suppose they come from the recursion
relations explained in the article.
--
Michel Talon