Are there tricks and tweaks for unevaluated integrals in SymPy?

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I try to verify some formula from a paper concerning mechanics by retracing their derivation, involving some integrals like these:

an example integral

The original source suggests that the solution is possible, but it seems quite tricky. I tried my best with SymPy (also with Maxima and Mathematica) fiddling around with different assumptions and simplifications, without any success:

import sympy as sp
from IPython.display import display, Math

x = sp.symbols('x', nonnegative=True, real=True, finite=True)
a = sp.symbols('a', positive=True, real=True, finite=True)
b = sp.symbols('b', positive=True, real=True, finite=True)
# print("assumptions for x:", x.assumptions0)
# print("assumptions for a, b:", r.assumptions0)

expr = x/(b + 2*(a-sp.sqrt(x)*sp.sqrt(2*a-x)))**3
expr = sp.expand(expr)
display(Math("e = "+sp.latex(expr)))

integ = sp.integrate(expr, (x, 0, a))
display(Math("integ = "+sp.latex(integ)))

SymPy returns my integrals unevaluated. Are there any other tricks and tweaks I could try to get a solution? The parameter a and b are physical dimensions (finite positive real numbers).

integral sympy python
1 Answers

The answer found using Rubi is

Mathematica graphics

(a^2*(2*a + b)^4*(Sqrt[2*a - Sqrt[-4*a - b]*Sqrt[b]]*Sqrt[2*a + Sqrt[-4*a - b]*Sqrt[b]]*
    (Sqrt[b]*(-2*a + b)*Sqrt[-(4*a + b)^2] + 6*a*Sqrt[-4*a - b]*(2*a + b)*ArcTan[(2*a)/(Sqrt[b]*Sqrt[4*a + b])]) - 
   6*a*(2*a + b)^2*Sqrt[4*a + b]*ArcTanh[Sqrt[2*a - Sqrt[-4*a - b]*Sqrt[b]]/Sqrt[2*a + Sqrt[-4*a - b]*Sqrt[b]]] + 
   6*a*(2*a + b)^2*Sqrt[4*a + b]*ArcTanh[Sqrt[2*a + Sqrt[-4*a - b]*Sqrt[b]]/Sqrt[2*a - Sqrt[-4*a - b]*Sqrt[b]]]))/
 (2*(2*a - Sqrt[-4*a - b]*Sqrt[b])^(5/2)*(2*a + Sqrt[-4*a - b]*Sqrt[b])^(5/2)*Sqrt[-4*a - b]*b^(5/2)*(4*a + b)^(5
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