Assume that $a,b>0$ and the following two integrals both exist. Prove that $$\int_0^{+\infty}f\left(ax+\frac{b}{x}\right){\rm d}x=\frac{1}{a}\int_0^{+\infty}f(\sqrt{x^2+4ab}){\rm d}x.$$
How to make the substitution for $x$? For exmpale, if we let $x=:\dfrac{b}{at}$, we can change $f\left(ax+\dfrac{b}{x}\right)$ into $f\left(at+\dfrac{b}{t}\right)$, but this does not work with $f(\sqrt{x^2+4ab})$. Can anyone help?
Hint. One may observe that $$ ax+\frac{b}{x}=\sqrt{\left(ax-\frac{b}{x}\right)^2+4ab},\qquad x>0, $$ then use http://mathworld.wolfram.com/GlassersMasterTheorem.html.
Observe that application $x\rightarrow ax+\frac bx$ isn't injective. Observe that $$ \min_{x>0}\left(ax+\frac bx\right)=\sqrt{4ab} $$ and becomes injective in $I_1=\left(0, \sqrt{\frac ba}\right)$ and in $I_2=\left(\sqrt{\frac ba}, +\infty\right)$.
Now let $$ ax+\frac bx=\sqrt{t^2+4ab}\Leftrightarrow ax^2-x\sqrt{t^2+4ab}+b=0 $$ so $$ x_1=\frac{\sqrt{t^2+4ab}-t}{2a}\\ x_2=\frac{\sqrt{t^2+4ab}+t}{2a} $$ observe that $x_1\in I_1$ and $x_2\in I_2$ for every $t\in(0, +\infty)$ and these maps are biijective because their derivative doesn't change sign and $$ \begin{align*} \lim_{t\to +\infty}\frac{\sqrt{t^2+4ab}-t}{2a}&=\lim_{t\to +\infty}\frac{4ab}{2a\left(\sqrt{t^2+4ab}+t\right)}=0\\ \lim_{t\to +\infty}\frac{\sqrt{t^2+4ab}+t}{2a}&=+\infty \end{align*} $$
We have $$ \int^{+\infty}_0f\left(ax+\frac bx\right)dx=\int^{\sqrt{\frac ba}}_0f\left(ax+\frac bx\right)dx+\int^{+\infty}_{\sqrt{\frac ba}}f\left(ax+\frac bx\right)dx\\ =\int^0_{+\infty}f\left(\sqrt{t^2+4ab}\right)\frac{t-\sqrt{t^2+4ab}}{2a\sqrt{t^2+4ab}}dx+\int^{+\infty}_{0}f\left(\sqrt{t^2+4ab}\right)\frac{t+\sqrt{t^2+4ab}}{2a\sqrt{t^2+4ab}}dx\\ =-\int^{+\infty}_0f\left(\sqrt{t^2+4ab}\right)\frac{t-\sqrt{t^2+4ab}}{2a\sqrt{t^2+4ab}}dx+\int^{+\infty}_{0}f\left(\sqrt{t^2+4ab}\right)\frac{t+\sqrt{t^2+4ab}}{2a\sqrt{t^2+4ab}}dx\\ =\frac 1a\int^{+\infty}_0f\left(\sqrt{t^2+4ab}\right)dx $$
