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
*****************************************************
Aleksas Domarkas solution:
(%i1) S:'integrate(sqrt(x^5)*sqrt(1-x),x);
(%o1) integrate(sqrt(1-x)*sqrt(x^5),x)
(%i2) changevar(S, 2*x-1=y, y, x),rootscontract;
(%o2) integrate(sqrt(1-y^2)*(y^2+2*y+1),y)/16
(%i3) ev(%, nouns);
(%o3)
((5*asin(y))/8-(y*(1-y^2)^(3/2))/4-(2*(1-y^2)^(3/2))/3+(5*y*sqrt(1-y^2))/8)/16
solution:
(%i4) sol:ratsimp(subst(y=2*x-1,%));
(%o4) (15*asin(2*x-1)+sqrt(4*x-4*x^2)*(48*x^3-8*x^2-10*x-15))/384
test of solution:
(%i5) diff(sol,x)-first(S)$ radcan(%);
(%o6) 0
Maxima version: 5.24.0
Maxima build date: 20:39 4/5/2011
Host type: i686-pc-mingw32
Lisp implementation type: GNU Common Lisp (GCL)
Lisp implementation version: GCL 2.6.8
Aleksas D