How to find: $\int_{-d}^d  \frac{z}{2\sqrt{z+c}\ \cdot \sqrt{d^2-z^2}} dz$ Let $c,d>0$. I do not know how to integrate the following:
$$\int_{-d}^d  \frac{z}{2\sqrt{z+c}\ \cdot \sqrt{d^2-z^2}}\ \text{d}z.$$

Sorry~~ it should be $z^2$. 
I think it can be simplified to the following: for $c\ge d>0$
$$\frac{\sqrt{d}}{2} \int_{-\pi/2}^{\pi/2} \frac{\sin y}{\sqrt{\sin
y+c/d}}\ \text{d}y.$$
Thanks again!
 A: I see somebody did it Weierstrass-style; I'll try out the Jacobi approach here for completeness.
With
$$\frac12\int_{-d}^d\frac{x}{\sqrt{(x+c)(d-x)(d+x)}}\mathrm dx$$
and the assumption $c \geq d > 0$, we use the substitution
$$x=d\left(1-2\mathrm{sn}^2\left(u\mid\frac{2d}{c+d}\right)\right)$$
which turns the integral into
$$\frac{d}{\sqrt{c+d}}\int_0^{K\left(\frac{2d}{c+d}\right)}\left(1-2\mathrm{sn}^2\left(u\mid\frac{2d}{c+d}\right)\right)\mathrm du$$
or, simplifying,
$$\frac{d}{\sqrt{c+d}}\left(K\left(\frac{2d}{c+d}\right)-2\int_0^{K\left(\frac{2d}{c+d}\right)}\mathrm{sn}^2\left(u\mid\frac{2d}{c+d}\right)\mathrm du\right)$$
We can use formula 22.16.15 in the DLMF to handle the remaining integral. Simplification yields the expression
$$\sqrt{c+d}E\left(\frac{2d}{c+d}\right)-\frac{c}{\sqrt{c+d}}K\left(\frac{2d}{c+d}\right)$$
A: $$\frac12\int_{-d}^d\frac{z}{\sqrt{(d^2-z^2)(c+z)}}\mathrm{d}z$$
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.
A: I wouldn't know how to do it manually, but contrarily to the case of yunone, Wolfram does yield an answer to me:
$$-\frac{\pi  d^2 \left(2\frac{d^2}{c^2}\right)}{8 c^{3/2}}$$
Under the conditions that $$\Re(c)>0\land d>0\land (c\notin \mathbb{R}\lor d\leq \Re(c))$$
