\(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^{5/2}} \, dx\) [236]

Optimal result

Integrand size = 28, antiderivative size = 174 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{5/2}} \, dx=\frac {\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{6 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{6 c^3 d^{5/2} \sqrt {a+b x+c x^2}} \] Output:

1/6*(2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^(1/2)/c^2/d^3-1/3*(c*x^2+b*x+a)^(3/2 
)/c/d/(2*c*d*x+b*d)^(3/2)-1/6*(-4*a*c+b^2)^(5/4)*(-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 
^3/d^(5/2)/(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 = 10.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.57 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{5/2}} \, dx=\frac {\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{4},\frac {1}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{24 c^2 d (d (b+2 c x))^{3/2} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(5/2),x]
 

Output:

((b^2 - 4*a*c)*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-3/2, -3/4, 1/4, (b 
 + 2*c*x)^2/(b^2 - 4*a*c)])/(24*c^2*d*(d*(b + 2*c*x))^(3/2)*Sqrt[(c*(a + x 
*(b + c*x)))/(-b^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1108, 1109, 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.

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

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 1109

\(\displaystyle \frac {\frac {\sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{3 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx}{6 c}}{2 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {\frac {\sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{3 c d}-\frac {\left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{6 c \sqrt {a+b x+c x^2}}}{2 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}\)

\(\Big \downarrow \) 1113

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

\(\Big \downarrow \) 762

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

Input:

Int[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(5/2),x]
 

Output:

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

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 1108

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 1))), x] - Si 
mp[b*(p/(d*e*(m + 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] && NeQ[m + 2*p + 
3, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0] 
) && IntegerQ[2*p]
 

rule 1109

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x 
] - Simp[d*p*((b^2 - 4*a*c)/(b*e*(m + 2*p + 1)))   Int[(d + e*x)^m*(a + b*x 
 + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[2*c*d - b* 
e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1 
)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p])) && RationalQ[m] && 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. \(572\) vs. \(2(146)=292\).

Time = 4.57 (sec) , antiderivative size = 573, normalized size of antiderivative = 3.29

method	result	size
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{48 c^{4} d^{3} \left (x +\frac {b}{2 c}\right )^{2}}+\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{12 d^{3} c^{2}}+\frac {2 \left (\frac {8 a c -b^{2}}{16 d^{2} c^{2}}-\frac {4 a c -b^{2}}{48 c^{2} d^{2}}-\frac {a c d +\frac {1}{2} b^{2} d}{12 d^{3} c^{2}}\right ) \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(573\)
default	\(\frac {\left (8 \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, a \,c^{2} x -2 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b^{2} c x +4 \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, a b c -\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b^{3}+4 c^{4} x^{4}+8 b \,c^{3} x^{3}+6 b^{2} c^{2} x^{2}+2 b^{3} c x -4 a^{2} c^{2}+2 c a \,b^{2}\right ) \sqrt {d \left (2 c x +b \right )}}{12 d^{3} \sqrt {c \,x^{2}+b x +a}\, \left (2 c x +b \right )^{2} c^{3}}\)	\(626\)
risch	\(\text {Expression too large to display}\)	\(1287\)

Input:

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

(d*(2*c*x+b)*(c*x^2+b*x+a))^(1/2)/(d*(2*c*x+b))^(1/2)/(c*x^2+b*x+a)^(1/2)* 
(-1/48*(4*a*c-b^2)/c^4/d^3*(2*c^2*d*x^3+3*b*c*d*x^2+2*a*c*d*x+b^2*d*x+a*b* 
d)^(1/2)/(x+1/2*b/c)^2+1/12/d^3/c^2*(2*c^2*d*x^3+3*b*c*d*x^2+2*a*c*d*x+b^2 
*d*x+a*b*d)^(1/2)+2*(1/16*(8*a*c-b^2)/d^2/c^2-1/48*(4*a*c-b^2)/c^2/d^2-1/1 
2/d^3/c^2*(a*c*d+1/2*b^2*d))*(1/2/c*(-b+(-4*a*c+b^2)^(1/2))+1/2*(b+(-4*a*c 
+b^2)^(1/2))/c)*((x+1/2*(b+(-4*a*c+b^2)^(1/2))/c)/(1/2/c*(-b+(-4*a*c+b^2)^ 
(1/2))+1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2)*((x+1/2*b/c)/(-1/2*(b+(-4*a*c+ 
b^2)^(1/2))/c+1/2*b/c))^(1/2)*((x-1/2/c*(-b+(-4*a*c+b^2)^(1/2)))/(-1/2*(b+ 
(-4*a*c+b^2)^(1/2))/c-1/2/c*(-b+(-4*a*c+b^2)^(1/2))))^(1/2)/(2*c^2*d*x^3+3 
*b*c*d*x^2+2*a*c*d*x+b^2*d*x+a*b*d)^(1/2)*EllipticF(((x+1/2*(b+(-4*a*c+b^2 
)^(1/2))/c)/(1/2/c*(-b+(-4*a*c+b^2)^(1/2))+1/2*(b+(-4*a*c+b^2)^(1/2))/c))^ 
(1/2),((-1/2*(b+(-4*a*c+b^2)^(1/2))/c-1/2/c*(-b+(-4*a*c+b^2)^(1/2)))/(-1/2 
*(b+(-4*a*c+b^2)^(1/2))/c+1/2*b/c))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{5/2}} \, dx=-\frac {\sqrt {2} {\left (b^{4} - 4 \, a b^{2} c + 4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 4 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) - 2 \, {\left (2 \, c^{4} x^{2} + 2 \, b c^{3} x + b^{2} c^{2} - 2 \, a c^{3}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{12 \, {\left (4 \, c^{6} d^{3} x^{2} + 4 \, b c^{5} d^{3} x + b^{2} c^{4} d^{3}\right )}} \] Input:

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(5/2),x, algorithm="fricas")
 

Output:

-1/12*(sqrt(2)*(b^4 - 4*a*b^2*c + 4*(b^2*c^2 - 4*a*c^3)*x^2 + 4*(b^3*c - 4 
*a*b*c^2)*x)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2* 
c*x + b)/c) - 2*(2*c^4*x^2 + 2*b*c^3*x + b^2*c^2 - 2*a*c^3)*sqrt(2*c*d*x + 
 b*d)*sqrt(c*x^2 + b*x + a))/(4*c^6*d^3*x^2 + 4*b*c^5*d^3*x + b^2*c^4*d^3)
 

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{5/2}} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((c*x**2+b*x+a)**(3/2)/(2*c*d*x+b*d)**(5/2),x)
 

Output:

Integral((a + b*x + c*x**2)**(3/2)/(d*(b + 2*c*x))**(5/2), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{5/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(5/2),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(5/2), x)
 

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{5/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(5/2),x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{5/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{5/2}} \,d x \] Input:

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(5/2),x)
 

Output:

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{5/2}} \, dx=\text {too large to display} \] Input:

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(5/2),x)
 

Output:

(sqrt(d)*( - 12*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*c + 4*sqrt(b + 
 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**2 + 12*sqrt(b + 2*c*x)*sqrt(a + b*x + 
c*x**2)*a*b*c*x + 12*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*c**2*x**2 - 
2*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**3*x - 2*sqrt(b + 2*c*x)*sqrt(a 
 + b*x + c*x**2)*b**2*c*x**2 + 288*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x 
**2)*x**2)/(6*a**2*b**3*c + 36*a**2*b**2*c**2*x + 72*a**2*b*c**3*x**2 + 48 
*a**2*c**4*x**3 - a*b**5 + 30*a*b**3*c**2*x**2 + 100*a*b**2*c**3*x**3 + 12 
0*a*b*c**4*x**4 + 48*a*c**5*x**5 - b**6*x - 7*b**5*c*x**2 - 18*b**4*c**2*x 
**3 - 20*b**3*c**3*x**4 - 8*b**2*c**4*x**5),x)*a**3*b**2*c**4 + 1152*int(( 
sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(6*a**2*b**3*c + 36*a**2*b**2 
*c**2*x + 72*a**2*b*c**3*x**2 + 48*a**2*c**4*x**3 - a*b**5 + 30*a*b**3*c** 
2*x**2 + 100*a*b**2*c**3*x**3 + 120*a*b*c**4*x**4 + 48*a*c**5*x**5 - b**6* 
x - 7*b**5*c*x**2 - 18*b**4*c**2*x**3 - 20*b**3*c**3*x**4 - 8*b**2*c**4*x* 
*5),x)*a**3*b*c**5*x + 1152*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x* 
*2)/(6*a**2*b**3*c + 36*a**2*b**2*c**2*x + 72*a**2*b*c**3*x**2 + 48*a**2*c 
**4*x**3 - a*b**5 + 30*a*b**3*c**2*x**2 + 100*a*b**2*c**3*x**3 + 120*a*b*c 
**4*x**4 + 48*a*c**5*x**5 - b**6*x - 7*b**5*c*x**2 - 18*b**4*c**2*x**3 - 2 
0*b**3*c**3*x**4 - 8*b**2*c**4*x**5),x)*a**3*c**6*x**2 - 192*int((sqrt(b + 
 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(6*a**2*b**3*c + 36*a**2*b**2*c**2*x 
+ 72*a**2*b*c**3*x**2 + 48*a**2*c**4*x**3 - a*b**5 + 30*a*b**3*c**2*x**...