$\frac12\int_{-d}^d\frac{z}{\sqrt{(d^2-z^2)(c+z)}}\mathrm{d}z$
is equivalent to (by depressing the cubic within the square root):
$=-\int_{-d-c/3}^{d-c/3}\frac{z+c/3}{\sqrt{\frac{8cd^2}{3}-\frac{8c^3}{27}-\left(\frac{4c^2}{3}+4d^2\right)z+4z^3}}\mathrm{d}z$
Using the substitution
$z=\wp\left(u;\frac{4c^2}{3}+4d^2,\frac{8c^3}{27}-\frac{8cd^2}{3}\right)$
where $\wp(u;g_2,g_3)$ is a Weierstrass elliptic function, and making use of the differential relation for the Weierstrass function:
${\wp^{\prime}}^2=4\wp^3-g_2\wp-g_3$
we have
$-\int_{\wp^{(-1)}(-d-c/3)}^{\wp^{(-1)}(d-c/3)}\left(\wp\left(u;\frac{4c^2}{3}+4d^2,\frac{8c^3}{27}-\frac{8cd^2}{3}\right)+\frac{c}{3}\right)\mathrm du$
whose integration I'll leave to the interested reader.