Yep. You can get that, but it is somewhat, say, artificial
jj = Integrate[x^3 (2 - x^2)^m, x] /. m -> 12
(jj /. (2 - x^2) -> p /. (2 + 13 x^2) -> q /.
q -> ((2 - x^2) (PolynomialQuotient[2 + 13 x^2, 2 - x^2, x] +
PolynomialRemainder[2 + 13 x^2, 2 - x^2,
x]/(2 - x^2))) /. (2 - x^2) -> p // Expand) /.
p -> 2 - x^2