  A knot in $S^3$ is topologically slice if it bounds a locally flat disk in
$B^4$. A knot in $S^3$ is rationally slice if it bounds a smooth disk in a
rational homology ball. We prove that the smooth concordance group of
topologically and rationally slice knots admits a $\mathbb{Z}^\infty$ subgroup.
All previously known examples of knots that are both topologically and
rationally slice were of order two. As a direct consequence, it follows that
there are infinitely many topologically slice knots that are strongly
rationally slice but not slice.
