# From the hruw paper:

# L= \partial^3 + \frac{21\left(x^2-x+1\right)}{25\,x^2\left(x^2-2\,x+1
# \right)}\partial - \frac{21\left(5\,x-2-3\,x^2+2\,x^3\right)}{50\,x^3\left(x^3
# -3\,x^2+3\,x-1\right)}

L:= DF^3 + (21*(x^2-x+1))/(25*x^2*(x^2-2*x+1))*DF
 - (21*(5*x-2-3*x^2+2*x^3))/(50*x^3*(x^3-3*x^2+3*x-1));

L:=collect(L,DF,factor);

# \begin{eqnarray*} && X^6 - 4\,\frac{(-1+2\,x)\,X^5}{x\,(x-1)} + \frac{
# \left(133\,x^2-133\,x+33\right)X^4}{5\,x^2\,(x-1)^2} \\ &&\quad\mbox{}
#  - \frac{12\,(7\,x-4)\,(-1+2\,x)\,(7\,x-3)\,X^3}{25\,x^3\,(x-1)^3}
#  \\ &&\quad\mbox{} + \frac{\left(351-11662\,x^3+5831\,x^4-2862\,x+8693
# \,x^2\right)X^2}{125\,x^4\,(x-1)^4} \\ &&\quad\mbox{} - \frac{4\,(-1+2
# \,x)\left(9604\,x^4-19208\,x^3+14275\,x^2-4671\,x+567\right)X}{3125\,x
# ^5\,(x-1)^5} \\ &&\quad\mbox{} + \frac{\left(16807\,x^4-33614\,x^3+
# 24907\,x^2-8100\,x+972\right)(-1+2\,x)^2}{12500\,x^6\,(x-1)^6}
# \end{eqnarray*} 

P:=
X^6 - 4*((-1+2*x)*X^5)/(x*(x-1)) + (
(133*x^2-133*x+33)*X^4)/(5*x^2*(x-1)^2) 
 - (12*(7*x-4)*(-1+2*x)*(7*x-3)*X^3)/(25*x^3*(x-1)^3)
 + ((351-11662*x^3+5831*x^4-2862*x+8693*x^2)*X^2)/(125*x^4*(x-1)^4)
 - (4*(-1+2*x)*(9604*x^4-19208*x^3+14275*x^2-4671*x+567)*X)/(3125*x^5*(x-1)^5)
+ ((16807*x^4-33614*x^3+24907*x^2-8100*x+972)*(-1+2*x)^2)/(12500*x^6*(x-1)^6) ;

with(algcurves);
N:=15;
r:=puiseux(numer(normal(P)),x=2,X,N);
r:=r[1];
Normalizer:=evala;
r:=series(exp(int(r,x)),x=2,N);
series( add(coeff(L,DF,i)*diff(r,x$i),i=1..3)+coeff(L,DF,0)*r ,x=2,N);
evala(Expand( convert(%,polynom) ));

# 0, so the example in our paper is correct.


# Note that RootOf(A,Y) is an algebraic solution of L, where A is:
A:=Y^12+5*x^4*(x-1)^4*Y^6-2*x^6*(x-1)^6*(x^2-x+1)*Y^2+5/4*x^8*(x-1)^8;

# The Galois group of this polynomial over C(x) has 60 elements.