Integrand size = 26, antiderivative size = 122 \[ \int (b+2 c x) (d+e x) \sqrt {a+b x+c x^2} \, dx=\frac {\left (b^2-4 a c\right ) e (b+2 c x) \sqrt {a+b x+c x^2}}{32 c^2}+\frac {(8 c d-b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 c}-\frac {\left (b^2-4 a c\right )^2 e \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{5/2}} \]
1/12*(6*c*e*x-b*e+8*c*d)*(c*x^2+b*x+a)^(3/2)/c-1/64*(-4*a*c+b^2)^2*e*arcta nh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)+1/32*(-4*a*c+b^2)*e* (2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^2
Time = 1.15 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.43 \[ \int (b+2 c x) (d+e x) \sqrt {a+b x+c x^2} \, dx=\frac {\left (64 a c^2 d+3 b^3 e-20 a b c e+64 b c^2 d x-2 b^2 c e x+24 a c^2 e x+64 c^3 d x^2+40 b c^2 e x^2+48 c^3 e x^3\right ) \sqrt {a+x (b+c x)}}{96 c^2}-\frac {\left (-b^2+4 a c\right )^2 e \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+b x+c x^2}}\right )}{32 c^{5/2} \sqrt {a+b x+c x^2}} \]
((64*a*c^2*d + 3*b^3*e - 20*a*b*c*e + 64*b*c^2*d*x - 2*b^2*c*e*x + 24*a*c^ 2*e*x + 64*c^3*d*x^2 + 40*b*c^2*e*x^2 + 48*c^3*e*x^3)*Sqrt[a + x*(b + c*x) ])/(96*c^2) - ((-b^2 + 4*a*c)^2*e*Sqrt[a + x*(b + c*x)]*ArcTanh[(Sqrt[c]*x )/(-Sqrt[a] + Sqrt[a + b*x + c*x^2])])/(32*c^(5/2)*Sqrt[a + b*x + c*x^2])
Time = 0.27 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1225, 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.
| \(\displaystyle \int (b+2 c x) (d+e x) \sqrt {a+b x+c x^2} \, dx\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {e \left (b^2-4 a c\right ) \int \sqrt {c x^2+b x+a}dx}{8 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (-b e+8 c d+6 c e x)}{12 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {e \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{8 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (-b e+8 c d+6 c e x)}{12 c}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {e \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c}\right )}{8 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (-b e+8 c d+6 c e x)}{12 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {e \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right )}{8 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (-b e+8 c d+6 c e x)}{12 c}\) |
((8*c*d - b*e + 6*c*e*x)*(a + b*x + c*x^2)^(3/2))/(12*c) + ((b^2 - 4*a*c)* e*(((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))))/(8*c)
3.16.49.3.1 Defintions of rubi rules used
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])
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])
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]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Time = 0.70 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.16
| method | result | size |
| risch | \(-\frac {\left (-48 c^{3} e \,x^{3}-40 b \,c^{2} e \,x^{2}-64 c^{3} d \,x^{2}-24 a \,c^{2} e x +2 b^{2} c e x -64 b d \,c^{2} x +20 a b c e -64 a \,c^{2} d -3 b^{3} e \right ) \sqrt {c \,x^{2}+b x +a}}{96 c^{2}}-\frac {e \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{64 c^{\frac {5}{2}}}\) | \(141\) |
| default | \(b d \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )+2 c e \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )+\left (b e +2 c d \right ) \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\) | \(353\) |
-1/96/c^2*(-48*c^3*e*x^3-40*b*c^2*e*x^2-64*c^3*d*x^2-24*a*c^2*e*x+2*b^2*c* e*x-64*b*c^2*d*x+20*a*b*c*e-64*a*c^2*d-3*b^3*e)*(c*x^2+b*x+a)^(1/2)-1/64*e *(16*a^2*c^2-8*a*b^2*c+b^4)/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^( 1/2))
Time = 0.44 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.81 \[ \int (b+2 c x) (d+e x) \sqrt {a+b x+c x^2} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} e \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, c^{4} e x^{3} + 64 \, a c^{3} d + 8 \, {\left (8 \, c^{4} d + 5 \, b c^{3} e\right )} x^{2} + {\left (3 \, b^{3} c - 20 \, a b c^{2}\right )} e + 2 \, {\left (32 \, b c^{3} d - {\left (b^{2} c^{2} - 12 \, a c^{3}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{3}}, \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} e \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (48 \, c^{4} e x^{3} + 64 \, a c^{3} d + 8 \, {\left (8 \, c^{4} d + 5 \, b c^{3} e\right )} x^{2} + {\left (3 \, b^{3} c - 20 \, a b c^{2}\right )} e + 2 \, {\left (32 \, b c^{3} d - {\left (b^{2} c^{2} - 12 \, a c^{3}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{192 \, c^{3}}\right ] \]
[1/384*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c)*e*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*(48*c^4 *e*x^3 + 64*a*c^3*d + 8*(8*c^4*d + 5*b*c^3*e)*x^2 + (3*b^3*c - 20*a*b*c^2) *e + 2*(32*b*c^3*d - (b^2*c^2 - 12*a*c^3)*e)*x)*sqrt(c*x^2 + b*x + a))/c^3 , 1/192*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-c)*e*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*(48*c^4*e*x^ 3 + 64*a*c^3*d + 8*(8*c^4*d + 5*b*c^3*e)*x^2 + (3*b^3*c - 20*a*b*c^2)*e + 2*(32*b*c^3*d - (b^2*c^2 - 12*a*c^3)*e)*x)*sqrt(c*x^2 + b*x + a))/c^3]
Leaf count of result is larger than twice the leaf count of optimal. 508 vs. \(2 (114) = 228\).
Time = 1.11 (sec) , antiderivative size = 508, normalized size of antiderivative = 4.16 \[ \int (b+2 c x) (d+e x) \sqrt {a+b x+c x^2} \, dx=\begin {cases} \sqrt {a + b x + c x^{2}} \left (\frac {c e x^{3}}{2} + \frac {x^{2} \cdot \left (\frac {5 b c e}{4} + 2 c^{2} d\right )}{3 c} + \frac {x \left (\frac {a c e}{2} + b^{2} e + 3 b c d - \frac {5 b \left (\frac {5 b c e}{4} + 2 c^{2} d\right )}{6 c}\right )}{2 c} + \frac {a b e + 2 a c d - \frac {2 a \left (\frac {5 b c e}{4} + 2 c^{2} d\right )}{3 c} + b^{2} d - \frac {3 b \left (\frac {a c e}{2} + b^{2} e + 3 b c d - \frac {5 b \left (\frac {5 b c e}{4} + 2 c^{2} d\right )}{6 c}\right )}{4 c}}{c}\right ) + \left (a b d - \frac {a \left (\frac {a c e}{2} + b^{2} e + 3 b c d - \frac {5 b \left (\frac {5 b c e}{4} + 2 c^{2} d\right )}{6 c}\right )}{2 c} - \frac {b \left (a b e + 2 a c d - \frac {2 a \left (\frac {5 b c e}{4} + 2 c^{2} d\right )}{3 c} + b^{2} d - \frac {3 b \left (\frac {a c e}{2} + b^{2} e + 3 b c d - \frac {5 b \left (\frac {5 b c e}{4} + 2 c^{2} d\right )}{6 c}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \cdot \left (\frac {2 c e \left (a + b x\right )^{\frac {7}{2}}}{7 b^{2}} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- 4 a c e + b^{2} e + 2 b c d\right )}{5 b^{2}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \cdot \left (2 a^{2} c e - a b^{2} e - 2 a b c d + b^{3} d\right )}{3 b^{2}}\right )}{b} & \text {for}\: b \neq 0 \\2 \sqrt {a} c \left (\frac {d x^{2}}{2} + \frac {e x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + b*x + c*x**2)*(c*e*x**3/2 + x**2*(5*b*c*e/4 + 2*c**2*d )/(3*c) + x*(a*c*e/2 + b**2*e + 3*b*c*d - 5*b*(5*b*c*e/4 + 2*c**2*d)/(6*c) )/(2*c) + (a*b*e + 2*a*c*d - 2*a*(5*b*c*e/4 + 2*c**2*d)/(3*c) + b**2*d - 3 *b*(a*c*e/2 + b**2*e + 3*b*c*d - 5*b*(5*b*c*e/4 + 2*c**2*d)/(6*c))/(4*c))/ c) + (a*b*d - a*(a*c*e/2 + b**2*e + 3*b*c*d - 5*b*(5*b*c*e/4 + 2*c**2*d)/( 6*c))/(2*c) - b*(a*b*e + 2*a*c*d - 2*a*(5*b*c*e/4 + 2*c**2*d)/(3*c) + b**2 *d - 3*b*(a*c*e/2 + b**2*e + 3*b*c*d - 5*b*(5*b*c*e/4 + 2*c**2*d)/(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)), Ne(c, 0)), (2*(2*c*e*(a + b*x)**(7/2)/(7*b**2) + (a + b*x)**(5/2)*(-4*a*c*e + b**2*e + 2*b*c*d)/(5*b**2) + (a + b*x)**(3/2) *(2*a**2*c*e - a*b**2*e - 2*a*b*c*d + b**3*d)/(3*b**2))/b, Ne(b, 0)), (2*s qrt(a)*c*(d*x**2/2 + e*x**3/3), True))
Exception generated. \[ \int (b+2 c x) (d+e x) \sqrt {a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.30 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.30 \[ \int (b+2 c x) (d+e x) \sqrt {a+b x+c x^2} \, dx=\frac {1}{96} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, c e x + \frac {8 \, c^{4} d + 5 \, b c^{3} e}{c^{3}}\right )} x + \frac {32 \, b c^{3} d - b^{2} c^{2} e + 12 \, a c^{3} e}{c^{3}}\right )} x + \frac {64 \, a c^{3} d + 3 \, b^{3} c e - 20 \, a b c^{2} e}{c^{3}}\right )} + \frac {{\left (b^{4} e - 8 \, a b^{2} c e + 16 \, a^{2} c^{2} e\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{64 \, c^{\frac {5}{2}}} \]
1/96*sqrt(c*x^2 + b*x + a)*(2*(4*(6*c*e*x + (8*c^4*d + 5*b*c^3*e)/c^3)*x + (32*b*c^3*d - b^2*c^2*e + 12*a*c^3*e)/c^3)*x + (64*a*c^3*d + 3*b^3*c*e - 20*a*b*c^2*e)/c^3) + 1/64*(b^4*e - 8*a*b^2*c*e + 16*a^2*c^2*e)*log(abs(2*( sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(5/2)
Time = 11.71 (sec) , antiderivative size = 395, normalized size of antiderivative = 3.24 \[ \int (b+2 c x) (d+e x) \sqrt {a+b x+c x^2} \, dx=\frac {e\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{2}-\frac {a\,e\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{2}-\frac {5\,b\,e\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{4}+\frac {d\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{8\,c^{3/2}}+\frac {d\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{12\,c}+b\,d\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {b\,e\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}+\frac {b\,d\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}+\frac {b\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}} \]
(e*x*(a + b*x + c*x^2)^(3/2))/2 - (a*e*((x/2 + b/(4*c))*(a + b*x + c*x^2)^ (1/2) + (log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4)) /(2*c^(3/2))))/2 - (5*b*e*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^ (1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x )*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/4 + (d*log((b + 2*c*x)/c^(1/2) + 2*( a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(8*c^(3/2)) + (d*(8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(12*c) + b*d*(x/2 + b/(4*c))* (a + b*x + c*x^2)^(1/2) + (b*e*(8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b* x + c*x^2)^(1/2))/(24*c^2) + (b*d*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x ^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2)) + (b*e*log((b + 2*c*x)/c^(1/2) + 2*( a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2))