How prove this integral equation $\int_{a}^{b}\frac{1}{\sqrt{|f(x)|}}dx=\int_{c}^{d}\frac{1}{\sqrt{|f(x)|}}dx$

let $a>b>c>d$,and $$f(x)=(x-a)(x-b)(x-c)(x-d)$$ show that $$\int_{a}^{b}\dfrac{1}{\sqrt{|f(x)|}}dx=\int_{c}^{d}\dfrac{1}{\sqrt{|f(x)|}}dx$$

my try: maybe let $$u=x+( )$$ such when $x=a,b$ then $u=c,d$?

if $$a+d=b+c$$ then we take $$y=a+d-x$$ then we have $$\int_{a}^{b}\dfrac{1}{\sqrt{|f(x)|}}dx=\int_{c}^{d}\dfrac{1}{\sqrt{|f(x)|}}dx$$

other case, I can't,Thank you

You need a fractional linear (a.k.a. projective linear) change of variable $$ x' = \frac{rx+s}{tx+u}, $$ with coefficients $r,s,t,u$ chosen so that the map $x \mapsto x'$ is an involution taking $a,b,c,d$ to $c,d,a,b$. Explicitly you can take $$ r = ac-bd, \quad s = bcd - acd + abd - abc, \quad t = a - b + c - d, \quad u = bd-ac $$ (note that $u = -r$, which is necessary and sufficient for a fractional linear transformation to be its own functional inverse). This transformation takes each of the integrals $\int_a^b dx \, / \sqrt{|f(x)|}$ and $\int_c^d dx \, / \sqrt{|f(x)|}$ to the other, proving that they are equal.

The general integral represents the Schwartz-Christoffel transformation of the upper halfplane to a rectangle, and the two integrals represent the differences between the images of the points $a$ and $b$ and $c$ and $d$ respectively. Since the opposite sides of a rectangle are equal, the two integral is the same (I am not sure why @Braindead, who is clearly aware of this fact, post this as an answer, but someone should :)