Compute $\int_0^1 \frac{\sqrt{t(1-t)}}{a+(t-b)^2} \ dt$ for $a,b>0$ I would like to compute the integral
$$\int_0^1 \frac{\sqrt{t(1-t)}}{a+(t-b)^2} \ dt$$
where $a$ and $b>0$ are positive parameters.
Wolfram Alpha is able to provide an answer for the indefinite integral, but is struggeling with computing the definite one
https://www.wolframalpha.com/input?i=integrate+sqrt%28t*%281-t%29%29%2F%28a%2B%28t-b%29%5E2%29+dt+
 A: Hint Here's a sketch of one method: The substitution $$t = \frac{u^2}{(1 + u^2)} , \qquad dt = \frac{2 u \,du}{(1 + u^2)^2}$$ transforms the original integral in $t$ to the improper integral
$$2 \int_0^\infty \frac{u^2 \,du}{(1 + u^2) (A u^4 + B u^2 + C)},$$
where
$$A := a + (b - 1)^2, \quad B := 2 [a + b(b - 1)], \quad C := a + b^2.$$
If we write $q(v) := A v^2 + B v + C$, so that $A u^4 + B u^2 + C = q(u^2)$, the discriminant of $q$ is $\Delta(q) = -4a < 0$, hence all the roots of the quartic are complex, and we can factor
$$A u^4 + B u^2 + C = (D u^2 - E u + F) (D u^2 + E u + F) ,$$ so that $$A = D^2, \qquad B = 2 DF - E^2, \qquad C = F^2 ,$$ where $D u^2 \pm E u + F$ are irreducible, equivalently, $E^2 - 4 D F < 0$.
At this point we can take either of the following approaches:

*

*Use the Method of Partial Fractions explicitly to find that our integrand is
$$\frac{D (D - F) u + EF}{2 E \Lambda} \left(\frac{1}{D u^2 - E u + F} - \frac{1}{D u^2 + E u + F}\right) - \frac{1}{\Lambda} \cdot \frac{1}{1 + u^2},$$ where $\Lambda := (D - F)^2 + E^2$, after which we can integrate each term as usual.


*Use the fact that the integrand of our new integral is even to write it as
$$\int_{-\infty}^\infty \frac{u^2 \,du}{(1 + u^2) (A u^4 + B u^2 + C)} ,$$ and then apply a standard Residue Theorem argument.
In several special cases the computation simplifies significantly:

*

*When $b \in \{0, 1\}$: $$\pi\left(\sqrt{\frac{1}{2}\left(1 + \sqrt{1 + \frac{1}{a}}\right)} - 1\right)$$

*When $b = \frac{1}{2}$, $$\pi \left(\sqrt{1 + \frac{1}{4 a}} - 1\right)$$

*In the limiting case $a = 0$, $$\pi \left(\frac{2 b - 1}{2 \sqrt{b(b - 1)}} - 1\right) ,$$ provided that $b \not\in [0, 1]$.

*As $a \to \infty$, the integral decays as $$\frac{\pi}{8 a} - \frac{(16 b^2 - 16 b + 5) \pi}{128 a^2} + O\left(\frac{1}{a^3}\right) .$$

*As $b \to \infty$, the integral decays as $$\frac{\pi}{8 b^2} + \frac{\pi}{8 b^3} + O\left(\frac{1}{b^4}\right) .$$
A: Use the shorthands $$ p = a+b^2, \>\>\>q = a+(1-b)^2,\>\>\>s=a-b(1-b)$$
to integrate
\begin{align}
\int_0^1 \frac{\sqrt{t(1-t)}}{a+(t-b)^2} \ dt
\overset{ t=\frac1{1+x^2}}= & \ \int_0^\infty \frac{2x^2}{(px^4+2sx^2+q)(1+x^2)}dx\\
=& \int_0^\infty \frac{2(px^2+ q)}{px^4+2sx^2+q} -\frac2{1+x^2} \ dx\\
= &\ \frac{\pi}{\sqrt2}\frac{\sqrt p+\sqrt q}{\sqrt{\sqrt{pq}+s}}-\pi
\end{align}
A: $$\int_{0}^{1}\frac{\sqrt{t(1-t)}}{a+(t-b)^2}\,dt \stackrel{t\mapsto\frac{1+u}{2}}{=}\int_{-1}^{1}\frac{\sqrt{1-u^2}}{4a+\left(1+u-2b\right)^2}\,du $$
equals
$$ \int_{0}^{1}\sqrt{1-u^2}\left(\frac{1}{4a+(1-2b)^2+u^2+u(1-2b)}+\frac{1}{4a+(1-2b)^2+u^2-u(1-2b)}\right)\,du $$
or
$$ 2\int_{0}^{1}\sqrt{1-u^2}\left(\frac{4a+(1-2b)^2+u^2}{(4a+(1-2b)^2+u^2)^2-u^2(1-2b)^2}\right)\,du $$
which reduces to the integral of a rational function via $u\mapsto\cos\arctan t=\frac{1}{\sqrt{1+t^2}}.$
