integrate(sqrt(x^2+y^2),x,0,1)$ integrate(%,y,0,1)$



> (%05)sqrt(a^2+(2-2*a)*sqrt(a)-1)

> Yes, as I pointed out in my email, that is a difficult case.

Hardly a method, but ...

 (%i1) load(to_poly_solver)$

 (%i2) ((sqrt(a)+1)*(sqrt(a)-1)^3)^(1/2)-w$

 (%i3) subst(a=b^2,%)$

 (%i4) to_poly(%,[w,b])$

 (%i5) elim_allbut(first(%),[b,w])$

 (%i6) first(first(%))$

 (%i7) solve(%,w)$

 (%i8) subst(b=sqrt(a),%)$

A Taylor expansion indicates (doesn't prove) that the last solution is correct:

 (%i9) facsum(%,sqrt(a-1));
 (%o9) [w=-%i*(a-2*sqrt(a)+1),w=%i*(a-2*sqrt(a)+1),w=-%i*(a+2*sqrt(a)+1),w=%i*(a+2*sqrt(a)+1),w=-(sqrt(a)+1)*sqrt(a-1),
              w=(sqrt(a)+1)*sqrt(a-1),w=-(sqrt(a)-1)*sqrt(a- 1),w=(sqrt(a)-1)*sqrt(a-1)]

--Barton