Symmetry of certain improper integral Let $a < b < c < d$ be real numbers. Is it true that the following identity holds?  $$\int_a^b \frac{{\rm d}x}{\sqrt{|x-a||x-b||x-c||x-d|}}= \int_c^d \frac{{\rm d}x}{\sqrt{|x-a||x-b||x-c||x-d|}}$$
A friend of mine told me a friend of his said that it was true. But alas, we don't have a clue of how to attack this. We made a couple of cases in Mathematica and it seems to be true, unless we missed something simple.
 A: 
There're similar results as in Giovanni Mingari Scarpello, Daniele Ritelli, The hyperelliptic integrals and π, Journal of Number Theory 129 (2009) 3094–3108.

\begin{align}
  \int_{d}^{c} \frac{dx}{\sqrt{(a-x)(b-x)(c-x)(x-d)}} &=
  \frac{2}{\sqrt{(a-c)(b-d)}}
  K\left( \sqrt{\frac{(a-b)(c-d)}{(a-c)(b-d)}}\, \right) \\
  \int_{b}^{a} \frac{dx}{\sqrt{(a-x)(b-x)(c-x)(x-d)}} &=
  \frac{2}{\sqrt{(a-c)(b-d)}}
  K\left( \sqrt{\frac{(a-b)(c-d)}{(a-c)(b-d)}}\, \right) \\ \\
\end{align}
with convention $a>b>c>d$.  These can also be verified by Alan Jeffrey,‎ Daniel Zwillinger, Table of Integrals, Series, and Products.  Also,
$$  \int_{c}^{b} \frac{dx}{\sqrt{(a-x)(b-x)(x-c)(x-d)}} =
  \frac{2}{\sqrt{(a-c)(b-d)}}
  K\left( \sqrt{\frac{(b-c)(a-d)}{(a-c)(b-d)}}\, \right)
$$
A: A low level verification.
From the fact that the cross-ratios $(a,b;c,d)$ and $(c,d;a,b)$ are equal, it follows that the homography $f(t)=\frac{pt+q}{rt+s}$ that maps $a$ to $c$, $b$ to $d$ and $c$ to $a$, will also map $d$ to $b$. Solving for the coefficients gives $$p=bd-ac,\; q=abc-abd+acd-bcd,\; r=d-c+b-a,\;s=-p$$
The resulting $f(t)$ is its own inverse and has domain and range $\Bbb R \setminus \{t_0\}$, where $t_0=-\frac sr$.
Now $t_0-b=\frac{(b-a)(c-b)}{d-c+b-a}>0$, so $f$ is continuous on $[a,b]$ and maps that interval to $[c,d]$ (and back).
Transforming the first integral in the question by applying the parameter change $x=f(y)$ has the second integral as result because $$\frac{dx}{dy}=\frac{ps-rq}{(ry+s)^2}\qquad \mathrm {with}\qquad ps-rq=(b-a)(d-c)(c-b)(d-a)$$
$$f(y)-a=\frac{(b-a)(d-a)(y-c)}{ry+s}$$
$$f(y)-b=\frac{(b-a)(c-b)(y-d)}{ry+s}$$
$$f(y)-c=-\frac{(c-b)(d-c)(y-a)}{ry+s}$$
$$f(y)-d=-\frac{(d-a)(d-c)(y-b)}{ry+s}$$
A: Since $a<b<c<d$, we have two cases:
1) If $a<x<b$, then 
$$|x-a||x-b||x-c||x-d| = (x-a)(b-x)(c-x)(d-x) = (a-x)(b-x)(c-x)(x-d).$$
2) If $c<x<d$, then
$$|x-a||x-b||x-c||x-d| = (x-a)(x-b)(x-c)(d-x) = (a-x)(b-x)(c-x)(x-d).$$
Therefore,
$$\int_a^b \frac{1}{\sqrt{|x-a||x-b||x-c||x-d|}} \ dx = \int_a^b \frac{1}{\sqrt{(a-x)(b-x)(c-x)(x-d)}} \ dx$$
and
$$\int_c^d \frac{1}{\sqrt{|x-a||x-b||x-c||x-d|}} \ dx = \int_c^d \frac{1}{\sqrt{(a-x)(b-x)(c-x)(x-d)}} \ dx.$$
Now use the answer given by @Ng Chung Tak to conclude these two integrals are equal.
