\(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{13/2}} \, dx\) [250]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 211 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=-\frac {5 \sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac {5 \sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{308 c^4 d^{13/2} \sqrt {a+b x+c x^2}} \] Output:

-5/308*(c*x^2+b*x+a)^(1/2)/c^3/d^5/(2*c*d*x+b*d)^(3/2)-5/154*(c*x^2+b*x+a) 
^(3/2)/c^2/d^3/(2*c*d*x+b*d)^(7/2)-1/11*(c*x^2+b*x+a)^(5/2)/c/d/(2*c*d*x+b 
*d)^(11/2)+5/308*(-4*a*c+b^2)^(1/4)*(-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^4/d^(13/2)/( 
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 = 10.07 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.52 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {5}{2},-\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{352 c^3 d^7 (b+2 c x)^6 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(13/2),x]
 

Output:

-1/352*((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeo 
metric2F1[-11/4, -5/2, -7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c^3*d^7*(b + 2 
*c*x)^6*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1108, 1108, 1108, 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.

\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {5 \int \frac {\left (c x^2+b x+a\right )^{3/2}}{(b d+2 c x d)^{9/2}}dx}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sqrt {c x^2+b x+a}}{(b d+2 c x d)^{5/2}}dx}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}\right )}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx}{6 c d^2}-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}\right )}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}\right )}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{6 c d^2 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}\right )}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}\right )}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}\)

\(\Big \downarrow \) 1113

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c^2 d^3 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}\right )}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}\right )}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c^2 d^{5/2} \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}\right )}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}\right )}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}\)

Input:

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(13/2),x]
 

Output:

-1/11*(a + b*x + c*x^2)^(5/2)/(c*d*(b*d + 2*c*d*x)^(11/2)) + (5*(-1/7*(a + 
 b*x + c*x^2)^(3/2)/(c*d*(b*d + 2*c*d*x)^(7/2)) + (3*(-1/3*Sqrt[a + b*x + 
c*x^2]/(c*d*(b*d + 2*c*d*x)^(3/2)) + ((b^2 - 4*a*c)^(1/4)*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*d^(5/2)*Sqrt[a + b*x + c*x^2])))/(14*c 
*d^2)))/(22*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 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. \(605\) vs. \(2(177)=354\).

Time = 8.89 (sec) , antiderivative size = 606, normalized size of antiderivative = 2.87

method result size
elliptic \(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{11264 c^{9} d^{7} \left (x +\frac {b}{2 c}\right )^{6}}-\frac {3 \left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{2464 c^{7} d^{7} \left (x +\frac {b}{2 c}\right )^{4}}-\frac {37 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{4928 d^{7} c^{5} \left (x +\frac {b}{2 c}\right )^{2}}+\frac {5 \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{308 d^{6} c^{3} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\) \(606\)
default \(\text {Expression too large to display}\) \(1035\)

Input:

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),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)* 
(-1/11264*(16*a^2*c^2-8*a*b^2*c+b^4)/c^9/d^7*(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)^6-3/2464/c^7*(4*a*c-b^2)/d^7*(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)^4-37/4928/d^ 
7/c^5*(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+5/308/d^6/c^3*(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 = 0.10 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.41 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\frac {5 \, \sqrt {2} {\left (64 \, c^{6} x^{6} + 192 \, b c^{5} x^{5} + 240 \, b^{2} c^{4} x^{4} + 160 \, b^{3} c^{3} x^{3} + 60 \, b^{4} c^{2} x^{2} + 12 \, b^{5} c x + b^{6}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) - 2 \, {\left (148 \, c^{6} x^{4} + 296 \, b c^{5} x^{3} + 5 \, b^{4} c^{2} + 10 \, a b^{2} c^{3} + 28 \, a^{2} c^{4} + 6 \, {\left (33 \, b^{2} c^{4} + 16 \, a c^{5}\right )} x^{2} + 2 \, {\left (25 \, b^{3} c^{3} + 48 \, a b c^{4}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{616 \, {\left (64 \, c^{11} d^{7} x^{6} + 192 \, b c^{10} d^{7} x^{5} + 240 \, b^{2} c^{9} d^{7} x^{4} + 160 \, b^{3} c^{8} d^{7} x^{3} + 60 \, b^{4} c^{7} d^{7} x^{2} + 12 \, b^{5} c^{6} d^{7} x + b^{6} c^{5} d^{7}\right )}} \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="fricas")
 

Output:

1/616*(5*sqrt(2)*(64*c^6*x^6 + 192*b*c^5*x^5 + 240*b^2*c^4*x^4 + 160*b^3*c 
^3*x^3 + 60*b^4*c^2*x^2 + 12*b^5*c*x + b^6)*sqrt(c^2*d)*weierstrassPInvers 
e((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c) - 2*(148*c^6*x^4 + 296*b*c^5*x^ 
3 + 5*b^4*c^2 + 10*a*b^2*c^3 + 28*a^2*c^4 + 6*(33*b^2*c^4 + 16*a*c^5)*x^2 
+ 2*(25*b^3*c^3 + 48*a*b*c^4)*x)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a) 
)/(64*c^11*d^7*x^6 + 192*b*c^10*d^7*x^5 + 240*b^2*c^9*d^7*x^4 + 160*b^3*c^ 
8*d^7*x^3 + 60*b^4*c^7*d^7*x^2 + 12*b^5*c^6*d^7*x + b^6*c^5*d^7)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\text {Timed out} \] Input:

integrate((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d)**(13/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {13}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(13/2), x)
 

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {13}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(13/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{13/2}} \,d x \] Input:

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(13/2),x)
 

Output:

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(13/2), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\text {too large to display} \] Input:

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(d)*( - 8*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**3*c**2 - 4*sqrt(b 
 + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**4 - 44*sqrt(b + 2*c*x)*sqrt(a + b*x 
+ c*x**2)*a*b**3*c*x - 132*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**2*c 
**2*x**2 - 176*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b*c**3*x**3 - 88*s 
qrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*c**4*x**4 + 2*sqrt(b + 2*c*x)*sqrt 
(a + b*x + c*x**2)*b**5*x + 6*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**4* 
c*x**2 + 8*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**3*c**2*x**3 + 4*sqrt( 
b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**2*c**3*x**4 + 4224*int((sqrt(b + 2*c* 
x)*sqrt(a + b*x + c*x**2)*x**2)/(22*a**2*b**7*c + 308*a**2*b**6*c**2*x + 1 
848*a**2*b**5*c**3*x**2 + 6160*a**2*b**4*c**4*x**3 + 12320*a**2*b**3*c**5* 
x**4 + 14784*a**2*b**2*c**6*x**5 + 9856*a**2*b*c**7*x**6 + 2816*a**2*c**8* 
x**7 - a*b**9 + 8*a*b**8*c*x + 246*a*b**7*c**2*x**2 + 1876*a*b**6*c**3*x** 
3 + 7448*a*b**5*c**4*x**4 + 17808*a*b**4*c**5*x**5 + 26656*a*b**3*c**6*x** 
6 + 24512*a*b**2*c**7*x**7 + 12672*a*b*c**8*x**8 + 2816*a*c**9*x**9 - b**1 
0*x - 15*b**9*c*x**2 - 98*b**8*c**2*x**3 - 364*b**7*c**3*x**4 - 840*b**6*c 
**4*x**5 - 1232*b**5*c**5*x**6 - 1120*b**4*c**6*x**7 - 576*b**3*c**7*x**8 
- 128*b**2*c**8*x**9),x)*a**4*b**6*c**5 + 50688*int((sqrt(b + 2*c*x)*sqrt( 
a + b*x + c*x**2)*x**2)/(22*a**2*b**7*c + 308*a**2*b**6*c**2*x + 1848*a**2 
*b**5*c**3*x**2 + 6160*a**2*b**4*c**4*x**3 + 12320*a**2*b**3*c**5*x**4 + 1 
4784*a**2*b**2*c**6*x**5 + 9856*a**2*b*c**7*x**6 + 2816*a**2*c**8*x**7 ...