Let us consider the definite integral

J:=int(abs(x-(-x^5+1)^(1/5)), x = 0 .. 1);

Maple fails with it, Mathematica 10.1 finds it in terms of special functions. Let us look at the integrand:

plot(x-(-x^5+1)^(1/5), x = 0 .. 1);

We see the expression under the modulus changes its sign at the unique point of RealRange(0,1). Therefore

solve(x-(-x^5+1)^(1/5));

Then

J:= int(-x+(-x^5+1)^(1/5), x = 0 .. (1/2)*2^(4/5))+int(x-(-x^5+1)^(1/5), x = (1/2)*2^(4/5) .. 1);

which outputs a complicated expression

(1/8)*2^(4/5)*(4*hypergeom([-1/5, 1/5], [6/5], 1/2)-2^(4/5))+(1/2)*2^(4/5)*((1/2)*2^(1/5)-(1/4)*2^(4/5))-(1/25)*Pi*csc((1/5)*Pi)*(-(25/2)*sin((1/5)*Pi)*GAMMA(4/5)*2^(4/5)*hypergeom([-1/5, 1/5], [6/5], 1/2)/Pi+(5/4)*sec((3/10)*Pi)*cos((1/10)*Pi)*2^(3/5)*Pi^(1/2)*csc((3/10)*Pi)/GAMMA(7/10))/GAMMA(4/5).

At the same time we have

int(abs(x-(-x^5+1)^(1/5)), x = 0 .. 1, numeric);

0.5000000000

How to obtain 1/2 symbolically?