# Approximate the relationship between 6 nonlinear functions involving elliptic integrals

I am trying to solve a physics-related problem, which results in approximating a relationship between 6 symbolic functions $F_1(\alpha,\beta ,x_0,y_0),F_2(\alpha,\beta ,x_0,y_0),...,F_6(\alpha,\beta ,x_0,y_0)$. The functions contain elliptic integral and their denominators are identical. My goal is to remove the variables $\alpha,\beta ,x_0,y_0$, and try to approximate the relationship with ideally a polynomial (example: $F_1^2 + F_2^2 + F_3^2 +F_4^2 + F_5^2 + F_6^2 = 1$), which can be used as a model to fit 6D data.

Following are two functions out of 6:

$F_1(\alpha,\beta ,x_0,y_0) = \int_{0}^{\frac{\pi}{2}}\frac{-c\sin(u)+c x_0\cos(u)\sin(u)-az_0 \cos(u)\sin(u)+c y_0\sin(u)^2-bz_0\sin(u)^2}{\sqrt{(bx_0-ay_0-b\cos(u)+a\sin(u))^2+(c+(-cx_0+az_0)\cos(u) + (-cy_0+bz_0)\sin(u))^2}}du\\ F_2(\alpha,\beta ,x_0,y_0) = \int_{0}^{\frac{\pi}{2}}\frac{bx_0 - ay_0 - b\cos(u)+ a\sin(u)}{\sqrt{(bx_0-ay_0-b\cos(u)+a\sin(u))^2+(c+(-cx_0+az_0)\cos(u) + (-cy_0+bz_0)\sin(u))^2}}du\\ \text{ where } a=\cos(\alpha)\cos(\beta), b= \cos(\alpha)\sin(\beta), c=\sin(\alpha), \alpha,\beta, x_0,y_0\in \mathbb{R}, z_0 =-\frac{y_0\sin(\beta) + x_0\cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}.$

Following is the code for the integrands of the six functions:

F1Integrand = (Sin[u]*(-c + (c*x0 - a*z0)*Cos[u] + (c*y0 - b*z0)*Sin[u]))/Sqrt[(b*x0 - a*y0 - b*Cos[u] + a*Sin[u])^2 + (c + ((-c)*x0 + a*z0)*Cos[u] + ((-c)*y0 + b*z0)*Sin[u])^2]
F2Integrand = (b*x0 - a*y0 - b*Cos[u] + a*Sin[u])/Sqrt[(b*x0 - a*y0 - b*Cos[u] + a*Sin[u])^2 + (c + ((-c)*x0 + a*z0)*Cos[u] + ((-c)*y0 + b*z0)*Sin[u])^2]
F3Integrand = -((Cos[u]*(-c + (c*x0 - a*z0)*Cos[u] + (c*y0 - b*z0)*Sin[u]))/Sqrt[(b*x0 - a*y0 - b*Cos[u] + a*Sin[u])^2 + (c + ((-c)*x0 + a*z0)*Cos[u] + ((-c)*y0 + b*z0)*Sin[u])^2])
F4Integrand = (Sin[u]*((-b)*x0 + a*y0 + b*Cos[u] - a*Sin[u]))/Sqrt[(b*x0 - a*y0 - b*Cos[u] + a*Sin[u])^2 + (c + ((-c)*x0 + a*z0)*Cos[u] + ((-c)*y0 + b*z0)*Sin[u])^2]
F5Integrand = -((Cos[u]*((-b)*x0 + a*y0 + b*Cos[u] - a*Sin[u]))/Sqrt[(b*x0 - a*y0 - b*Cos[u] + a*Sin[u])^2 + (c + ((-c)*x0 + a*z0)*Cos[u] + ((-c)*y0 + b*z0)*Sin[u])^2])
F6Integrand = (-c + (c*x0 - a*z0)*Cos[u] + (c*y0 - b*z0)*Sin[u])/Sqrt[(b*x0 - a*y0 - b*Cos[u] + a*Sin[u])^2 + (c + ((-c)*x0 + a*z0)*Cos[u] + ((-c)*y0 + b*z0)*Sin[u])^2]
a = Cos[alpha]*Cos[beta]
b = Cos[alpha]*Sin[beta]
c = Sin[alpha]
z0 = -(y0*Sin[alpha] + x0*Cos[alpha]*Sin[beta])/(Cos[alpha]*Cos[beta])


What I have tried so far, is to:

1. Solve the elliptic integral symbolically, then approximate the elliptic integrals.

1) I used tangent half-angle substitution and solved the elliptic integral based on the paper from Calson. The result involves all three kinds of incomplete elliptic integrals.

2) I tried to approximate them based on Taylor expansion and Pade approximation as suggested here. However, the approximation is only accurate around one point and the relationship differs a lot than the original ones.

2. I compute the lower bounds of $F_1,F_2,...F_6$, such that no elliptic integrals are involved.

1) Use the inequality: $\sqrt{a^2+b^2+c^2} \leq |a|+|b|+|c|$

$F_{1\text{(lowerBound)}} =\int_{0}^{\frac{\pi}{2}}\frac{-c\sin(u)+c x_0\cos(u)\sin(u)-az_0 \cos(u)\sin(u)+c y_0\sin(u)^2-bz_0\sin(u)^2}{|bx_0 - ay_0 - b\cos(u)+ a\sin(u)| + |c + (-cx_0 + az_0)\cos(u)+ (-cy_0 + b z_0)\sin(u)|}du$

After tangent half-angle substituion with $t = \tan(\frac{u}{2}),\sin(u)=\frac{2t}{1+t^2},\cos(u)=\frac{1-t^2}{1+t^2}$:

$F_{1\text{(lowerBound)}} = \int_{0}^{1}\frac{t(4c - 4cx_0+4az_0)+t^2(-8cy_0+8bz_0)+t^3(4c+4cx_0-4az_0)}{(1+t^2)^2(|b(-1+x_0+t^2(1+x_0))-a(y_0+t(-2+ty_0))|+ |c-cx_0+ct(t+tx_0-2y_0)+(a+2bt-at^2)z_0|)}dt$

The code for the integrands of the lowerbound after substitution:

F1LBIntegrand = -((4*t*(c - c*x0 + c*t*(t + t*x0 - 2*y0) + (a + 2*b*t - a*t^2)*z0))/((1 + t^2)^2*(Abs[b*(-1 + x0 + t^2*(1 + x0)) - a*(y0 + t*(-2 + t*y0))] + Abs[c - c*x0 + c*t*(t + t*x0 - 2*y0) + (a + 2*b*t -a*t^2)*z0])))
F2LBIntegrand = (2*(b*(-1 + x0 + t^2*(1 + x0)) - a*(y0 + t*(-2 + t*y0))))/((1 + t^2)*(Abs[b*(-1 + x0 + t^2*(1 + x0)) - a*(y0 + t*(-2 + t*y0))] + Abs[c - c*x0 + c*t*(t + t*x0 - 2*y0) + (a + 2*b*t -a*t^2)*z0]))
F3LBIntegrand = -((2*(-1 + t^2)*(c - c*x0 + c*t*(t + t*x0 - 2*y0) + (a + 2*b*t - a*t^2)*z0))/((1 + t^2)^2*(Abs[b*(-1 + x0 + t^2*(1 + x0)) - a*(y0 + t*(-2 + t*y0))] + Abs[c - c*x0 + c*t*(t + t*x0 - 2*y0) + (a + 2*b*t - a*t^2)*z0])))
F4LBIntegrand = (4*t*((-b)*(-1 + x0 + t^2*(1 + x0)) + a*(y0 + t*(-2 + t*y0))))/((1 + t^2)^2*(Abs[b*(-1 + x0 + t^2*(1 + x0)) - a*(y0 + t*(-2 + t*y0))] + Abs[c - c*x0 + c*t*(t + t*x0 - 2*y0) + (a + 2*b*t -a*t^2)*z0]))
F5LBIntegrand = -((2*(-1 + t^2)*(b*(-1 + x0 + t^2*(1 + x0)) - a*(y0 + t*(-2 + t*y0))))/((1 + t^2)^2*(Abs[b*(-1 + x0 + t^2*(1 + x0)) - a*(y0 + t*(-2 + t*y0))] + Abs[c - c*x0 + c*t*(t + t*x0 - 2*y0) + (a + 2*b*t - a*t^2)*z0])))
F6LBIntegrand = (2*(-c + c*x0 - c*t*(t + t*x0 - 2*y0) - 2*b*t*z0 + a*(-1 + t^2)*z0))/((1 + t^2)*(Abs[b*(-1 + x0 + t^2*(1 + x0)) - a*(y0 + t*(-2 + t*y0))] + Abs[c - c*x0 + c*t*(t + t*x0 - 2*y0) + (a + 2*b*t -a*t^2)*z0]))


2) Following are two indefinite integrals with one of the four cases when removing the abs.

F1LB=-4*(((-b)*c*x0 - a*c*t*x0 - c^2*x0^2 + a*c*y0 - b*c*t*y0 - c^2*y0^2 + a^2*t*z0 + b^2*t*z0 + 2*a*c*x0*z0 + 2*b*c*y0*z0 - a^2*z0^2 - b^2*z0^2)/(2*(1 + t^2)*(a^2 + b^2 + 2*b*c*x0 + c^2*x0^2 - 2*a*c*y0 + c^2*y0^2 - 2*a*c*x0*z0 - 2*b*c*y0*z0 + a^2*z0^2 + b^2*z0^2)) + ((a^3*c + a*b^2*c - 2*a*b^2*c*x0^2 - a*c^3*x0^2 - 2*a*b*c^2*x0^3 - 2*a^2*c^2*y0 - 2*b^2*c^2*y0 + 3*a^2*b*c*x0*y0 - b^3*c*x0*y0 - 2*b*c^3*x0*y0 +2*a^2*c^2*x0^2*y0 + b*c^3*x0^3*y0 - a^3*c*y0^2 + a*b^2*c*y0^2 +a*c^3*y0^2 - 2*a*b*c^2*x0*y0^2 - a*c^3*x0^2*y0^2 + 2*a^2*c^2*y0^3 +b*c^3*x0*y0^3 - a*c^3*y0^4 + 2*a^2*b*c*z0 + 2*b^3*c*z0 + a^2*b^2*x0*z0 +b^4*x0*z0 + 2*a^2*c^2*x0*z0 + 2*b^2*c^2*x0*z0 + 4*a^2*b*c*x0^2*z0 -b^2*c^2*x0^3*z0 - a^3*b*y0*z0 - a*b^3*y0*z0 - 4*a^3*c*x0*y0*z0 +4*a*b^2*c*x0*y0*z0 - a*b*c^2*x0^2*y0*z0 - 4*a^2*b*c*y0^2*z0 +2*a^2*c^2*x0*y0^2*z0 - 3*b^2*c^2*x0*y0^2*z0 + 3*a*b*c^2*y0^3*z0 -a^3*c*z0^2 - a*b^2*c*z0^2 - 2*a^3*b*x0*z0^2 - 2*a*b^3*x0*z0^2 +2*a*b^2*c*x0^2*z0^2 + 2*a^4*y0*z0^2 + 2*a^2*b^2*y0*z0^2 a^2*b*c*x0*y0*z0^2 + 3*b^3*c*x0*y0*z0^2 - a^3*c*y0^2*z0^2 - 3*a*b^2*c*y0^2*z0^2 - a^2*b^2*x0*z0^3 - b^4*x0*z0^3 + a^3*b*y0*z0^3 + a*b^3*y0*z0^3)*ArcTan[t])/(2*(a^2 + b^2 + 2*b*c*x0 + c^2*x0^2 - 2*a*c*y0 + c^2*y0^2 - 2*a*c*x0*z0 - 2*b*c*y0*z0 + a^2*z0^2 + b^2*z0^2)^2) + (((-a^3)*c^2 - a*b^2*c^2 - 2*a^3*b*c*x0 - 2*a*b^3*c*x0 - a^3*c^2*x0^2 - a*b^2*c^2*x0^2 + a*c^4*x0^2 + 2*a*b^3*c*x0^3 + 2*a*b^2*c^2*x0^4 - a*c^4*x0^4 + a^4*c*y0 - b^4*c*y0 + 2*a^2*c^3*y0 + 2*b^2*c^3*y0 + 2*b*c^4*x0*y0 - 5*a^2*b^2*c*x0^2*y0 + b^4*c*x0^2*y0 - a^2*c^3*x0^2*y0 - b^2*c^3*x0^2*y0 - 4*a^2*b*c^2*x0^3*y0 - 2*b*c^4*x0^3*y0 - b^2*c^3*x0^4*y0 - a^3*c^2*y0^2 - a*b^2*c^2*y0^2 - a*c^4*y0^2 + 4*a^3*b*c*x0*y0^2 - 2*a*b^3*c*x0*y0^2 + 2*a^3*c^2*x0^2*y0^2 + 2*a*b^2*c^2*x0^2*y0^2 + 2*a*b*c^3*x0^3*y0^2 - a^4*c*y0^3 + a^2*b^2*c*y0^3 - a^2*c^3*y0^3 - b^2*c^3*y0^3 - 4*a^2*b*c^2*x0*y0^3 - 2*b*c^4*x0*y0^3 - a^2*c^3*x0^2*y0^3 - b^2*c^3*x0^2*y0^3 + 2*a^3*c^2*y0^4 + a*c^4*y0^4 + 2*a*b*c^3*x0*y0^4 - a^2*c^3*y0^5 + a^4*b*z0 + 2*a^2*b^3*z0 + b^5*z0 - 2*a^2*b*c^2*z0 - 2*b^3*c^2*z0 + 2*a^4*c*x0*z0 + 2*a^2*b^2*c*x0*z0 - 2*a^2*c^3*x0*z0 - 2*b^2*c^3*x0*z0 - a^2*b^3*x0^2*z0 - b^5*x0^2*z0 + a^2*b*c^2*x0^2*z0 + b^3*c^2*x0^2*z0 - 4*a^2*b^2*c*x0^3*z0 + 4*a^2*c^3*x0^3*z0 + 2*b^2*c^3*x0^3*z0 + b^3*c^2*x0^4*z0 + 2*a^3*b*c*y0*z0 + 2*a*b^3*c*y0*z0 + 2*a^3*b^2*x0*y0*z0 + 2*a*b^4*x0*y0*z0 + 2*a^3*c^2*x0*y0*z0 + 2*a*b^2*c^2*x0*y0*z0 + 8*a^3*b*c*x0^2*y0*z0 - 4*a*b^3*c*x0^2*y0*z0 + 6*a*b*c^3*x0^2*y0*z0 - a^4*b*y0^2*z0 - a^2*b^3*y0^2*z0 + 3*a^2*b*c^2*y0^2*z0 + 3*b^3*c^2*y0^2*z0 - 4*a^4*c*x0*y0^2*z0 + 8*a^2*b^2*c*x0*y0^2*z0 + 6*b^2*c^3*x0*y0^2*z0 - 3*a^2*b*c^2*x0^2*y0^2*z0 + 3*b^3*c^2*x0^2*y0^2*z0 - 4*a^3*b*c*y0^3*z0 - 2*a*b*c^3*y0^3*z0 + 2*a^3*c^2*x0*y0^3*z0 - 6*a*b^2*c^2*x0*y0^3*z0 + 3*a^2*b*c^2*y0^4*z0 - a^5*z0^2 - 2*a^3*b^2*z0^2 - a*b^4*z0^2 + a^3*c^2*z0^2 + a*b^2*c^2*z0^2 - 2*a^3*b*c*x0*z0^2 - 2*a*b^3*c*x0*z0^2 + 2*a^3*b^2*x0^2*z0^2 + 2*a*b^4*x0^2*z0^2 - 6*a^3*c^2*x0^2*z0^2 - 6*a*b^2*c^2*x0^2*z0^2 - 2*a*b^3*c*x0^3*z0^2 - a^4*c*y0*z0^2 - 4*a^2*b^2*c*y0*z0^2 - 3*b^4*c*y0*z0^2 - 4*a^4*b*x0*y0*z0^2 - 4*a^2*b^3*x0*y0*z0^2 - 6*a^2*b*c^2*x0*y0*z0^2 - 6*b^3*c^2*x0*y0*z0^2 + 3*a^2*b^2*c*x0^2*y0*z0^2 - 3*b^4*c*x0^2*y0*z0^2 + 2*a^5*y0^2*z0^2 + 2*a^3*b^2*y0^2*z0^2 + 6*a*b^3*c*x0*y0^2*z0^2 - a^4*c*y0^3*z0^2 - 3*a^2*b^2*c*y0^3*z0^2 + a^4*b*z0^3 + 2*a^2*b^3*z0^3 + b^5*z0^3 + 4*a^4*c*x0*z0^3 + 6*a^2*b^2*c*x0*z0^3 + 2*b^4*c*x0*z0^3 + a^2*b^3*x0^2*z0^3 + b^5*x0^2*z0^3 + 2*a^3*b*c*y0*z0^3 + 2*a*b^3*c*y0*z0^3 - 2*a^3*b^2*x0*y0*z0^3 - 2*a*b^4*x0*y0*z0^3 + a^4*b*y0^2*z0^3 + a^2*b^3*y0^2*z0^3 - a^5*z0^4 - 2*a^3*b^2*z0^4 - a*b^4*z0^4)*ArcTan[(a + b*t + c*t + b*t*x0 + c*t*x0 -c*y0 - a*t*y0 + b*z0 - a*t*z0)/Sqrt[-a^2 - b^2 + c^2 + b^2*x0^2 -c^2*x0^2- 2*a*b*x0*y0 + a^2*y0^2 - c^2*y0^2 + 2*a*c*x0*z0 + 2*b*c*y0*z0 -a^2*z0^2 - b^2*z0^2]])/(2*Sqrt[-a^2 - b^2 + c^2 + b^2*x0^2 - c^2*x0^2 -2*a*b*x0*y0 + a^2*y0^2 - c^2*y0^2 + 2*a*c*x0*z0 + 2*b*c*y0*z0 - a^2*z0^2- b^2*z0^2]*(a^2 + b^2 + 2*b*c*x0 + c^2*x0^2 - 2*a*c*y0 +c^2*y0^2- 2*a*c*x0*z0 - 2*b*c*y0*z0 + a^2*z0^2 + b^2*z0^2)^2) + (((a^2)*b*c - b^3*c - 2*a^2*c^2*x0 - 2*b^2*c^2*x0 - a^2*b*c*x0^2 +b^3*c*x0^2 - b*c^3*x0^2 + 2*b^2*c^2*x0^3 + b*c^3*x0^4 + a^3*c*x0*y0 -3*a*b^2*c*x0*y0 + 2*a*c^3*x0*y0 - 2*a*b*c^2*x0^2*y0 - a*c^3*x0^3*y0 +2*a^2*b*c*y0^2 + b*c^3*y0^2 + 2*b^2*c^2*x0*y0^2 + b*c^3*x0^2*y0^2 -2*a*b*c^2*y0^3 - a*c^3*x0*y0^3 + 2*a^3*c*z0 + 2*a*b^2*c*z0 +a^3*b*x0*z0+a*b^3*x0*z0 - 4*a*b^2*c*x0^2*z0 - 3*a*b*c^2*x0^3*z0 -a^4*y0*z0 - a^2*b^2*y0*z0 - 2*a^2*c^2*y0*z0 - 2*b^2*c^2*y0*z0 +4*a^2*b*c*x0*y0*z0 - 4*b^3*c*x0*y0*z0 + 3*a^2*c^2*x0^2*y0*z0 -2*b^2*c^2*x0^2*y0*z0 + 4*a*b^2*c*y0^2*z0 + a*b*c^2*x0*y0^2*z0 +a^2*c^2*y0^3*z0 + a^2*b*c*z0^2 + b^3*c*z0^2 + 2*a^2*b^2*x0*z0^2 +2*b^4*x0*z0^2 + 3*a^2*b*c*x0^2*z0^2 + b^3*c*x0^2*z0^2 - 2*a^3*b*y0*z0^2-2*a*b^3*y0*z0^2 - 3*a^3*c*x0*y0*z0^2 + a*b^2*c*x0*y0*z0^2 -2*a^2*b*c*y0^2*z0^2 - a^3*b*x0*z0^3 - a*b^3*x0*z0^3 + a^4*y0*z0^3 +a^2*b^2*y0*z0^3)*Log[1 + t^2])/(4*(a^2 + b^2 + 2*b*c*x0 + c^2*x0^2 -2*a*c*y0 + c^2*y0^2 - 2*a*c*x0*z0 - 2*b*c*y0*z0 + a^2*z0^2 +b^2*z0^2)^2)+ ((a^2*b*c + b^3*c + 2*a^2*c^2*x0 + 2*b^2*c^2*x0 +a^2*b*c*x0^2 - b^3*c*x0^2 + b*c^3*x0^2 - 2*b^2*c^2*x0^3 - b*c^3*x0^4 -a^3*c*x0*y0 + 3*a*b^2*c*x0*y0 - 2*a*c^3*x0*y0 + 2*a*b*c^2*x0^2*y0 + a*c^3*x0^3*y0 - 2*a^2*b*c*y0^2 - b*c^3*y0^2 - 2*b^2*c^2*x0*y0^2 -b*c^3*x0^2*y0^2 + 2*a*b*c^2*y0^3 + a*c^3*x0*y0^3 - 2*a^3*c*z0 -2*a*b^2*c*z0 - a^3*b*x0*z0 - a*b^3*x0*z0 + 4*a*b^2*c*x0^2*z0 + 3*a*b*c^2*x0^3*z0 + a^4*y0*z0 + a^2*b^2*y0*z0 + 2*a^2*c^2*y0*z0 +2*b^2*c^2*y0*z0 - 4*a^2*b*c*x0*y0*z0 + 4*b^3*c*x0*y0*z0 -3*a^2*c^2*x0^2*y0*z0 + 2*b^2*c^2*x0^2*y0*z0 - 4*a*b^2*c*y0^2*z0 - a*b*c^2*x0*y0^2*z0 - a^2*c^2*y0^3*z0 - a^2*b*c*z0^2 - b^3*c*z0^2 -2*a^2*b^2*x0*z0^2 - 2*b^4*x0*z0^2 - 3*a^2*b*c*x0^2*z0^2 -b^3*c*x0^2*z0^2 + 2*a^3*b*y0*z0^2 + 2*a*b^3*y0*z0^2 + 3*a^3*c*x0*y0*z0^2 -a*b^2*c*x0*y0*z0^2 + 2*a^2*b*c*y0^2*z0^2 + a^3*b*x0*z0^3 + a*b^3*x0*z0^3- a^4*y0*z0^3 - a^2*b^2*y0*z0^3)*Log[-b + c + 2*a*t + b*t^2 + c*t^2 + b*x0- c*x0 + b*t^2*x0 + c*t^2*x0 - a*y0 - 2*c*t*y0 - a*t^2*y0 + a*z0 + 2*b*t*z0- a*t^2*z0])/(4*(a^2 + b^2 + 2*b*c*x0 + c^2*x0^2 - 2*a*c*y0 +c^2*y0^2 - 2*a*c*x0*z0 - 2*b*c*y0*z0 + a^2*z0^2 + b^2*z0^2)^2))
F2LB=(-2*(a^2 + b*(b + c*x0) - a*c*y0)*ArcTan[t] + (2*(a^3*y0*z0^2 - b*(c^2*x0*(-1 + x0^2 + y0^2) + b^2*x0*z0^2 + b*c*(-1 + x0^2 -2*x0*y0*z0)) - a^2*(b*x0*z0^2 + c*(-1 + y0^2 + 2*x0*y0*z0)) + a*(c^2*y0*(-1 + x0^2 + y0^2) + b^2*y0*z0^2 + 2*b*c*(x0*y0 + x0^2*z0 - y0^2*z0)))*ArcTan[(a + c*(t + t*x0 - y0) + b*(t + t*x0 + z0) - a*t*(y0 + z0))/Sqrt[(-c^2)*(-1 + x0^2 + y0^2) + 2*b*c*y0*z0 + a*(-2*b*x0*y0 + 2*c*x0*z0) + b^2*(-1 + x0^2 - z0^2) + a^2*(-1 + y0^2 - z0^2)]])/Sqrt[(-c^2)*(-1 + x0^2 + y0^2) + 2*b*c*y0*z0 + a*(-2*b*x0*y0 + 2*c*x0*z0) + b^2*(-1 + x0^2 - z0^2) + a^2*(-1 + y0^2 - z0^2)] + (a*c*x0 - a^2*z0 + b*(c*y0 - b*z0))*Log[1 + t^2] + ((-a)*c*x0 + a^2*z0 + b*((-c)*y0 + b*z0))*Log[c*(1 - x0 + t^2*(1 + x0) - 2*t*y0) + b*(-1 + x0 + t^2*(1 + x0) + 2*t*z0) - a*(-2*t + y0 - z0 + t^2*(y0 + z0))])/(c^2*(x0^2 + y0^2) - 2*a*c*(y0 + x0*z0) + 2*b*c*(x0 - y0*z0) + a^2*(1 + z0^2) + b^2*(1 + z0^2))
b - 2*a*t - b*t^2 - b*x0 - b*t^2*x0 + a*y0 + a*t^2*y0 > 0 && -c - c*t^2 + c*x0 - c*t^2*x0 + 2*c*t*y0 - a*z0 - 2*b*t*z0 + a*t^2*z0 > 0


To have an idea how the relationship should look like, I sampled the variables $\alpha,\beta ,x_0,y_0$ and compute the integral numerically. This is some examples of the relationship from 3 out of 6 functions, where blue dots are the integral of the original functions and orange dots are the lower bound.

Is there any way to further approximate the functions or the integrals? Or is it possible to approximate the relationship of the functions as a polynomial based on some observations of the integrated functions?

Thank you very much!

• We do not see the code, only the pictures. Can you publish the code? – Alex Trounev Jul 24 '18 at 14:20
• Could you provide the code for Figure 4? This looks like a TIE fighter. – DrLaplace Jul 24 '18 at 15:19