Evaluation of:
$$U'_{\rm rad} = {U_{\rm rad} \gamma^2 \over 2} \int_{\theta = 0}^\pi (1 + \beta \cos \theta)^2 \sin \theta d \theta$$

Substitute $z \equiv \cos \theta$ so $d z = - \sin \theta d \theta$ and eliminate $\beta$ in favor of $\gamma$:
$$U'_{\rm rad} = { U_{\rm rad} \gamma^2 \over 2} \int_1^{-1} (1 + \beta z)^2 (-1) d z$$
$$U'_{\rm rad} = {U_{\rm rad} \gamma^2 \over 2} \int_{-1}^1 (1 + 2 \beta z + \beta^2 z^2) d z$$
$$U'_{\rm rad} = { U_{\rm rad} \gamma^2 \over 2} \biggl[ z + \beta z^2 + {\beta^2 z^3 \over 3} \biggl]_{-1}^1$$
$$U'_{\rm rad} = { U_{\rm rad} \gamma^2 \over 2} \biggl( 1 + \beta + {\beta^2 \over 3} + 1 - \beta + {\beta^2 \over 3} \biggr)$$
$$U'_{\rm rad} = U_{\rm rad} \gamma^2 (1 + \beta^2 / 3)$$
$$U'_{\rm rad} = U_{\rm rad} \biggl[ \gamma^2 + {\gamma^2 \over 3} - \biggl( {\gamma^2 \over 3} - {\gamma^2 \beta^2 \over 3} \biggr) \biggr]$$

The final result is
$$U'_{\rm rad} = U_{\rm rad} \biggl[ {4 \gamma^2 \over 3} -{1\over 3} \gamma^2 (1 - \beta^2) \biggr]$$