Variation of load cable

import sympy as sp

q, p, H, dH, q1, q2, L = sp.symbols('q p H dH q1 q2 L',positive=True,real=True)
x = sp.symbols('x')
C1, C2, C3, C4 = sp.symbols('C1 C2 C3 C4')
q1 = q + p
q2 = q - p

z1 = sp.nsimplify((-1/(H+dH))*((1/2)*q1*x**2+C1+C2*x))
z2 = sp.nsimplify((-1/(H+dH))*((1/2)*q2*x**2+C3+C4*x))

V1 = (H+dH)*sp.diff(z1,x)
V2 = (H+dH)*sp.diff(z2,x)
eq1= sp.Eq(z1.subs(x,0),0)
eq2= sp.Eq(z1.subs(x,L/2),z2.subs(x,L/2))
eq3= sp.Eq(V1.subs(x,L/2),V2.subs(x,L/2))
eq4= sp.Eq(z2.subs(x,L),0)
sol = sp.solve((eq1,eq2,eq3,eq4),(C1,C2,C3,C4))
display(sol)

{C1: 0, C2: -L*p/4 - L*q/2, C3: -L**2*p/4, C4: 3*L*p/4 - L*q/2}

z1 = z1.subs(sol)
z2 = z2.subs(sol)
display(z1)
display(z2)

\[\displaystyle - \frac{x^{2} \left(\frac{p}{2} + \frac{q}{2}\right) + x \left(- \frac{L p}{4} - \frac{L q}{2}\right)}{H + dH}\]

\[\displaystyle - \frac{- \frac{L^{2} p}{4} + x^{2} \left(- \frac{p}{2} + \frac{q}{2}\right) + x \left(\frac{3 L p}{4} - \frac{L q}{2}\right)}{H + dH}\]

eq5 = sp.Eq(sp.integrate(1+sp.diff(z1,x)**2/2,(x,0,L/2)) + sp.integrate(1+sp.diff(z2,x)**2/2,(x,L/2,L)),L+q**2*L**3/(24*H**2))
dH_sol = sp.solve(eq5,dH)[0]
dH_sol

\[\displaystyle \frac{- 2 H q + H \sqrt{p^{2} + 4 q^{2}}}{2 q}\]

beta=dH_sol/H
sp.simplify(beta)

\[\displaystyle \frac{- q + \frac{\sqrt{p^{2} + 4 q^{2}}}{2}}{q}\]

beta.subs(p,1/4*q)

\[\displaystyle 0.00778221853731864\]

zk1 = z1
pp1 = zk1.subs([(x,L/4),(dH,dH_sol),(p,q/4)])/((3/32)*q*L**2/H)
sp.simplify(pp1.evalf())

\[\displaystyle 1.07496769977314\]