\(\int x (a+b x+c x^2)^{3/2} \, dx\) [77]

Optimal result

Integrand size = 16, antiderivative size = 136 \[ \int x \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {3 b \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^3}-\frac {b (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c}-\frac {3 b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2}} \] Output:

3/128*b*(-4*a*c+b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^3-1/16*b*(2*c*x+b)*(c 
*x^2+b*x+a)^(3/2)/c^2+1/5*(c*x^2+b*x+a)^(5/2)/c-3/256*b*(-4*a*c+b^2)^2*arc 
tanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.94 \[ \int x \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {\sqrt {a+x (b+c x)} \left (15 b^4-10 b^3 c x+128 c^2 \left (a+c x^2\right )^2+4 b^2 c \left (-25 a+2 c x^2\right )+8 b c^2 x \left (7 a+22 c x^2\right )\right )}{640 c^3}+\frac {3 b \left (b^2-4 a c\right )^2 \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{256 c^{7/2}} \] Input:

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

Output:

(Sqrt[a + x*(b + c*x)]*(15*b^4 - 10*b^3*c*x + 128*c^2*(a + c*x^2)^2 + 4*b^ 
2*c*(-25*a + 2*c*x^2) + 8*b*c^2*x*(7*a + 22*c*x^2)))/(640*c^3) + (3*b*(b^2 
 - 4*a*c)^2*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(256*c^(7/2) 
)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1160, 1087, 1087, 1092, 219}

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.

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1087

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {c x^2+b x+a}dx}{16 c}\right )}{2 c}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{16 c}\right )}{2 c}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c}\right )}{16 c}\right )}{2 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right )}{16 c}\right )}{2 c}\)

Input:

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

Output:

(a + b*x + c*x^2)^(5/2)/(5*c) - (b*(((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/ 
(8*c) - (3*(b^2 - 4*a*c)*(((b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(4*c) - ((b^ 
2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(3 
/2))))/(16*c)))/(2*c)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\)	\(128\)
risch	\(\frac {\left (128 c^{4} x^{4}+176 b \,c^{3} x^{3}+256 a \,c^{3} x^{2}+8 b^{2} c^{2} x^{2}+56 a b \,c^{2} x -10 b^{3} c x +128 a^{2} c^{2}-100 c a \,b^{2}+15 b^{4}\right ) \sqrt {c \,x^{2}+b x +a}}{640 c^{3}}-\frac {3 b \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {7}{2}}}\)	\(142\)

Input:

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

Output:

1/5*(c*x^2+b*x+a)^(5/2)/c-1/2*b/c*(1/8*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c+3/1 
6*(4*a*c-b^2)/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/ 
2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.51 \[ \int x \left (a+b x+c x^2\right )^{3/2} \, dx=\left [\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2560 \, c^{4}}, \frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1280 \, c^{4}}\right ] \] Input:

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

Output:

[1/2560*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(c)*log(-8*c^2*x^2 - 8*b* 
c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(128* 
c^5*x^4 + 176*b*c^4*x^3 + 15*b^4*c - 100*a*b^2*c^2 + 128*a^2*c^3 + 8*(b^2* 
c^3 + 32*a*c^4)*x^2 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a)) 
/c^4, 1/1280*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(-c)*arctan(1/2*sqrt 
(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(128*c 
^5*x^4 + 176*b*c^4*x^3 + 15*b^4*c - 100*a*b^2*c^2 + 128*a^2*c^3 + 8*(b^2*c 
^3 + 32*a*c^4)*x^2 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/ 
c^4]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (128) = 256\).

Time = 0.39 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.52 \[ \int x \left (a+b x+c x^2\right )^{3/2} \, dx=\begin {cases} \left (- \frac {a \left (\frac {47 a b}{40} - \frac {5 b \left (\frac {6 a c}{5} + \frac {3 b^{2}}{80}\right )}{6 c}\right )}{2 c} - \frac {b \left (a^{2} - \frac {2 a \left (\frac {6 a c}{5} + \frac {3 b^{2}}{80}\right )}{3 c} - \frac {3 b \left (\frac {47 a b}{40} - \frac {5 b \left (\frac {6 a c}{5} + \frac {3 b^{2}}{80}\right )}{6 c}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {a + b x + c x^{2}} \cdot \left (\frac {11 b x^{3}}{40} + \frac {c x^{4}}{5} + \frac {x^{2} \cdot \left (\frac {6 a c}{5} + \frac {3 b^{2}}{80}\right )}{3 c} + \frac {x \left (\frac {47 a b}{40} - \frac {5 b \left (\frac {6 a c}{5} + \frac {3 b^{2}}{80}\right )}{6 c}\right )}{2 c} + \frac {a^{2} - \frac {2 a \left (\frac {6 a c}{5} + \frac {3 b^{2}}{80}\right )}{3 c} - \frac {3 b \left (\frac {47 a b}{40} - \frac {5 b \left (\frac {6 a c}{5} + \frac {3 b^{2}}{80}\right )}{6 c}\right )}{4 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (- \frac {a \left (a + b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a + b x\right )^{\frac {7}{2}}}{7}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {a^{\frac {3}{2}} x^{2}}{2} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise(((-a*(47*a*b/40 - 5*b*(6*a*c/5 + 3*b**2/80)/(6*c))/(2*c) - b*(a* 
*2 - 2*a*(6*a*c/5 + 3*b**2/80)/(3*c) - 3*b*(47*a*b/40 - 5*b*(6*a*c/5 + 3*b 
**2/80)/(6*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c 
*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c 
) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + sqrt(a + b*x + c*x**2)*(11*b*x** 
3/40 + c*x**4/5 + x**2*(6*a*c/5 + 3*b**2/80)/(3*c) + x*(47*a*b/40 - 5*b*(6 
*a*c/5 + 3*b**2/80)/(6*c))/(2*c) + (a**2 - 2*a*(6*a*c/5 + 3*b**2/80)/(3*c) 
 - 3*b*(47*a*b/40 - 5*b*(6*a*c/5 + 3*b**2/80)/(6*c))/(4*c))/c), Ne(c, 0)), 
 (2*(-a*(a + b*x)**(5/2)/5 + (a + b*x)**(7/2)/7)/b**2, Ne(b, 0)), (a**(3/2 
)*x**2/2, True))
 

Maxima [F(-2)]

Exception generated. \[ \int x \left (a+b x+c x^2\right )^{3/2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.15 \[ \int x \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {1}{640} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c x + 11 \, b\right )} x + \frac {b^{2} c^{3} + 32 \, a c^{4}}{c^{4}}\right )} x - \frac {5 \, b^{3} c^{2} - 28 \, a b c^{3}}{c^{4}}\right )} x + \frac {15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3}}{c^{4}}\right )} + \frac {3 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {7}{2}}} \] Input:

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

Output:

1/640*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*c*x + 11*b)*x + (b^2*c^3 + 32*a*c^ 
4)/c^4)*x - (5*b^3*c^2 - 28*a*b*c^3)/c^4)*x + (15*b^4*c - 100*a*b^2*c^2 + 
128*a^2*c^3)/c^4) + 3/256*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*log(abs(2*(sqrt 
(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2)
 

Mupad [B] (verification not implemented)

Time = 11.31 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.46 \[ \int x \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{5\,c}-\frac {b\,\left (\frac {3\,a\,\left (\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{4\,c}\right )}{4}+\frac {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4}+\frac {b\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{8\,c}-\frac {3\,b^2\,\left (\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{4\,c}\right )}{16\,c}\right )}{2\,c} \] Input:

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

Output:

(a + b*x + c*x^2)^(5/2)/(5*c) - (b*((3*a*(log((b/2 + c*x)/c^(1/2) + (a + b 
*x + c*x^2)^(1/2))*(a/(2*c^(1/2)) - b^2/(8*c^(3/2))) + ((b + 2*c*x)*(a + b 
*x + c*x^2)^(1/2))/(4*c)))/4 + (x*(a + b*x + c*x^2)^(3/2))/4 + (b*(a + b*x 
 + c*x^2)^(3/2))/(8*c) - (3*b^2*(log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^ 
2)^(1/2))*(a/(2*c^(1/2)) - b^2/(8*c^(3/2))) + ((b + 2*c*x)*(a + b*x + c*x^ 
2)^(1/2))/(4*c)))/(16*c)))/(2*c)
 

Reduce [B] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.33 \[ \int x \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {256 \sqrt {c \,x^{2}+b x +a}\, a^{2} c^{3}-200 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} c^{2}+112 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{3} x +512 \sqrt {c \,x^{2}+b x +a}\, a \,c^{4} x^{2}+30 \sqrt {c \,x^{2}+b x +a}\, b^{4} c -20 \sqrt {c \,x^{2}+b x +a}\, b^{3} c^{2} x +16 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{3} x^{2}+352 \sqrt {c \,x^{2}+b x +a}\, b \,c^{4} x^{3}+256 \sqrt {c \,x^{2}+b x +a}\, c^{5} x^{4}-240 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a^{2} b \,c^{2}+120 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{3} c -15 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{5}}{1280 c^{4}} \] Input:

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

Output:

(256*sqrt(a + b*x + c*x**2)*a**2*c**3 - 200*sqrt(a + b*x + c*x**2)*a*b**2* 
c**2 + 112*sqrt(a + b*x + c*x**2)*a*b*c**3*x + 512*sqrt(a + b*x + c*x**2)* 
a*c**4*x**2 + 30*sqrt(a + b*x + c*x**2)*b**4*c - 20*sqrt(a + b*x + c*x**2) 
*b**3*c**2*x + 16*sqrt(a + b*x + c*x**2)*b**2*c**3*x**2 + 352*sqrt(a + b*x 
 + c*x**2)*b*c**4*x**3 + 256*sqrt(a + b*x + c*x**2)*c**5*x**4 - 240*sqrt(c 
)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a 
**2*b*c**2 + 120*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x 
)/sqrt(4*a*c - b**2))*a*b**3*c - 15*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + 
c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**5)/(1280*c**4)