Integrand size = 20, antiderivative size = 135 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=-\frac {2 (b d-2 a e+(2 c d-b e) x)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}+\frac {16 (2 c d-b e) (b+2 c x)}{15 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}}-\frac {128 c (2 c d-b e) (b+2 c x)}{15 \left (b^2-4 a c\right )^3 \sqrt {a+b x+c x^2}} \]
-2/5*(b*d-2*a*e+(-b*e+2*c*d)*x)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(5/2)+16/15*(-b *e+2*c*d)*(2*c*x+b)/(-4*a*c+b^2)^2/(c*x^2+b*x+a)^(3/2)-128/15*c*(-b*e+2*c* d)*(2*c*x+b)/(-4*a*c+b^2)^3/(c*x^2+b*x+a)^(1/2)
Time = 2.51 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.59 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\frac {-2 b^5 (3 d+5 e x)-64 c^2 \left (-3 a^3 e+15 a^2 c d x+20 a c^2 d x^3+8 c^3 d x^5\right )-32 b c^2 \left (15 a^2 (d-e x)-8 c^2 x^4 (-5 d+e x)-20 a c x^2 (-3 d+e x)\right )+32 b^2 c \left (3 a^2 e-15 a c x (d-2 e x)+10 c^2 x^3 (-3 d+2 e x)\right )+80 b^3 c \left (2 c x^2 (-d+3 e x)+a (d+3 e x)\right )+b^4 (-4 a e+20 c x (d+4 e x))}{15 \left (b^2-4 a c\right )^3 (a+x (b+c x))^{5/2}} \]
(-2*b^5*(3*d + 5*e*x) - 64*c^2*(-3*a^3*e + 15*a^2*c*d*x + 20*a*c^2*d*x^3 + 8*c^3*d*x^5) - 32*b*c^2*(15*a^2*(d - e*x) - 8*c^2*x^4*(-5*d + e*x) - 20*a *c*x^2*(-3*d + e*x)) + 32*b^2*c*(3*a^2*e - 15*a*c*x*(d - 2*e*x) + 10*c^2*x ^3*(-3*d + 2*e*x)) + 80*b^3*c*(2*c*x^2*(-d + 3*e*x) + a*(d + 3*e*x)) + b^4 *(-4*a*e + 20*c*x*(d + 4*e*x)))/(15*(b^2 - 4*a*c)^3*(a + x*(b + c*x))^(5/2 ))
Time = 0.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1159, 1089, 1088}
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.
| \(\displaystyle \int \frac {d+e x}{\left (a+b x+c x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1159 |
| \(\displaystyle -\frac {8 (2 c d-b e) \int \frac {1}{\left (c x^2+b x+a\right )^{5/2}}dx}{5 \left (b^2-4 a c\right )}-\frac {2 (-2 a e+x (2 c d-b e)+b d)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 1089 |
| \(\displaystyle -\frac {8 (2 c d-b e) \left (-\frac {8 c \int \frac {1}{\left (c x^2+b x+a\right )^{3/2}}dx}{3 \left (b^2-4 a c\right )}-\frac {2 (b+2 c x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\right )}{5 \left (b^2-4 a c\right )}-\frac {2 (-2 a e+x (2 c d-b e)+b d)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 1088 |
| \(\displaystyle -\frac {2 (-2 a e+x (2 c d-b e)+b d)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}-\frac {8 \left (\frac {16 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (b+2 c x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\right ) (2 c d-b e)}{5 \left (b^2-4 a c\right )}\) |
(-2*(b*d - 2*a*e + (2*c*d - b*e)*x))/(5*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(5 /2)) - (8*(2*c*d - b*e)*((-2*(b + 2*c*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^ 2)^(3/2)) + (16*c*(b + 2*c*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2]))) /(5*(b^2 - 4*a*c))
3.10.78.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[((b*d - 2*a*e + (2*c*d - b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b* x + c*x^2)^(p + 1), x] - Simp[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 - 4*a* c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] & & LtQ[p, -1] && NeQ[p, -3/2]
Leaf count of result is larger than twice the leaf count of optimal. \(257\) vs. \(2(123)=246\).
Time = 0.38 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.91
| method | result | size |
| default | \(d \left (\frac {\frac {4 c x}{5}+\frac {2 b}{5}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}+\frac {16 c \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{5 \left (4 a c -b^{2}\right )}\right )+e \left (-\frac {1}{5 c \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}-\frac {b \left (\frac {\frac {4 c x}{5}+\frac {2 b}{5}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}+\frac {16 c \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{5 \left (4 a c -b^{2}\right )}\right )}{2 c}\right )\) | \(258\) |
| trager | \(-\frac {2 \left (128 b \,c^{4} e \,x^{5}-256 c^{5} d \,x^{5}+320 b^{2} c^{3} e \,x^{4}-640 b \,c^{4} d \,x^{4}+320 a b \,c^{3} e \,x^{3}-640 a \,c^{4} d \,x^{3}+240 b^{3} c^{2} e \,x^{3}-480 b^{2} c^{3} d \,x^{3}+480 a \,b^{2} c^{2} e \,x^{2}-960 a b \,c^{3} d \,x^{2}+40 b^{4} c e \,x^{2}-80 b^{3} c^{2} d \,x^{2}+240 a^{2} b \,c^{2} e x -480 a^{2} c^{3} d x +120 a \,b^{3} c e x -240 a \,b^{2} c^{2} d x -5 b^{5} e x +10 b^{4} c d x +96 a^{3} c^{2} e +48 c e \,b^{2} a^{2}-240 a^{2} b \,c^{2} d -2 e \,b^{4} a +40 a \,b^{3} c d -3 b^{5} d \right )}{15 \left (4 a c -b^{2}\right )^{3} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}\) | \(266\) |
| gosper | \(-\frac {2 \left (128 b \,c^{4} e \,x^{5}-256 c^{5} d \,x^{5}+320 b^{2} c^{3} e \,x^{4}-640 b \,c^{4} d \,x^{4}+320 a b \,c^{3} e \,x^{3}-640 a \,c^{4} d \,x^{3}+240 b^{3} c^{2} e \,x^{3}-480 b^{2} c^{3} d \,x^{3}+480 a \,b^{2} c^{2} e \,x^{2}-960 a b \,c^{3} d \,x^{2}+40 b^{4} c e \,x^{2}-80 b^{3} c^{2} d \,x^{2}+240 a^{2} b \,c^{2} e x -480 a^{2} c^{3} d x +120 a \,b^{3} c e x -240 a \,b^{2} c^{2} d x -5 b^{5} e x +10 b^{4} c d x +96 a^{3} c^{2} e +48 c e \,b^{2} a^{2}-240 a^{2} b \,c^{2} d -2 e \,b^{4} a +40 a \,b^{3} c d -3 b^{5} d \right )}{15 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}\) | \(288\) |
d*(2/5*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(5/2)+16/5*c/(4*a*c-b^2)*(2/3*( 2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c *x^2+b*x+a)^(1/2)))+e*(-1/5/c/(c*x^2+b*x+a)^(5/2)-1/2*b/c*(2/5*(2*c*x+b)/( 4*a*c-b^2)/(c*x^2+b*x+a)^(5/2)+16/5*c/(4*a*c-b^2)*(2/3*(2*c*x+b)/(4*a*c-b^ 2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c*x^2+b*x+a)^(1/2)) ))
Leaf count of result is larger than twice the leaf count of optimal. 550 vs. \(2 (123) = 246\).
Time = 3.05 (sec) , antiderivative size = 550, normalized size of antiderivative = 4.07 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (128 \, {\left (2 \, c^{5} d - b c^{4} e\right )} x^{5} + 320 \, {\left (2 \, b c^{4} d - b^{2} c^{3} e\right )} x^{4} + 80 \, {\left (2 \, {\left (3 \, b^{2} c^{3} + 4 \, a c^{4}\right )} d - {\left (3 \, b^{3} c^{2} + 4 \, a b c^{3}\right )} e\right )} x^{3} + 40 \, {\left (2 \, {\left (b^{3} c^{2} + 12 \, a b c^{3}\right )} d - {\left (b^{4} c + 12 \, a b^{2} c^{2}\right )} e\right )} x^{2} + {\left (3 \, b^{5} - 40 \, a b^{3} c + 240 \, a^{2} b c^{2}\right )} d + 2 \, {\left (a b^{4} - 24 \, a^{2} b^{2} c - 48 \, a^{3} c^{2}\right )} e - 5 \, {\left (2 \, {\left (b^{4} c - 24 \, a b^{2} c^{2} - 48 \, a^{2} c^{3}\right )} d - {\left (b^{5} - 24 \, a b^{3} c - 48 \, a^{2} b c^{2}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{15 \, {\left (a^{3} b^{6} - 12 \, a^{4} b^{4} c + 48 \, a^{5} b^{2} c^{2} - 64 \, a^{6} c^{3} + {\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} x^{6} + 3 \, {\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} x^{5} + 3 \, {\left (b^{8} c - 11 \, a b^{6} c^{2} + 36 \, a^{2} b^{4} c^{3} - 16 \, a^{3} b^{2} c^{4} - 64 \, a^{4} c^{5}\right )} x^{4} + {\left (b^{9} - 6 \, a b^{7} c - 24 \, a^{2} b^{5} c^{2} + 224 \, a^{3} b^{3} c^{3} - 384 \, a^{4} b c^{4}\right )} x^{3} + 3 \, {\left (a b^{8} - 11 \, a^{2} b^{6} c + 36 \, a^{3} b^{4} c^{2} - 16 \, a^{4} b^{2} c^{3} - 64 \, a^{5} c^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{7} - 12 \, a^{3} b^{5} c + 48 \, a^{4} b^{3} c^{2} - 64 \, a^{5} b c^{3}\right )} x\right )}} \]
-2/15*(128*(2*c^5*d - b*c^4*e)*x^5 + 320*(2*b*c^4*d - b^2*c^3*e)*x^4 + 80* (2*(3*b^2*c^3 + 4*a*c^4)*d - (3*b^3*c^2 + 4*a*b*c^3)*e)*x^3 + 40*(2*(b^3*c ^2 + 12*a*b*c^3)*d - (b^4*c + 12*a*b^2*c^2)*e)*x^2 + (3*b^5 - 40*a*b^3*c + 240*a^2*b*c^2)*d + 2*(a*b^4 - 24*a^2*b^2*c - 48*a^3*c^2)*e - 5*(2*(b^4*c - 24*a*b^2*c^2 - 48*a^2*c^3)*d - (b^5 - 24*a*b^3*c - 48*a^2*b*c^2)*e)*x)*s qrt(c*x^2 + b*x + a)/(a^3*b^6 - 12*a^4*b^4*c + 48*a^5*b^2*c^2 - 64*a^6*c^3 + (b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*x^6 + 3*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*x^5 + 3*(b^8*c - 11*a*b^6 *c^2 + 36*a^2*b^4*c^3 - 16*a^3*b^2*c^4 - 64*a^4*c^5)*x^4 + (b^9 - 6*a*b^7* c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*x^3 + 3*(a*b^8 - 11* a^2*b^6*c + 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^4)*x^2 + 3*(a^2*b^7 - 12*a^3*b^5*c + 48*a^4*b^3*c^2 - 64*a^5*b*c^3)*x)
Timed out. \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 436 vs. \(2 (123) = 246\).
Time = 0.29 (sec) , antiderivative size = 436, normalized size of antiderivative = 3.23 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left ({\left (8 \, {\left (2 \, {\left (4 \, {\left (\frac {2 \, {\left (2 \, c^{5} d - b c^{4} e\right )} x}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}} + \frac {5 \, {\left (2 \, b c^{4} d - b^{2} c^{3} e\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x + \frac {5 \, {\left (6 \, b^{2} c^{3} d + 8 \, a c^{4} d - 3 \, b^{3} c^{2} e - 4 \, a b c^{3} e\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x + \frac {5 \, {\left (2 \, b^{3} c^{2} d + 24 \, a b c^{3} d - b^{4} c e - 12 \, a b^{2} c^{2} e\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x - \frac {5 \, {\left (2 \, b^{4} c d - 48 \, a b^{2} c^{2} d - 96 \, a^{2} c^{3} d - b^{5} e + 24 \, a b^{3} c e + 48 \, a^{2} b c^{2} e\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x + \frac {3 \, b^{5} d - 40 \, a b^{3} c d + 240 \, a^{2} b c^{2} d + 2 \, a b^{4} e - 48 \, a^{2} b^{2} c e - 96 \, a^{3} c^{2} e}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )}}{15 \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \]
-2/15*((8*(2*(4*(2*(2*c^5*d - b*c^4*e)*x/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^ 2 - 64*a^3*c^3) + 5*(2*b*c^4*d - b^2*c^3*e)/(b^6 - 12*a*b^4*c + 48*a^2*b^2 *c^2 - 64*a^3*c^3))*x + 5*(6*b^2*c^3*d + 8*a*c^4*d - 3*b^3*c^2*e - 4*a*b*c ^3*e)/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))*x + 5*(2*b^3*c^2*d + 24*a*b*c^3*d - b^4*c*e - 12*a*b^2*c^2*e)/(b^6 - 12*a*b^4*c + 48*a^2*b^2 *c^2 - 64*a^3*c^3))*x - 5*(2*b^4*c*d - 48*a*b^2*c^2*d - 96*a^2*c^3*d - b^5 *e + 24*a*b^3*c*e + 48*a^2*b*c^2*e)/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 6 4*a^3*c^3))*x + (3*b^5*d - 40*a*b^3*c*d + 240*a^2*b*c^2*d + 2*a*b^4*e - 48 *a^2*b^2*c*e - 96*a^3*c^2*e)/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c ^3))/(c*x^2 + b*x + a)^(5/2)
Time = 10.80 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.93 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\frac {x\,\left (\frac {4\,c^2\,d}{5\,\left (4\,a\,c^2-b^2\,c\right )}-\frac {2\,b\,c\,e}{5\,\left (4\,a\,c^2-b^2\,c\right )}\right )-\frac {4\,a\,c\,e}{5\,\left (4\,a\,c^2-b^2\,c\right )}+\frac {2\,b\,c\,d}{5\,\left (4\,a\,c^2-b^2\,c\right )}}{{\left (c\,x^2+b\,x+a\right )}^{5/2}}-\frac {x\,\left (\frac {2\,c^2\,\left (20\,b\,e-32\,c\,d\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}-\frac {8\,b\,c^2\,e}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}\right )+\frac {b\,c\,\left (20\,b\,e-32\,c\,d\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}-\frac {16\,a\,c^2\,e}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}}+\frac {\frac {b\,c\,\left (256\,c^2\,d-128\,b\,c\,e\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,{\left (4\,a\,c-b^2\right )}^2}+\frac {2\,c^2\,x\,\left (256\,c^2\,d-128\,b\,c\,e\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,{\left (4\,a\,c-b^2\right )}^2}}{\sqrt {c\,x^2+b\,x+a}}-\frac {4\,e}{\left (60\,a\,c-15\,b^2\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \]
(x*((4*c^2*d)/(5*(4*a*c^2 - b^2*c)) - (2*b*c*e)/(5*(4*a*c^2 - b^2*c))) - ( 4*a*c*e)/(5*(4*a*c^2 - b^2*c)) + (2*b*c*d)/(5*(4*a*c^2 - b^2*c)))/(a + b*x + c*x^2)^(5/2) - (x*((2*c^2*(20*b*e - 32*c*d))/(15*(4*a*c^2 - b^2*c)*(4*a *c - b^2)) - (8*b*c^2*e)/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2))) + (b*c*(20* b*e - 32*c*d))/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2)) - (16*a*c^2*e)/(15*(4* a*c^2 - b^2*c)*(4*a*c - b^2)))/(a + b*x + c*x^2)^(3/2) + ((b*c*(256*c^2*d - 128*b*c*e))/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2)^2) + (2*c^2*x*(256*c^2*d - 128*b*c*e))/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2)^2))/(a + b*x + c*x^2)^( 1/2) - (4*e)/((60*a*c - 15*b^2)*(a + b*x + c*x^2)^(3/2))