Question about radcan



I found radexpand option variable in the reference manual:
> radexpand
> Option variable
> Default value: true
> radexpand controls some simpli?cations of radicals.
> When radexpand is all, causes nth roots of factors of a product which 
> are powers
> of n to be pulled outside of the radical. E.g. if radexpand is all, 
> sqrt (16*x^2)
> simpli?es to 4*x.
> More particularly, consider sqrt (x^2).
> ? If radexpand is all or assume (x > 0) has been executed, sqrt(x^2) 
> simpli?es
> to x.
> ? If radexpand is true and domain is real (its default), sqrt(x^2) 
> simpli?es to
> abs(x).
> ? If radexpand is false, or radexpand is true and domain is complex, 
> sqrt(x^2)
> is not simpli?ed.
> Note that domain only matters when radexpand is true.

But it does not work for me(version of maxima is 5.25.1):

(%i1) radexpand:true;
(%o1) true
(%i2) b: ((1-x)/sqrt(1-2*x+x^2));
1 - x
(%o2) ------------------
2
sqrt(x - 2 x + 1)
(%i3) radcan(b);
(%o3) - 1
(%i4) radexpand: false;
(%o4) false
(%i5) radcan(b);
(%o5) - 1


any ideas?

> We have a question at
> http://ask.sagemath.org/question/767/simplification-errors-in-simple-expressions
> which seems to find a bug in radcan.  Here is the relevant Maxima
> session.
>
>
> (%i3) b:((1-x)/sqrt(1-2*x+x^2));
>                                      1 - x
> (%o3)                         ------------------
>                                      2
>                                sqrt(x  - 2 x + 1)
> (%i4) radcan(b);
> (%o4)                                 - 1
>
> which seems a little aggressive. But
>
> (%i7) d:((1-x)/sqrt((1-x)^2));
>                                      1 - x
> (%o7)                             ----------
>                                    abs(x - 1)
>
>
> Radcan's documentation is a little confusing.
>
>     radicals, by converting it into a form which is canonical over a
>       large class of expressions and a given ordering of variables; that
>       is, all functionally equivalent forms are mapped into a unique
>       form.  For a somewhat larger class of expressions, `radcan'
>
> doesn't give very many details. But I'm certainly not an expert in
> making expressions canonical :)
>
> Thanks for any help!
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