Is there a closed form expression for the definite integral $$I=\int_b^c \frac{x^2}{\sqrt{(x-a)(x-b)(c-x)(d-x)}} dx$$ for $a<b<c<d$?
Mathematica 9.0 can do it for special cases using elliptic integrals, so I suspect that there may be a more general closed form solution. Some examples:
For $a=1$, $b=2$, $c=3$, $d=4$: $$I=-2 E\left(\frac34\right) - K\left( \frac34 \right) + 32 K\left( -3 \right) - 20\, \Pi(-1,-3)$$ For $a=1$, $b=2$, $c=3$, $d=5$: $$I=-\sqrt{6}\, E\left(\frac23\right) - \sqrt{6}\, K\left( \frac34 \right) + 25\sqrt{2}\, K\left( -2 \right) - \frac{33}{\sqrt{2}}\, \Pi\left(-\frac12,-2\right)$$ For $a=1$, $b=2$, $c=5$, $d=7$: $$I=-\sqrt{20}\, E\left(\frac9{10}\right) - \sqrt{5}\, K\left( \frac9{10} \right) + 49\sqrt{2}\, K\left( -9 \right) - \frac{75}{\sqrt{2}}\, \Pi\left(-\frac32,-9\right)$$ For $a=1$, $b=2$, $c=7$, $d=11$: $$I=-\sqrt{54}\, E\left(\frac{25}{27}\right) + \frac{103\sqrt{2}}{\sqrt{27}}\, K\left( \frac{25}{27} \right) - \frac{42\sqrt{2}}{\sqrt{27}}\, \Pi\left(-\frac59,\frac{25}{27}\right)$$
For $a < b < c < d$, let $I^{(n)}(a,b,c,d)$ and $J^{(n)}(a,d)$ be the integrals defined by:
$$ \begin{align} I^{(n)}(a,b,c,d) &= \int_b^c \frac{x^n dx}{\sqrt{(x-a)(x-b)(x-c)(x-d)}}\\ J^{(n)}(a,d) &= I^{(n)}(a,-1,1,d) \end{align} $$ By a change of variable $x = \left(\frac{c-b}{2}\right)\left( u + \frac{c+b}{c-b}\right)$, it is easy to check
$$I^{(n)}(a,b,c,d) = \left(\frac{c-b}{2}\right)^{n-1} \sum_{\ell=0}^n \binom{n}{\ell} J^{(n-\ell)}(\tilde{a},\tilde{d}) \left(\frac{c+b}{c-b}\right)^\ell $$
where $\tilde{a} = \frac{2a - (b+c)}{c-b}$ and $\tilde{d} = \frac{2d-(b+c)}{c-b}$. To get an expression for the integral we want, i.e. that for $I^{(2)}(a,b,c,d)$, we just need to figure out what $J^{(\ell)}(\tilde{a},\tilde{d})$ are for $\ell = 0, 1, 2$.
Consider another change of variables
$$x = \frac{y+\lambda}{1+\lambda y} \quad\longleftrightarrow\quad y = \frac{x-\lambda}{1-\lambda x}$$ where $\lambda$ is a root of following equation with $|\lambda| \le 1$:
$$\frac{\tilde{a} - \lambda}{1 - \lambda \tilde{a}} + \frac{\tilde{d} - \lambda}{1 - \lambda \tilde{d}} = 0 \quad\iff\quad \lambda^2 - 2\left(\frac{\tilde{a}\tilde{d}+1}{\tilde{a}+\tilde{d}}\right)\lambda + 1 = 0 $$ For the parameter ranges relevant to this question, we have $\lambda = \frac{\Lambda}{1+\sqrt{1-\Lambda^2}}$ where $\Lambda = \frac{\tilde{a}+\tilde{d}}{1+\tilde{a}\tilde{d}}$. Next, let $\;\displaystyle\mu = \frac{\tilde{d} - \lambda}{1 - \lambda \tilde{d}}\;$, we have $$ \tilde{a} = \frac{-\mu + \lambda}{1 -\mu\lambda}\quad\text{ and }\quad \tilde{d} = \frac{\mu + \lambda}{1 +\mu\lambda}$$ It is easy to check
$$\begin{align} x - 1 =& \frac{y + \lambda}{1+\lambda y}-1 = \frac{(y-1)(1-\lambda)}{1+\lambda y}\\ x + 1 =& \frac{y + \lambda}{1+\lambda y}+1 = \frac{(y+1)(1+\lambda)}{1+\lambda y}\\ x - \tilde{a} =& \frac{y + \lambda}{1+\lambda y}-\frac{-\mu + \lambda}{1 -\lambda\mu} = \frac{(y+\mu)(1-\lambda^2)}{(1+\lambda y)(1-\lambda\mu)}\\ x - \tilde{d} =& \frac{y + \lambda}{1+\lambda y}-\frac{\mu + \lambda}{1 + \lambda\mu} = \frac{(y-\mu)(1-\lambda^2)}{(1+\lambda y)(1+\lambda\mu)}\\ \end{align}$$ Since $dx = \frac{(1-\lambda^2)dy}{(1+\lambda y)^2}$, we get
$$\frac{dx}{\sqrt{(x-\tilde{a})(x^2-1)(x-\tilde{d})}} = \sqrt{\frac{1-\lambda^2\mu^2}{1-\lambda^2}}\frac{dy}{\sqrt{(1-y^2)(\mu^2-y^2)}} $$
Let $k = \frac{1}{\mu}$, $A(\lambda,\mu) = \frac{1}{\mu}\sqrt{\frac{1-\lambda^2\mu^2}{1-\lambda^2}} = \sqrt{\frac{k^2-\lambda^2}{1-\lambda^2}}$ and $P(y,k) = (1-y^2)(1-k^2y^2)$, we have
$$\begin{align} J^{(0)}(\tilde{a},\tilde{d}) = & 2A(\lambda,\mu)\int_0^1 \frac{dy}{\sqrt{P(y,k)}}\\ J^{(1)}(\tilde{a},\tilde{d}) = & A(\lambda,\mu)\int_0^1\frac{dy}{\sqrt{P(y,k)}} \left[ \frac{ y+\lambda}{1+\lambda y} +\frac{-y+\lambda}{1-\lambda y} \right]\\ J^{(2)}(\tilde{a},\tilde{d}) = & A(\lambda,\mu)\int_0^1\frac{dy}{\sqrt{P(y,k)}} \left[ \left(\frac{ y+\lambda}{1+\lambda y}\right)^2 +\left(\frac{-y+\lambda}{1-\lambda y}\right)^2 \right]\\ \end{align}$$
Let $\eta = \lambda^2$. In terms of following complete elliptic integrals, \begin{align} K(k) &= \int_0^1 \frac{dz}{\sqrt{(1-z^2)(1-k^2z^2)}}\\ E(k) &= \int_0^1 \sqrt{\frac{1-k^2 z^2}{1-z^2}} dz\\ \Pi(\eta,k) &= \int_0^1 \frac{dz}{(1 - \eta z^2)\sqrt{(1-z^2)(1-k^2 z^2)}} \end{align}
It is easy to see $$J^{(0}(\tilde{a},\tilde{d}) = 2A(\lambda,\mu) K(k).$$
Notice
$$\frac{y+\lambda}{1+\lambda y} +\frac{-y+\lambda}{1-\lambda y} = \frac{2\lambda(1-y^2)}{1-\lambda^2 y^2} = \frac{2}{\lambda}\left[1 - \frac{1-\lambda^2}{1-\lambda^2y^2}\right]$$ We get
$$J^{(1)}(\tilde{a},\tilde{d}) = \frac{2A(\lambda,\mu)}{\lambda}(K(k) - (1-\eta)\Pi(\eta,k))$$
Similarly, we have $$\left(\frac{ y+\lambda}{1+\lambda y}\right)^2 +\left(\frac{-y+\lambda}{1-\lambda y}\right)^2 = \frac{2}{\lambda^2}\left\{ 1 - \frac{(1-\eta)(3-\eta)}{1-\eta y^2} + 2\frac{(1-\eta)^2}{(1-\eta y^2)^2} \right\}\\ = \frac{2}{\lambda^2}\left\{ 1 - (1-\eta)\left[ (1+\eta)-2\eta(1-\eta)\frac{\partial}{\partial\eta} \right] \frac{1}{1-\eta y^2} \right\} $$ Form this, we find
$$\begin{align} J^{(2)}(\tilde{a},\tilde{d}) = &\frac{2A(\lambda,\mu)}{\lambda^2}\left[ K(k) - (1-\eta)\left( 1+\eta - 2\eta(1-\eta)\frac{\partial}{\partial\eta}\right)\Pi(\eta,k) \right]\\ = & 2A(\lambda,\mu)\left[ K(k) + \frac{1-\eta}{\eta-k^2}\left( E(k) - (1-k^2)\Pi(\eta,k) \right) \right] \end{align}$$
In the last expression, we have used following identity $$ \frac{\partial\Pi(\eta,k)}{\partial \eta} = \frac{1}{2\eta(k^2 - \eta)(\eta-1)}\left(\eta E(k) + (k^2 -\eta)K(k) + (\eta^2 - k^2)\Pi(\eta,k)\right) $$ to get rid of the derivatives of $\Pi(\eta,k)$.
Collect everything, we have
$$I^{(2)}(a,b,c,d) = (c-b)\sqrt{\frac{k^2-\lambda^2}{1-\lambda^2}} \left\{ \begin{align} & K(k) \left(\frac{c+b}{c-b}\right)^2\\ + & \frac{2}{\lambda}(K(k) - (1-\lambda^2)\Pi(\lambda^2,k)) \left(\frac{c+b}{c-b}\right)\\ + & K(k) + \frac{1-\lambda^2}{\lambda^2-k^2}\left( E(k) - (1-k^2)\Pi(\lambda^2,k) \right) \end{align} \right\} $$ Please note that in the case where $a+d = b+c$, $\tilde{a} + \tilde{d} = 0 \implies \lambda = 0$. Even though there is a term proportional to $\frac{2}{\lambda}$ in the second line of above expression, it won't cause any problem. This is because for small $\lambda$, $\Pi(\lambda^2,k) = K(k) + O(\lambda^2)$. That part actually vanishes in the limit of small $\lambda$.
I suspect above expression can be further simplified but I haven't figure out a good way to do that yet.