\(\int \frac {(b d+2 c d x)^{7/2}}{(a+b x+c x^2)^{5/2}} \, dx\) [287]

Optimal result

Integrand size = 28, antiderivative size = 165 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {20 c d^3 \sqrt {b d+2 c d x}}{3 \sqrt {a+b x+c x^2}}+\frac {40 c \sqrt [4]{b^2-4 a c} d^{7/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 \sqrt {a+b x+c x^2}} \] Output:

-2/3*d*(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^(3/2)-20/3*c*d^3*(2*c*d*x+b*d)^(1 
/2)/(c*x^2+b*x+a)^(1/2)+40/3*c*(-4*a*c+b^2)^(1/4)*d^(7/2)*(-c*(c*x^2+b*x+a 
)/(-4*a*c+b^2))^(1/2)*EllipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^( 
1/2),I)/(c*x^2+b*x+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.17 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.74 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d^3 \sqrt {d (b+2 c x)} \left (b^2+14 b c x+2 c \left (5 a+7 c x^2\right )-20 c (a+x (b+c x)) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{3 (a+x (b+c x))^{3/2}} \] Input:

Integrate[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(-2*d^3*Sqrt[d*(b + 2*c*x)]*(b^2 + 14*b*c*x + 2*c*(5*a + 7*c*x^2) - 20*c*( 
a + x*(b + c*x))*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric 
2F1[1/4, 1/2, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(3*(a + x*(b + c*x))^(3/ 
2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1110, 1110, 1115, 1113, 762}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1110

\(\displaystyle \frac {10}{3} c d^2 \int \frac {(b d+2 c x d)^{3/2}}{\left (c x^2+b x+a\right )^{3/2}}dx-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1110

\(\displaystyle \frac {10}{3} c d^2 \left (2 c d^2 \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {10}{3} c d^2 \left (\frac {2 c d^2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{\sqrt {a+b x+c x^2}}-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1113

\(\displaystyle \frac {10}{3} c d^2 \left (\frac {4 d \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{\sqrt {a+b x+c x^2}}-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {10}{3} c d^2 \left (\frac {4 d^{3/2} \sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\sqrt {a+b x+c x^2}}-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

Input:

Int[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(-2*d*(b*d + 2*c*d*x)^(5/2))/(3*(a + b*x + c*x^2)^(3/2)) + (10*c*d^2*((-2* 
d*Sqrt[b*d + 2*c*d*x])/Sqrt[a + b*x + c*x^2] + (4*(b^2 - 4*a*c)^(1/4)*d^(3 
/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqrt[b*d 
 + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/Sqrt[a + b*x + c*x^2]))/3
 

Defintions of rubi rules used

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 1110

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), 
x] - Simp[d*e*((m - 1)/(b*(p + 1)))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x^ 
2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0] && N 
eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
 

rule 1113

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_ 
Symbol] :> Simp[(4/e)*Sqrt[-c/(b^2 - 4*a*c)]   Subst[Int[1/Sqrt[Simp[1 - b^ 
2*(x^4/(d^2*(b^2 - 4*a*c))), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, 
 c, d, e}, x] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]
 

rule 1115

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sym 
bol] :> Simp[Sqrt[(-c)*((a + b*x + c*x^2)/(b^2 - 4*a*c))]/Sqrt[a + b*x + c* 
x^2]   Int[(d + e*x)^m/Sqrt[(-a)*(c/(b^2 - 4*a*c)) - b*c*(x/(b^2 - 4*a*c)) 
- c^2*(x^2/(b^2 - 4*a*c))], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c* 
d - b*e, 0] && EqQ[m^2, 1/4]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(478\) vs. \(2(137)=274\).

Time = 7.99 (sec) , antiderivative size = 479, normalized size of antiderivative = 2.90

method	result	size
default	\(\frac {2 \left (10 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) c^{2} x^{2}+10 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b c x +10 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) a c -28 c^{3} x^{3}-42 b \,c^{2} x^{2}-20 a \,c^{2} x -16 b^{2} c x -10 a b c -b^{3}\right ) d^{3} \sqrt {d \left (2 c x +b \right )}}{3 \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(479\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \sqrt {d \left (2 c x +b \right )}\, \left (\frac {2 \left (4 a c -b^{2}\right ) d^{3} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{3 c^{2} \left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right )^{2}}-\frac {28 \left (2 c^{2} d x +d b c \right ) d^{3}}{3 \sqrt {\left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right ) \left (2 c^{2} d x +d b c \right )}}+\frac {40 c^{2} d^{4} \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{3 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, d}\)	\(544\)

Input:

int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/3*(10*(-4*a*c+b^2)^(1/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2) 
+b))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/ 
2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*2^(1/2)*(1/(-4*a*c+b^2)^(1/2 
)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2),2^(1/2))*c^2*x^2+10*(-4*a*c+b^2)^(1/ 
2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+b)/( 
-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2) 
)^(1/2)*EllipticF(1/2*2^(1/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1 
/2)+b))^(1/2),2^(1/2))*b*c*x+10*(-4*a*c+b^2)^(1/2)*(1/(-4*a*c+b^2)^(1/2)*( 
2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*( 
(-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*2^(1 
/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2),2^(1/2))*a*c 
-28*c^3*x^3-42*b*c^2*x^2-20*a*c^2*x-16*b^2*c*x-10*a*b*c-b^3)*d^3*(d*(2*c*x 
+b))^(1/2)/(2*c*x+b)/(c*x^2+b*x+a)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.12 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (10 \, \sqrt {2} {\left (c^{2} d^{3} x^{4} + 2 \, b c d^{3} x^{3} + 2 \, a b d^{3} x + {\left (b^{2} + 2 \, a c\right )} d^{3} x^{2} + a^{2} d^{3}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) - {\left (14 \, c^{2} d^{3} x^{2} + 14 \, b c d^{3} x + {\left (b^{2} + 10 \, a c\right )} d^{3}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \] Input:

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

2/3*(10*sqrt(2)*(c^2*d^3*x^4 + 2*b*c*d^3*x^3 + 2*a*b*d^3*x + (b^2 + 2*a*c) 
*d^3*x^2 + a^2*d^3)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 
1/2*(2*c*x + b)/c) - (14*c^2*d^3*x^2 + 14*b*c*d^3*x + (b^2 + 10*a*c)*d^3)* 
sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x 
+ (b^2 + 2*a*c)*x^2 + a^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((2*c*d*x+b*d)**(7/2)/(c*x**2+b*x+a)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {7}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate((2*c*d*x + b*d)^(7/2)/(c*x^2 + b*x + a)^(5/2), x)
 

Giac [F]

\[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {7}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")
 

Output:

integrate((2*c*d*x + b*d)^(7/2)/(c*x^2 + b*x + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \] Input:

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(5/2),x)
 

Output:

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {too large to display} \] Input:

int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

(2*sqrt(d)*d**3*(16*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**2*c - 16*s 
qrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b*c**2*x - 16*sqrt(b + 2*c*x)*sqrt 
(a + b*x + c*x**2)*a*c**3*x**2 + sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b* 
*4 + 24*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**3*c*x + 24*sqrt(b + 2*c* 
x)*sqrt(a + b*x + c*x**2)*b**2*c**2*x**2 + 160*int((sqrt(b + 2*c*x)*sqrt(a 
 + b*x + c*x**2)*x**2)/(2*a**4*b*c + 4*a**4*c**2*x - 3*a**3*b**3 + 18*a**3 
*b*c**2*x**2 + 12*a**3*c**3*x**3 - 9*a**2*b**4*x - 21*a**2*b**3*c*x**2 + 6 
*a**2*b**2*c**2*x**3 + 30*a**2*b*c**3*x**4 + 12*a**2*c**4*x**5 - 9*a*b**5* 
x**2 - 34*a*b**4*c*x**3 - 35*a*b**3*c**2*x**4 + 14*a*b*c**4*x**6 + 4*a*c** 
5*x**7 - 3*b**6*x**3 - 15*b**5*c*x**4 - 27*b**4*c**2*x**5 - 21*b**3*c**3*x 
**6 - 6*b**2*c**4*x**7),x)*a**5*c**5 - 320*int((sqrt(b + 2*c*x)*sqrt(a + b 
*x + c*x**2)*x**2)/(2*a**4*b*c + 4*a**4*c**2*x - 3*a**3*b**3 + 18*a**3*b*c 
**2*x**2 + 12*a**3*c**3*x**3 - 9*a**2*b**4*x - 21*a**2*b**3*c*x**2 + 6*a** 
2*b**2*c**2*x**3 + 30*a**2*b*c**3*x**4 + 12*a**2*c**4*x**5 - 9*a*b**5*x**2 
 - 34*a*b**4*c*x**3 - 35*a*b**3*c**2*x**4 + 14*a*b*c**4*x**6 + 4*a*c**5*x* 
*7 - 3*b**6*x**3 - 15*b**5*c*x**4 - 27*b**4*c**2*x**5 - 21*b**3*c**3*x**6 
- 6*b**2*c**4*x**7),x)*a**4*b**2*c**4 + 320*int((sqrt(b + 2*c*x)*sqrt(a + 
b*x + c*x**2)*x**2)/(2*a**4*b*c + 4*a**4*c**2*x - 3*a**3*b**3 + 18*a**3*b* 
c**2*x**2 + 12*a**3*c**3*x**3 - 9*a**2*b**4*x - 21*a**2*b**3*c*x**2 + 6*a* 
*2*b**2*c**2*x**3 + 30*a**2*b*c**3*x**4 + 12*a**2*c**4*x**5 - 9*a*b**5*...