I am attempting to simplify a horribly complicated equation that can be expressed in the form $\frac{ay^2}{(b(x-c))^d)}=z$. By plugging in x and y values I am able to find points to set up a system of equations to solve for a, b, c, and d. Simplifying the original equation via normal methods is not viable. Some points on the function: (0,25,30.8493);(0,50,123.3972);(0.75,100,18.2265);(0.75,125,28.4790);(1,75,2.5545);(1,125,7.0588); (1,25,0.2838). I have tried simplifying many times, most of which ended up in $0=1$, however, I have had 1 successful attempt, which resulted in $d=\frac{ln{\frac{z_1}{z_2}}}{ln{\frac{1-c}{c}}}$, where $z_1$ is the z-value of (0,25,30.8493) and $z_2$ is the z-value of (1,25,0.2838).
Therefore, I now have a triple system, but one that seems almost unsimplifiable, at least at my level, algebraically. I don't have access to graphing software capable of solving this graphically. This is a physics model, so I know that by physical contraints that $a$, $b$, and $d$ must be positive, and $c$ must be a negative number greater than -2.
The original equation is $f\left(d,i\right)=\frac{4\left(100i\right)^2}{10^7}\int_0^{\pi}\frac{\left(\frac{d}{0.25}\cos^2x\ +\left(1+\frac{d^2}{0.25^2}\right)\cos x-\frac{3d}{0.25}\right)T}{\sqrt{\left(\sqrt{1+\left(\frac{d}{0.25}\right)^2-\frac{2d}{0.25}\cos x}\right)^7}}dx\ +\frac{\left(100i\right)^2}{10^7}\int_0^{\pi}\frac{\left(\left(1+\frac{d}{0.25}\cos x\right)\left(\frac{d}{0.25}-\cos x\right)\left(1+0.24^2-\left(1+\left(\frac{d}{r}\right)^2-\frac{2d}{r}\cos x\right)\right)\right)O}{\sqrt{\left(\sqrt{1+\left(\frac{d}{0.25}\right)^2-\frac{2d}{0.25}\cos x}\right)^9}}dx$, where $T=\frac{1}{k}\left(\frac{\left(2-k^2\right)}{2}(\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1-k^2\sin^2a}}da)-(\int_0^{\frac{\pi}{2}}\sqrt{1-k^2\sin^2a}da)\right)$, $O=k\left(\frac{\left(2-k^2\right)}{2\left(1-k^2\right)}(\int_0^{\frac{\pi}{2}}\sqrt{1-k^2\sin^2a}da)-(\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1-k^2\sin^2a}}da)\right)$, and $k=\sqrt{\frac{4\sqrt{1+\left(\frac{d}{r}\right)^2-\frac{2d}{r}\cos x}}{\left(1+\sqrt{1+\left(\frac{d}{r}\right)^2-\frac{2d}{r}\cos x}\right)^2+0.24^2}}$.