Consider a rectangle with edges $2a$ and $2b$ in the $xy$-plane with its centre (the point of intersection of diagonals) at the origin. The problem is to find a closed form for the solid angle subtended by the rectangular surface at the point $(0,0,h)$. I tried and reduced the problem to the problem down to solving $\Omega=4\displaystyle\int_0^a\int_0^b\dfrac{h}{(x^2+y^2+h^2)^{3/2}}$d$x$d$y$ .
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If a parametrization of the surface $S$ is:
$$ \mathbf{s}(u,v)=(u,v,0),\qquad(u,v)\in D:=[-a,a]\times[-b,b] $$
the radius vector with respect to $P(0,0,h)$ is:
$$ \mathbf{r}(u,v)=\mathbf{s}(u,v)-P=(u,v,-h) $$
and the solid angle $\Omega$ for $S$ subtended at $P$ is:
$$ \begin{aligned} \Omega &=\iint\limits_S\frac{\mathbf{r}(u,v)}{||\mathbf{r}(u,v)||^3}\cdot\mathbf{\hat{n}}(u,v)\,\text{d}S\\ &=\iint\limits_D\frac{(u,v,-h)}{\sqrt{\left(u^2+v^2+h^2\right)^3}}\cdot\left(\mathbf{s}_v(u,v)\land\mathbf{s}_u(u,v)\right)\,\text{d}u\,\text{d}v\\ &=4\int_0^a\left(\int_0^b\frac{h}{\sqrt{\left(u^2+v^2+h^2\right)^3}}\,\text{d}v\right)\text{d}u\\ &=4\int_0^a\frac{bh}{\left(u^2+h^2\right)\sqrt{u^2+b^2+h^2}}\,\text{d}u\\ &=4\left(\arctan\left(\frac{h}{b}\right)-\arctan\left(\frac{a^2+h^2-a\sqrt{a^2+b^2+h^2}}{bh}\right)\right)\\ &=\boxed{4\arctan\left(\frac{ab}{h\sqrt{a^2+b^2+h^2}}\right)} \end{aligned} $$
which is exactly what was requested.