int sqrt(1-sqrt(1-x^2)) "d"x.
Note that
(1/sqrt(2) (sqrt(1-x)-sqrt(1+x)))^2=1-sqrt(1-x^2).
(I found this by setting (a+bsqrt(1-x^2))^2=1-sqrt(1-x^2), expanding the left hand side, equating coefficients and solving for a and b).
Then,
sqrt(1-sqrt(1-x^2))=1/sqrt(2) abs(sqrt(1+x)-sqrt(1-x)).
int sqrt(1-sqrt(1-x^2)) "d"x= 1/sqrt(2) int abs(sqrt(1+x)-sqrt(1-x)) "d"x.
The function sqrt(1+x)-sqrt(1-x) is real only for -1<=x<=1.
Clearly 1+x>1-x for 0<x<=1. As the square root function is increasing, sqrt(1+x)>sqrt(1-x) or sqrt(1+x)-sqrt(1-x)>0 for 0<x<=1.
1-x>1+x for -1<=x<0. As the square root function is increasing, sqrt(1-x)>sqrt(1+x) or sqrt(1-x)-sqrt(1+x)>0 for -1<=x<0.
We conclude,
sqrt(1-sqrt(1-x^2)) = { (1/sqrt(2)(sqrt(1+x)-sqrt(1-x)), 0<=x<=1), (1/sqrt(2)(sqrt(1-x)-sqrt(1+x)), -1<=x<=0) :},
int sqrt(1-sqrt(1-x^2)) "d"x = { (sqrt(2)/3((1+x)^(3/2)+(1-x)^(3/2))+C, 0<x<=1), (-sqrt(2)/3((1-x)^(3/2)+(1+x)^(3/2))+C, -1<=x<0) :}
Note that the result is the same but positive for x>0 and negative for x<0. We can define the function "sign"(x) to return +1 for x>0, -1 for x<0 and be undefined at 0. We then conclude,
int sqrt(1-sqrt(1-x^2)) "d"x = "sign"(x) sqrt(2)/3 ((1+x)^(3/2)+(1-x)^(3/2))+C
If you don't like this "sign"(x) function you can alternatively define it as sqrt(x^2)/x and conclude,
int sqrt(1-sqrt(1-x^2)) "d"x = sqrt(2x^2)/(3x)((1+x)sqrt(1+x)+(1-x)sqrt(1-x))+C
