Can an apparently intractable integral be made tractable in this way? Say there are two integrals:
$$F(\beta)=\int_{b_0}^{\beta}f(b)\: db$$
$$H(\rho)=\int_{r_0}^{\rho}h(r)\: dr$$
where $F(\beta)$ is (apparently) intractable while $H(\rho)$ is tractable (expressible in elementary functions).
And we know that $b$ is a function of $r$ and that
$$f(b(r))=h(r)$$
Can this info be used to make $F(\beta)$ expressible in terms of elementary functions?  How?
 A: Like this?
$h(r) = e^r$ has elementary integral
$$
\int e^r\;dr
$$
and $f(b) = e^{e^b}$ has non-elementary integral
$$
\int e^{e^b} db
$$
despite the fact that $b(r) = \log r$ gives us $f(b(r))=h(r)$.
A: Each elementary function is differentiable, and its derivative is again an elementary function. $H$ is elementary means therefore, also $h$ is elementary. 
Because $h$ is elementary and $f\circ b=h$, also $f\circ b$ is elementary.
$\ $
$$F(\beta)=\int_{b_0}^{\beta}f(b)db$$
Use that $b$ is a function of $r$: change the integration variable:
$$b_0=b(r_0),\ \beta=b(r_1),\ \frac{db(r)}{dr}=b'(r)\ \curvearrowright\ db=b'(r)dr$$
$$F(r_1)=\int_{r_0}^{r_1}f(b(r))b'(r)dr$$
Use that $f(b(r))=h(r)$: Restrict $f$ to domains $D_i$ where the restriction $f_i=f|_{D_i}$ is bijective, and consider each $f_i$. The $f_i^{-1}$ are the partial (compositional) inverses of $f$.
$$f_i(b(r))=h(r),\ \curvearrowright\ b(r)=f_i^{-1}(h(r))$$
$$F(r_1)=\int_{r_0}^{r_1}f(f_i^{-1}(h(r)))b'(r)dr$$
$$F(r_1)=\int_{r_0}^{r_1}h(r)b'(r)dr$$
Even if the integrand of $F$ is elmentary, $F$ can be an elementary integral or a non-elementary integral.
Take e.g. the example of GEdgar's answer.
