= Solution
By <spherical symmetry>, rotate a <geodesic> into the equatorial plane $\theta=\pi/2$. The <geodesic conserved quantities from Killing vectors> are
$$
E=V\dot t,\qquad L=R^2\dot\varphi,
$$
where the dot denotes a suitable <affine parameter>. Normalize the tangent by $g(\dot x,\dot x)=-\epsilon$, with $\epsilon=1$ for timelike proper time, $\epsilon=-1$ for spacelike proper length, and $\epsilon=0$ for null geodesics. Then
$$
-\epsilon=-\frac{E^2}{V}+\frac{\dot r^2}{V}+\frac{L^2}{R^2},\qquad
\dot r^2=E^2-V\left(\epsilon+\frac{L^2}{R^2}\right).
$$
Therefore the <geodesic potential of the magnetic dilaton black hole> is
$$
\boxed{\widetilde V(r)=\frac12\left[\left(1-\frac{r_+}{r}\right)\left(\epsilon+\frac{L^2}{r(r-r_-)}\right)-E^2\right].}
$$
Together with the two conserved first integrals, this reduces the radial motion to $\dot r^2/2+\widetilde V=0$. To cover turning points as well, the radial geodesic equation itself gives
$$
\ddot r=-\frac{V'}{2V}(E^2-\dot r^2)+\frac{VL^2R'}{R^3}
=-\frac12\left[V'\left(\epsilon+\frac{L^2}{R^2}\right)-\frac{2VL^2R'}{R^3}\right]=-\widetilde V'.
$$
Thus a constant-radius geodesic must satisfy both $\widetilde V=0$ and $\widetilde V'=0$. The intermediate static-coordinate formulas apply where $V\ne0$; the radial equation extends through a regular horizon in regular coordinates.
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