Bug fix: Integrals involving fractional powers

Maple has trouble with some integrals involving products of half-integer powers of linear terms. In some cases it seems the problem has to do with taking different branches of multivalued functions, and the antiderivative may be correct in some regions but not in others. 

> f1:= (2*x-3)^(-3/2)*x^(1/2);

f1 := sqrt(x)/((2*x-3)^(3/2))

> F1:= int(f1,x);

F1 := -1/2*sqrt(2)*(1/3*I*sqrt(6)*sqrt(Pi)*sqrt(x)*...

> N:= simplify(diff(F1,x)-f1);

N := -(4*I*sqrt(6*x-9)*x^2-12*I*sqrt(6*x-9)*x+9*I*s...
N := -(4*I*sqrt(6*x-9)*x^2-12*I*sqrt(6*x-9)*x+9*I*s...

This complicated function turns out to be zero (as it should be) when x < 3/2 .

> assume(x<3/2); simplify(N);

0

> assume(x>3/2); simplify(N);

-2/3*sqrt(6*x-9)*sqrt(x)*sqrt(3)/((2*x-3)^2)

Unfortunately we are probably most interested in the case x > 3/2 , as that is when the terms in f1 are both real.

In other cases the result is wrong for all values of x .

> x:= 'x':
f2:=(2*x+3)^(1/2)*x^(-5/2);
F2:=int(f2,x);

f2 := sqrt(2*x+3)/(x^(5/2))

F2 := -2/3*sqrt(2*x+3)/(x^(3/2))-2/9*sqrt(2*x+3)/(s...

> simplify(diff(F2,x) - f2);

-1/3*1/(x^(3/2)*sqrt(2*x+3))

A work-around is to express the integrand as a single square root. Note that this is not quite equivalent to the original integrand (although it is equivalent when at least one of the linear terms is positive).

> f1p:= sqrt((2*x-3)^(-3)*x);
F1p:= simplify(int(f1p,x));

f1p := sqrt(x/((2*x-3)^3))

F1p := -1/4*sqrt(x/((2*x-3)^3))*(8*sqrt((2*x-3)*x)*...
F1p := -1/4*sqrt(x/((2*x-3)^3))*(8*sqrt((2*x-3)*x)*...
F1p := -1/4*sqrt(x/((2*x-3)^3))*(8*sqrt((2*x-3)*x)*...
F1p := -1/4*sqrt(x/((2*x-3)^3))*(8*sqrt((2*x-3)*x)*...

> normal(diff(F1p,x)-f1p);

0

> f2p:= sqrt((2*x+3)/x^5);
F2p:= simplify(int(f2p,x));

f2p := sqrt((2*x+3)/(x^5))

F2p := -2/9*sqrt((2*x+3)/(x^5))*x*(2*x+3)

> normal(diff(F2p,x)-f2p);

0

In general it is prudent to check the correctness of integrals, either by comparing definite integrals to their floating-point approximations or by comparing the derivative of an indefinite integral to the integrand.

See also: Errors in symbolic integration , int

Maple Advisor Database R. Israel, 1998