<s>[INST] Closed-form expression for $F(x,y) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+x)^2 (t+y)} \mathrm{d}t$? I am considering the following function
$$F(x,y) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+x)^2 (t+y)} \mathrm{d}t,$$
which is well-defined for any $x > 0$ and $y \geq 0$.
Is there a hope to obtain a closed form formula with respect to $x$ and $y$? 
For instance, according to Mathematical, we have that 
$$F(x,0) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+x)^2 t} \mathrm{d}t = \frac{\pi}{x \sqrt{x (x+2)}}.$$
Remark: To give a bit of context, the function $F$ appears when I consider the quadratic optimization problem of the form $\min_{\mathbf{x} \in \mathrm{R}^N} \lVert \mathbf{A} \mathbf{x} - \mathbf{y} \rVert_2^2 + \lambda \lVert \mathbf{x} \rVert_2^2$ and I try to understand the behavior of $\lVert \widehat{\mathbf{x}} - \mathbf{x}_0 \rVert_2^2$ with $\widehat{\mathbf{x}}$ the unique optimizer and $\mathbf{x}_0$ the vector we aim at recovering, with $\mathbf{y} = \mathbf{A} \mathbf{x}_0 + \mathbf{n} \in \mathbb{R}^M$ and $\mathbf{n}$ an i.i.d. Gaussian vector. The values $x$ and $y$ above appear as functions of $\lambda$ and $\gamma = \lim M/N$ when $N\rightarrow \infty$ when the matrix $\mathbf{A}$ is i.i.d. Gaussian and its spectrum behaves according to the Marchenko-Pastur law.
 [/INST] Let's change notation to $\displaystyle F(a,b) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+a)^2 (t+b)}\, \mathrm{d}t$ to emphasise the constant variables. 
Let $\displaystyle t = \frac{u-b}{u+a} \implies \mathrm{d}t = \frac{a+b}{(u+a)^2}\,\mathrm{d}u$; moreover, this maps $[0, 1]$ to $[b, \infty)$. We've
$$\displaystyle F(a,b) = (a+b)\sqrt{a+b}\int_b^{\infty} \frac{\sqrt{u-b}}{(a b + b u - b + u) (a^2 + a u - b + u)^2}\,\mathrm{d}u $$
Now, letting $u = v^2+b$ we obtain integral of a rational function
$$\displaystyle F(a,b) ={2(a+b)\sqrt{a+b}}\int_0^{\infty} \frac{v^2}{(a^2 + a b + v^2 + a v^2)^2 (a b + b^2 + v^2 + b v^2)}\mathrm{d}v $$
So using partial fractions on the integrand we find
$$\displaystyle \frac{F(a,b)}{2(a+b)\sqrt{a+b}} = \frac{a}{a-b}I+\frac{(a+1)b}{(a-b)^2(a+b)}J -\frac{b(b+1)}{(a-b)^2(a+b)}K ~~~~~~~~~ (1)$$
where $I, J, K$ are the integrals $$\displaystyle I= \int_0^\infty \frac{1}{(a^2+ab+av^2+v^2)^2} \,\mathrm{d}v = \frac{\pi}{2} \cdot\frac{\sqrt{a (a + b)}}{2 a^2 \sqrt{1 + a} (a + b)^2}$$ 
$$J = \int_0^\infty \frac{1}{(a^2+ab+av^2+v^2)} \,\mathrm{d}v  = \frac{\pi}{2} \cdot \frac{1}{\sqrt{a (1 + a) (a + b)}}.$$ 
$$K = \int_0^\infty \frac{1}{(ab+b^2+bv^2+v^2)} \, \mathrm{d}v = \frac{\pi}{2} \cdot \frac{1}{\sqrt{b (1 + b) (a + b)}}. $$
Which routinely fall-out as standard arctangent integrals. Putting these values in $(1)$ we get: 
$$ F(a,b) = \frac{\pi}{2}\cdot \frac{a + b + 2 a b - 2 \sqrt{ab (1 + a) (1 + b)}}{2 \sqrt{a (1 + a)} (a - b)^2}.$$
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