Don't the expressions sqrt(1/x) and 1/sqrt(x) denote the same analytic
function over C-{0}?
It is only when we try to restrict to a particular sheet that we run into
problems, right?
So shouldn't the default be radexpand:all?
-s
On Sat, Apr 2, 2011 at 11:39, Dieter Kaiser <drdieterkaiser at web.de> wrote:
> Am Samstag, den 02.04.2011, 07:20 +0200 schrieb Rene Grothmann:
> > On of the previous versions of Maxima knew how to find
> >
> > integrate(sqrt(x^5)*sqrt(1-x),x)
> >
> > But the current version cannot do this. Is there a way to get it?
>
> I think this has changed because expression like sqrt(1/x) and 1/sqrt(x)
> are no longer treated as equivalent.
>
> It is the flag radexpand, which switches on simplifications like
> sqrt(1/x) -> 1/sqrt(x). Therefore we get:
>
> (%i1) integrate(sqrt(x^5)*sqrt(1-x),x), radexpand:all;
>
> (%o1)
> -(-15*sqrt(1-x)/sqrt(x)+73*(1-x)^(3/2)/x^(3/2)+55*(1-x)^(5/2)/x^(5/2)
> +15*(1-x)^(7/2)/x^(7/2))
> /(768*(1-x)/x+1152*(1-x)^2/x^2+768*(1-x)^3/x^3+192*(1-x)^4/x^4
> +192)
> -5*atan(sqrt(1-x)/sqrt(x))/64
>
> Another possibility is to assume the variable x to be positive:
>
> (%i1) assume(x>0);
> (%o1) [x > 0]
>
> (%i2) integrate(sqrt(x^5)*sqrt(1-x),x);
> (%o2)
> -(-15*sqrt(1-x)/sqrt(x)+73*(1-x)^(3/2)/x^(3/2)+55*(1-x)^(5/2)/x^(5/2)
> +15*(1-x)^(7/2)/x^(7/2))
> /(768*(1-x)/x+1152*(1-x)^2/x^2+768*(1-x)^3/x^3+192*(1-x)^4/x^4
> +192)
> -5*atan(sqrt(1-x)/sqrt(x))/64
>
> Dieter Kaiser
>
>
> _______________________________________________
> Maxima mailing list
> Maxima at math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima
>