Integrand size = 24, antiderivative size = 161 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {d (B d-A e) (c d-b e)^2 x}{e^5}-\frac {(B d-A e) (c d-b e)^2 x^2}{2 e^4}-\frac {\left (A c e (c d-2 b e)-B (c d-b e)^2\right ) x^3}{3 e^3}-\frac {c (B c d-2 b B e-A c e) x^4}{4 e^2}+\frac {B c^2 x^5}{5 e}-\frac {d^2 (B d-A e) (c d-b e)^2 \log (d+e x)}{e^6} \]
d*(-A*e+B*d)*(-b*e+c*d)^2*x/e^5-1/2*(-A*e+B*d)*(-b*e+c*d)^2*x^2/e^4-1/3*(A *c*e*(-2*b*e+c*d)-B*(-b*e+c*d)^2)*x^3/e^3-1/4*c*(-A*c*e-2*B*b*e+B*c*d)*x^4 /e^2+1/5*B*c^2*x^5/e-d^2*(-A*e+B*d)*(-b*e+c*d)^2*ln(e*x+d)/e^6
Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {60 d e (B d-A e) (c d-b e)^2 x+30 e^2 (-B d+A e) (c d-b e)^2 x^2+20 e^3 \left (B (c d-b e)^2+A c e (-c d+2 b e)\right ) x^3+15 c e^4 (-B c d+2 b B e+A c e) x^4+12 B c^2 e^5 x^5-60 d^2 (B d-A e) (c d-b e)^2 \log (d+e x)}{60 e^6} \]
(60*d*e*(B*d - A*e)*(c*d - b*e)^2*x + 30*e^2*(-(B*d) + A*e)*(c*d - b*e)^2* x^2 + 20*e^3*(B*(c*d - b*e)^2 + A*c*e*(-(c*d) + 2*b*e))*x^3 + 15*c*e^4*(-( B*c*d) + 2*b*B*e + A*c*e)*x^4 + 12*B*c^2*e^5*x^5 - 60*d^2*(B*d - A*e)*(c*d - b*e)^2*Log[d + e*x])/(60*e^6)
Time = 0.42 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1195, 2009}
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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^2}{d+e x} \, dx\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \int \left (-\frac {d^2 (B d-A e) (c d-b e)^2}{e^5 (d+e x)}+\frac {d (B d-A e) (c d-b e)^2}{e^5}+\frac {x (A e-B d) (b e-c d)^2}{e^4}+\frac {x^2 \left (B (c d-b e)^2-A c e (c d-2 b e)\right )}{e^3}+\frac {c x^3 (A c e+2 b B e-B c d)}{e^2}+\frac {B c^2 x^4}{e}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^2 (B d-A e) (c d-b e)^2 \log (d+e x)}{e^6}+\frac {d x (B d-A e) (c d-b e)^2}{e^5}-\frac {x^2 (B d-A e) (c d-b e)^2}{2 e^4}-\frac {x^3 \left (A c e (c d-2 b e)-B (c d-b e)^2\right )}{3 e^3}-\frac {c x^4 (-A c e-2 b B e+B c d)}{4 e^2}+\frac {B c^2 x^5}{5 e}\) |
(d*(B*d - A*e)*(c*d - b*e)^2*x)/e^5 - ((B*d - A*e)*(c*d - b*e)^2*x^2)/(2*e ^4) - ((A*c*e*(c*d - 2*b*e) - B*(c*d - b*e)^2)*x^3)/(3*e^3) - (c*(B*c*d - 2*b*B*e - A*c*e)*x^4)/(4*e^2) + (B*c^2*x^5)/(5*e) - (d^2*(B*d - A*e)*(c*d - b*e)^2*Log[d + e*x])/e^6
3.12.19.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 0.21 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.73
| method | result | size |
| norman | \(\frac {\left (A \,b^{2} e^{3}-2 A b c d \,e^{2}+A \,c^{2} d^{2} e -B \,b^{2} d \,e^{2}+2 B b c \,d^{2} e -B \,c^{2} d^{3}\right ) x^{2}}{2 e^{4}}+\frac {\left (2 A b c \,e^{2}-A \,c^{2} d e +B \,b^{2} e^{2}-2 B b c d e +B \,c^{2} d^{2}\right ) x^{3}}{3 e^{3}}+\frac {B \,c^{2} x^{5}}{5 e}+\frac {c \left (A c e +2 B b e -B c d \right ) x^{4}}{4 e^{2}}-\frac {d \left (A \,b^{2} e^{3}-2 A b c d \,e^{2}+A \,c^{2} d^{2} e -B \,b^{2} d \,e^{2}+2 B b c \,d^{2} e -B \,c^{2} d^{3}\right ) x}{e^{5}}+\frac {d^{2} \left (A \,b^{2} e^{3}-2 A b c d \,e^{2}+A \,c^{2} d^{2} e -B \,b^{2} d \,e^{2}+2 B b c \,d^{2} e -B \,c^{2} d^{3}\right ) \ln \left (e x +d \right )}{e^{6}}\) | \(279\) |
| default | \(-\frac {-\frac {1}{5} B \,c^{2} x^{5} e^{4}-\frac {1}{4} A \,c^{2} e^{4} x^{4}-\frac {1}{2} B b c \,e^{4} x^{4}+\frac {1}{4} B \,c^{2} d \,e^{3} x^{4}-\frac {2}{3} A b c \,e^{4} x^{3}+\frac {1}{3} A \,c^{2} d \,e^{3} x^{3}-\frac {1}{3} B \,b^{2} e^{4} x^{3}+\frac {2}{3} B b c d \,e^{3} x^{3}-\frac {1}{3} B \,c^{2} d^{2} e^{2} x^{3}-\frac {1}{2} A \,b^{2} e^{4} x^{2}+A b c d \,e^{3} x^{2}-\frac {1}{2} A \,c^{2} d^{2} e^{2} x^{2}+\frac {1}{2} B \,b^{2} d \,e^{3} x^{2}-B b c \,d^{2} e^{2} x^{2}+\frac {1}{2} B \,c^{2} d^{3} e \,x^{2}+A \,b^{2} d \,e^{3} x -2 A b c \,d^{2} e^{2} x +A \,c^{2} d^{3} e x -B \,b^{2} d^{2} e^{2} x +2 B b c \,d^{3} e x -B \,c^{2} d^{4} x}{e^{5}}+\frac {d^{2} \left (A \,b^{2} e^{3}-2 A b c d \,e^{2}+A \,c^{2} d^{2} e -B \,b^{2} d \,e^{2}+2 B b c \,d^{2} e -B \,c^{2} d^{3}\right ) \ln \left (e x +d \right )}{e^{6}}\) | \(330\) |
| risch | \(\frac {B \,c^{2} x^{5}}{5 e}+\frac {A \,c^{2} x^{4}}{4 e}+\frac {B \,b^{2} x^{3}}{3 e}+\frac {A \,b^{2} x^{2}}{2 e}+\frac {2 d^{4} \ln \left (e x +d \right ) B b c}{e^{5}}-\frac {2 d^{3} \ln \left (e x +d \right ) A b c}{e^{4}}-\frac {2 B b c d \,x^{3}}{3 e^{2}}-\frac {A b c d \,x^{2}}{e^{2}}+\frac {B b c \,d^{2} x^{2}}{e^{3}}+\frac {2 A b c \,d^{2} x}{e^{3}}-\frac {2 B b c \,d^{3} x}{e^{4}}+\frac {B \,c^{2} d^{4} x}{e^{5}}+\frac {d^{2} \ln \left (e x +d \right ) A \,b^{2}}{e^{3}}+\frac {d^{4} \ln \left (e x +d \right ) A \,c^{2}}{e^{5}}-\frac {d^{3} \ln \left (e x +d \right ) B \,b^{2}}{e^{4}}-\frac {d^{5} \ln \left (e x +d \right ) B \,c^{2}}{e^{6}}+\frac {B b c \,x^{4}}{2 e}-\frac {B \,c^{2} d \,x^{4}}{4 e^{2}}+\frac {2 A b c \,x^{3}}{3 e}-\frac {A \,c^{2} d \,x^{3}}{3 e^{2}}+\frac {B \,c^{2} d^{2} x^{3}}{3 e^{3}}+\frac {A \,c^{2} d^{2} x^{2}}{2 e^{3}}-\frac {B \,b^{2} d \,x^{2}}{2 e^{2}}-\frac {B \,c^{2} d^{3} x^{2}}{2 e^{4}}-\frac {A \,b^{2} d x}{e^{2}}-\frac {A \,c^{2} d^{3} x}{e^{4}}+\frac {B \,b^{2} d^{2} x}{e^{3}}\) | \(369\) |
| parallelrisch | \(\frac {-60 B \ln \left (e x +d \right ) c^{2} d^{5}+15 A \,x^{4} c^{2} e^{5}+20 B \,x^{3} b^{2} e^{5}+30 A \,x^{2} b^{2} e^{5}+12 B \,x^{5} c^{2} e^{5}-120 A \ln \left (e x +d \right ) b c \,d^{3} e^{2}+120 B \ln \left (e x +d \right ) b c \,d^{4} e -120 B x b c \,d^{3} e^{2}+120 A x b c \,d^{2} e^{3}+60 B \,x^{2} b c \,d^{2} e^{3}-60 A \,x^{2} b c d \,e^{4}-40 B \,x^{3} b c d \,e^{4}+40 A \,x^{3} b c \,e^{5}-20 A \,x^{3} c^{2} d \,e^{4}+20 B \,x^{3} c^{2} d^{2} e^{3}+30 A \,x^{2} c^{2} d^{2} e^{3}-30 B \,x^{2} b^{2} d \,e^{4}-30 B \,x^{2} c^{2} d^{3} e^{2}-60 A x \,b^{2} d \,e^{4}-60 A x \,c^{2} d^{3} e^{2}+60 B x \,b^{2} d^{2} e^{3}+60 B x \,c^{2} d^{4} e +60 A \ln \left (e x +d \right ) b^{2} d^{2} e^{3}+60 A \ln \left (e x +d \right ) c^{2} d^{4} e -60 B \ln \left (e x +d \right ) b^{2} d^{3} e^{2}+30 B \,x^{4} b c \,e^{5}-15 B \,x^{4} c^{2} d \,e^{4}}{60 e^{6}}\) | \(370\) |
1/2/e^4*(A*b^2*e^3-2*A*b*c*d*e^2+A*c^2*d^2*e-B*b^2*d*e^2+2*B*b*c*d^2*e-B*c ^2*d^3)*x^2+1/3/e^3*(2*A*b*c*e^2-A*c^2*d*e+B*b^2*e^2-2*B*b*c*d*e+B*c^2*d^2 )*x^3+1/5*B*c^2*x^5/e+1/4*c/e^2*(A*c*e+2*B*b*e-B*c*d)*x^4-d*(A*b^2*e^3-2*A *b*c*d*e^2+A*c^2*d^2*e-B*b^2*d*e^2+2*B*b*c*d^2*e-B*c^2*d^3)/e^5*x+d^2*(A*b ^2*e^3-2*A*b*c*d*e^2+A*c^2*d^2*e-B*b^2*d*e^2+2*B*b*c*d^2*e-B*c^2*d^3)/e^6* ln(e*x+d)
Time = 0.27 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.76 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {12 \, B c^{2} e^{5} x^{5} - 15 \, {\left (B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 20 \, {\left (B c^{2} d^{2} e^{3} - {\left (2 \, B b c + A c^{2}\right )} d e^{4} + {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} - 30 \, {\left (B c^{2} d^{3} e^{2} - A b^{2} e^{5} - {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} + {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 60 \, {\left (B c^{2} d^{4} e - A b^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x - 60 \, {\left (B c^{2} d^{5} - A b^{2} d^{2} e^{3} - {\left (2 \, B b c + A c^{2}\right )} d^{4} e + {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2}\right )} \log \left (e x + d\right )}{60 \, e^{6}} \]
1/60*(12*B*c^2*e^5*x^5 - 15*(B*c^2*d*e^4 - (2*B*b*c + A*c^2)*e^5)*x^4 + 20 *(B*c^2*d^2*e^3 - (2*B*b*c + A*c^2)*d*e^4 + (B*b^2 + 2*A*b*c)*e^5)*x^3 - 3 0*(B*c^2*d^3*e^2 - A*b^2*e^5 - (2*B*b*c + A*c^2)*d^2*e^3 + (B*b^2 + 2*A*b* c)*d*e^4)*x^2 + 60*(B*c^2*d^4*e - A*b^2*d*e^4 - (2*B*b*c + A*c^2)*d^3*e^2 + (B*b^2 + 2*A*b*c)*d^2*e^3)*x - 60*(B*c^2*d^5 - A*b^2*d^2*e^3 - (2*B*b*c + A*c^2)*d^4*e + (B*b^2 + 2*A*b*c)*d^3*e^2)*log(e*x + d))/e^6
Time = 0.33 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.74 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {B c^{2} x^{5}}{5 e} - \frac {d^{2} \left (- A e + B d\right ) \left (b e - c d\right )^{2} \log {\left (d + e x \right )}}{e^{6}} + x^{4} \left (\frac {A c^{2}}{4 e} + \frac {B b c}{2 e} - \frac {B c^{2} d}{4 e^{2}}\right ) + x^{3} \cdot \left (\frac {2 A b c}{3 e} - \frac {A c^{2} d}{3 e^{2}} + \frac {B b^{2}}{3 e} - \frac {2 B b c d}{3 e^{2}} + \frac {B c^{2} d^{2}}{3 e^{3}}\right ) + x^{2} \left (\frac {A b^{2}}{2 e} - \frac {A b c d}{e^{2}} + \frac {A c^{2} d^{2}}{2 e^{3}} - \frac {B b^{2} d}{2 e^{2}} + \frac {B b c d^{2}}{e^{3}} - \frac {B c^{2} d^{3}}{2 e^{4}}\right ) + x \left (- \frac {A b^{2} d}{e^{2}} + \frac {2 A b c d^{2}}{e^{3}} - \frac {A c^{2} d^{3}}{e^{4}} + \frac {B b^{2} d^{2}}{e^{3}} - \frac {2 B b c d^{3}}{e^{4}} + \frac {B c^{2} d^{4}}{e^{5}}\right ) \]
B*c**2*x**5/(5*e) - d**2*(-A*e + B*d)*(b*e - c*d)**2*log(d + e*x)/e**6 + x **4*(A*c**2/(4*e) + B*b*c/(2*e) - B*c**2*d/(4*e**2)) + x**3*(2*A*b*c/(3*e) - A*c**2*d/(3*e**2) + B*b**2/(3*e) - 2*B*b*c*d/(3*e**2) + B*c**2*d**2/(3* e**3)) + x**2*(A*b**2/(2*e) - A*b*c*d/e**2 + A*c**2*d**2/(2*e**3) - B*b**2 *d/(2*e**2) + B*b*c*d**2/e**3 - B*c**2*d**3/(2*e**4)) + x*(-A*b**2*d/e**2 + 2*A*b*c*d**2/e**3 - A*c**2*d**3/e**4 + B*b**2*d**2/e**3 - 2*B*b*c*d**3/e **4 + B*c**2*d**4/e**5)
Time = 0.20 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.75 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {12 \, B c^{2} e^{4} x^{5} - 15 \, {\left (B c^{2} d e^{3} - {\left (2 \, B b c + A c^{2}\right )} e^{4}\right )} x^{4} + 20 \, {\left (B c^{2} d^{2} e^{2} - {\left (2 \, B b c + A c^{2}\right )} d e^{3} + {\left (B b^{2} + 2 \, A b c\right )} e^{4}\right )} x^{3} - 30 \, {\left (B c^{2} d^{3} e - A b^{2} e^{4} - {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{2} + {\left (B b^{2} + 2 \, A b c\right )} d e^{3}\right )} x^{2} + 60 \, {\left (B c^{2} d^{4} - A b^{2} d e^{3} - {\left (2 \, B b c + A c^{2}\right )} d^{3} e + {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{2}\right )} x}{60 \, e^{5}} - \frac {{\left (B c^{2} d^{5} - A b^{2} d^{2} e^{3} - {\left (2 \, B b c + A c^{2}\right )} d^{4} e + {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2}\right )} \log \left (e x + d\right )}{e^{6}} \]
1/60*(12*B*c^2*e^4*x^5 - 15*(B*c^2*d*e^3 - (2*B*b*c + A*c^2)*e^4)*x^4 + 20 *(B*c^2*d^2*e^2 - (2*B*b*c + A*c^2)*d*e^3 + (B*b^2 + 2*A*b*c)*e^4)*x^3 - 3 0*(B*c^2*d^3*e - A*b^2*e^4 - (2*B*b*c + A*c^2)*d^2*e^2 + (B*b^2 + 2*A*b*c) *d*e^3)*x^2 + 60*(B*c^2*d^4 - A*b^2*d*e^3 - (2*B*b*c + A*c^2)*d^3*e + (B*b ^2 + 2*A*b*c)*d^2*e^2)*x)/e^5 - (B*c^2*d^5 - A*b^2*d^2*e^3 - (2*B*b*c + A* c^2)*d^4*e + (B*b^2 + 2*A*b*c)*d^3*e^2)*log(e*x + d)/e^6
Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (153) = 306\).
Time = 0.27 (sec) , antiderivative size = 338, normalized size of antiderivative = 2.10 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {12 \, B c^{2} e^{4} x^{5} - 15 \, B c^{2} d e^{3} x^{4} + 30 \, B b c e^{4} x^{4} + 15 \, A c^{2} e^{4} x^{4} + 20 \, B c^{2} d^{2} e^{2} x^{3} - 40 \, B b c d e^{3} x^{3} - 20 \, A c^{2} d e^{3} x^{3} + 20 \, B b^{2} e^{4} x^{3} + 40 \, A b c e^{4} x^{3} - 30 \, B c^{2} d^{3} e x^{2} + 60 \, B b c d^{2} e^{2} x^{2} + 30 \, A c^{2} d^{2} e^{2} x^{2} - 30 \, B b^{2} d e^{3} x^{2} - 60 \, A b c d e^{3} x^{2} + 30 \, A b^{2} e^{4} x^{2} + 60 \, B c^{2} d^{4} x - 120 \, B b c d^{3} e x - 60 \, A c^{2} d^{3} e x + 60 \, B b^{2} d^{2} e^{2} x + 120 \, A b c d^{2} e^{2} x - 60 \, A b^{2} d e^{3} x}{60 \, e^{5}} - \frac {{\left (B c^{2} d^{5} - 2 \, B b c d^{4} e - A c^{2} d^{4} e + B b^{2} d^{3} e^{2} + 2 \, A b c d^{3} e^{2} - A b^{2} d^{2} e^{3}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{6}} \]
1/60*(12*B*c^2*e^4*x^5 - 15*B*c^2*d*e^3*x^4 + 30*B*b*c*e^4*x^4 + 15*A*c^2* e^4*x^4 + 20*B*c^2*d^2*e^2*x^3 - 40*B*b*c*d*e^3*x^3 - 20*A*c^2*d*e^3*x^3 + 20*B*b^2*e^4*x^3 + 40*A*b*c*e^4*x^3 - 30*B*c^2*d^3*e*x^2 + 60*B*b*c*d^2*e ^2*x^2 + 30*A*c^2*d^2*e^2*x^2 - 30*B*b^2*d*e^3*x^2 - 60*A*b*c*d*e^3*x^2 + 30*A*b^2*e^4*x^2 + 60*B*c^2*d^4*x - 120*B*b*c*d^3*e*x - 60*A*c^2*d^3*e*x + 60*B*b^2*d^2*e^2*x + 120*A*b*c*d^2*e^2*x - 60*A*b^2*d*e^3*x)/e^5 - (B*c^2 *d^5 - 2*B*b*c*d^4*e - A*c^2*d^4*e + B*b^2*d^3*e^2 + 2*A*b*c*d^3*e^2 - A*b ^2*d^2*e^3)*log(abs(e*x + d))/e^6
Time = 10.73 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.91 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{d+e x} \, dx=x^4\,\left (\frac {A\,c^2+2\,B\,b\,c}{4\,e}-\frac {B\,c^2\,d}{4\,e^2}\right )+x^3\,\left (\frac {B\,b^2+2\,A\,c\,b}{3\,e}-\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e}-\frac {B\,c^2\,d}{e^2}\right )}{3\,e}\right )+x^2\,\left (\frac {A\,b^2}{2\,e}-\frac {d\,\left (\frac {B\,b^2+2\,A\,c\,b}{e}-\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e}-\frac {B\,c^2\,d}{e^2}\right )}{e}\right )}{2\,e}\right )-\frac {\ln \left (d+e\,x\right )\,\left (B\,b^2\,d^3\,e^2-A\,b^2\,d^2\,e^3-2\,B\,b\,c\,d^4\,e+2\,A\,b\,c\,d^3\,e^2+B\,c^2\,d^5-A\,c^2\,d^4\,e\right )}{e^6}-\frac {d\,x\,\left (\frac {A\,b^2}{e}-\frac {d\,\left (\frac {B\,b^2+2\,A\,c\,b}{e}-\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e}-\frac {B\,c^2\,d}{e^2}\right )}{e}\right )}{e}\right )}{e}+\frac {B\,c^2\,x^5}{5\,e} \]
x^4*((A*c^2 + 2*B*b*c)/(4*e) - (B*c^2*d)/(4*e^2)) + x^3*((B*b^2 + 2*A*b*c) /(3*e) - (d*((A*c^2 + 2*B*b*c)/e - (B*c^2*d)/e^2))/(3*e)) + x^2*((A*b^2)/( 2*e) - (d*((B*b^2 + 2*A*b*c)/e - (d*((A*c^2 + 2*B*b*c)/e - (B*c^2*d)/e^2)) /e))/(2*e)) - (log(d + e*x)*(B*c^2*d^5 - A*c^2*d^4*e - A*b^2*d^2*e^3 + B*b ^2*d^3*e^2 - 2*B*b*c*d^4*e + 2*A*b*c*d^3*e^2))/e^6 - (d*x*((A*b^2)/e - (d* ((B*b^2 + 2*A*b*c)/e - (d*((A*c^2 + 2*B*b*c)/e - (B*c^2*d)/e^2))/e))/e))/e + (B*c^2*x^5)/(5*e)