Integrand size = 24, antiderivative size = 238 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=-\frac {c (4 B c d-2 b B e-A c e) x}{e^5}+\frac {B c^2 x^2}{2 e^4}+\frac {d^2 (B d-A e) (c d-b e)^2}{3 e^6 (d+e x)^3}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{2 e^6 (d+e x)^2}-\frac {A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )}{e^6 (d+e x)}-\frac {\left (2 A c e (2 c d-b e)-B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )\right ) \log (d+e x)}{e^6} \]
-c*(-A*c*e-2*B*b*e+4*B*c*d)*x/e^5+1/2*B*c^2*x^2/e^4+1/3*d^2*(-A*e+B*d)*(-b *e+c*d)^2/e^6/(e*x+d)^3-1/2*d*(-b*e+c*d)*(B*d*(-3*b*e+5*c*d)-2*A*e*(-b*e+2 *c*d))/e^6/(e*x+d)^2+(-A*e*(b^2*e^2-6*b*c*d*e+6*c^2*d^2)+B*d*(3*b^2*e^2-12 *b*c*d*e+10*c^2*d^2))/e^6/(e*x+d)-(2*A*c*e*(-b*e+2*c*d)-B*(b^2*e^2-8*b*c*d *e+10*c^2*d^2))*ln(e*x+d)/e^6
Time = 0.10 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {6 c e (-4 B c d+2 b B e+A c e) x+3 B c^2 e^2 x^2+\frac {2 d^2 (B d-A e) (c d-b e)^2}{(d+e x)^3}-\frac {3 d (c d-b e) (B d (5 c d-3 b e)+2 A e (-2 c d+b e))}{(d+e x)^2}-\frac {6 \left (B d \left (-10 c^2 d^2+12 b c d e-3 b^2 e^2\right )+A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )\right )}{d+e x}+6 \left (2 A c e (-2 c d+b e)+B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )\right ) \log (d+e x)}{6 e^6} \]
(6*c*e*(-4*B*c*d + 2*b*B*e + A*c*e)*x + 3*B*c^2*e^2*x^2 + (2*d^2*(B*d - A* e)*(c*d - b*e)^2)/(d + e*x)^3 - (3*d*(c*d - b*e)*(B*d*(5*c*d - 3*b*e) + 2* A*e*(-2*c*d + b*e)))/(d + e*x)^2 - (6*(B*d*(-10*c^2*d^2 + 12*b*c*d*e - 3*b ^2*e^2) + A*e*(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2)))/(d + e*x) + 6*(2*A*c*e*( -2*c*d + b*e) + B*(10*c^2*d^2 - 8*b*c*d*e + b^2*e^2))*Log[d + e*x])/(6*e^6 )
Time = 0.50 (sec) , antiderivative size = 238, 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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \int \left (\frac {B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )-2 A c e (2 c d-b e)}{e^5 (d+e x)}+\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{e^5 (d+e x)^2}-\frac {d^2 (B d-A e) (c d-b e)^2}{e^5 (d+e x)^4}+\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^5 (d+e x)^3}+\frac {c (A c e+2 b B e-4 B c d)}{e^5}+\frac {B c^2 x}{e^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{e^6 (d+e x)}-\frac {\log (d+e x) \left (2 A c e (2 c d-b e)-B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )\right )}{e^6}+\frac {d^2 (B d-A e) (c d-b e)^2}{3 e^6 (d+e x)^3}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{2 e^6 (d+e x)^2}-\frac {c x (-A c e-2 b B e+4 B c d)}{e^5}+\frac {B c^2 x^2}{2 e^4}\) |
-((c*(4*B*c*d - 2*b*B*e - A*c*e)*x)/e^5) + (B*c^2*x^2)/(2*e^4) + (d^2*(B*d - A*e)*(c*d - b*e)^2)/(3*e^6*(d + e*x)^3) - (d*(c*d - b*e)*(B*d*(5*c*d - 3*b*e) - 2*A*e*(2*c*d - b*e)))/(2*e^6*(d + e*x)^2) - (A*e*(6*c^2*d^2 - 6*b *c*d*e + b^2*e^2) - B*d*(10*c^2*d^2 - 12*b*c*d*e + 3*b^2*e^2))/(e^6*(d + e *x)) - ((2*A*c*e*(2*c*d - b*e) - B*(10*c^2*d^2 - 8*b*c*d*e + b^2*e^2))*Log [d + e*x])/e^6
3.12.22.3.1 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 = 0.23 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.23
| method | result | size |
| default | \(\frac {c \left (\frac {1}{2} B c e \,x^{2}+A c e x +2 B b e x -4 B c d x \right )}{e^{5}}-\frac {d^{2} \left (A \,b^{2} e^{3}-2 A b c d \,e^{2}+A \,c^{2} d^{2} e -B \,b^{2} d \,e^{2}+2 B b c \,d^{2} e -B \,c^{2} d^{3}\right )}{3 e^{6} \left (e x +d \right )^{3}}-\frac {A \,b^{2} e^{3}-6 A b c d \,e^{2}+6 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+12 B b c \,d^{2} e -10 B \,c^{2} d^{3}}{e^{6} \left (e x +d \right )}+\frac {\left (2 A b c \,e^{2}-4 A \,c^{2} d e +B \,b^{2} e^{2}-8 B b c d e +10 B \,c^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{6}}+\frac {d \left (2 A \,b^{2} e^{3}-6 A b c d \,e^{2}+4 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+8 B b c \,d^{2} e -5 B \,c^{2} d^{3}\right )}{2 e^{6} \left (e x +d \right )^{2}}\) | \(292\) |
| norman | \(\frac {-\frac {d^{2} \left (2 A \,b^{2} e^{3}-22 A b c d \,e^{2}+44 A \,c^{2} d^{2} e -11 B \,b^{2} d \,e^{2}+88 B b c \,d^{2} e -110 B \,c^{2} d^{3}\right )}{6 e^{6}}-\frac {\left (A \,b^{2} e^{3}-6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -30 B \,c^{2} d^{3}\right ) x^{2}}{e^{4}}+\frac {B \,c^{2} x^{5}}{2 e}+\frac {c \left (2 A c e +4 B b e -5 B c d \right ) x^{4}}{2 e^{2}}-\frac {d \left (2 A \,b^{2} e^{3}-18 A b c d \,e^{2}+36 A \,c^{2} d^{2} e -9 B \,b^{2} d \,e^{2}+72 B b c \,d^{2} e -90 B \,c^{2} d^{3}\right ) x}{2 e^{5}}}{\left (e x +d \right )^{3}}+\frac {\left (2 A b c \,e^{2}-4 A \,c^{2} d e +B \,b^{2} e^{2}-8 B b c d e +10 B \,c^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{6}}\) | \(292\) |
| risch | \(\frac {B \,c^{2} x^{2}}{2 e^{4}}+\frac {c^{2} A x}{e^{4}}+\frac {2 c B b x}{e^{4}}-\frac {4 c^{2} B d x}{e^{5}}+\frac {\left (-A \,b^{2} e^{4}+6 A b c d \,e^{3}-6 A \,c^{2} d^{2} e^{2}+3 B \,b^{2} d \,e^{3}-12 B b c \,d^{2} e^{2}+10 B \,c^{2} d^{3} e \right ) x^{2}-\frac {d \left (2 A \,b^{2} e^{3}-18 A b c d \,e^{2}+20 A \,c^{2} d^{2} e -9 B \,b^{2} d \,e^{2}+40 B b c \,d^{2} e -35 B \,c^{2} d^{3}\right ) x}{2}-\frac {d^{2} \left (2 A \,b^{2} e^{3}-22 A b c d \,e^{2}+26 A \,c^{2} d^{2} e -11 B \,b^{2} d \,e^{2}+52 B b c \,d^{2} e -47 B \,c^{2} d^{3}\right )}{6 e}}{e^{5} \left (e x +d \right )^{3}}+\frac {2 \ln \left (e x +d \right ) A b c}{e^{4}}-\frac {4 \ln \left (e x +d \right ) A \,c^{2} d}{e^{5}}+\frac {b^{2} B \ln \left (e x +d \right )}{e^{4}}-\frac {8 \ln \left (e x +d \right ) B b c d}{e^{5}}+\frac {10 \ln \left (e x +d \right ) B \,c^{2} d^{2}}{e^{6}}\) | \(324\) |
| parallelrisch | \(\frac {110 B \,c^{2} d^{5}+180 B \ln \left (e x +d \right ) x \,c^{2} d^{4} e -72 A \ln \left (e x +d \right ) x \,c^{2} d^{3} e^{2}+18 B \ln \left (e x +d \right ) x \,b^{2} d^{2} e^{3}-24 A \ln \left (e x +d \right ) x^{3} c^{2} d \,e^{4}+60 B \ln \left (e x +d \right ) x^{3} c^{2} d^{2} e^{3}+12 A \ln \left (e x +d \right ) x^{3} b c \,e^{5}-72 A \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{3}+18 B \ln \left (e x +d \right ) x^{2} b^{2} d \,e^{4}+180 B \ln \left (e x +d \right ) x^{2} c^{2} d^{3} e^{2}+60 B \ln \left (e x +d \right ) c^{2} d^{5}+6 A \,x^{4} c^{2} e^{5}-6 A \,x^{2} b^{2} e^{5}+3 B \,x^{5} c^{2} e^{5}+12 A \ln \left (e x +d \right ) b c \,d^{3} e^{2}-48 B \ln \left (e x +d \right ) b c \,d^{4} e -216 B x b c \,d^{3} e^{2}+54 A x b c \,d^{2} e^{3}-144 B \,x^{2} b c \,d^{2} e^{3}+36 A \,x^{2} b c d \,e^{4}-2 A \,b^{2} d^{2} e^{3}-44 A \,c^{2} d^{4} e +11 B \,b^{2} d^{3} e^{2}-48 B \ln \left (e x +d \right ) x^{3} b c d \,e^{4}+22 A b c \,d^{3} e^{2}-88 B b c \,d^{4} e +6 B \ln \left (e x +d \right ) x^{3} b^{2} e^{5}-144 B \ln \left (e x +d \right ) x^{2} b c \,d^{2} e^{3}+36 A \ln \left (e x +d \right ) x^{2} b c d \,e^{4}+36 A \ln \left (e x +d \right ) x b c \,d^{2} e^{3}-144 B \ln \left (e x +d \right ) x b c \,d^{3} e^{2}-72 A \,x^{2} c^{2} d^{2} e^{3}+18 B \,x^{2} b^{2} d \,e^{4}+180 B \,x^{2} c^{2} d^{3} e^{2}-6 A x \,b^{2} d \,e^{4}-108 A x \,c^{2} d^{3} e^{2}+27 B x \,b^{2} d^{2} e^{3}+270 B x \,c^{2} d^{4} e -24 A \ln \left (e x +d \right ) c^{2} d^{4} e +6 B \ln \left (e x +d \right ) b^{2} d^{3} e^{2}+12 B \,x^{4} b c \,e^{5}-15 B \,x^{4} c^{2} d \,e^{4}}{6 e^{6} \left (e x +d \right )^{3}}\) | \(642\) |
c/e^5*(1/2*B*c*e*x^2+A*c*e*x+2*B*b*e*x-4*B*c*d*x)-1/3*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/(e*x+d)^3-1/ e^6*(A*b^2*e^3-6*A*b*c*d*e^2+6*A*c^2*d^2*e-3*B*b^2*d*e^2+12*B*b*c*d^2*e-10 *B*c^2*d^3)/(e*x+d)+1/e^6*(2*A*b*c*e^2-4*A*c^2*d*e+B*b^2*e^2-8*B*b*c*d*e+1 0*B*c^2*d^2)*ln(e*x+d)+1/2*d/e^6*(2*A*b^2*e^3-6*A*b*c*d*e^2+4*A*c^2*d^2*e- 3*B*b^2*d*e^2+8*B*b*c*d^2*e-5*B*c^2*d^3)/(e*x+d)^2
Leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (231) = 462\).
Time = 0.27 (sec) , antiderivative size = 505, normalized size of antiderivative = 2.12 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {3 \, B c^{2} e^{5} x^{5} + 47 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 26 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e + 11 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} - 3 \, {\left (5 \, B c^{2} d e^{4} - 2 \, {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} - 9 \, {\left (7 \, B c^{2} d^{2} e^{3} - 2 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4}\right )} x^{3} - 3 \, {\left (3 \, B c^{2} d^{3} e^{2} + 2 \, A b^{2} e^{5} + 6 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} - 6 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 3 \, {\left (27 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 18 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + 9 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x + 6 \, {\left (10 \, B c^{2} d^{5} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e + {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + {\left (10 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} + {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} + 3 \, {\left (10 \, B c^{2} d^{3} e^{2} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} + {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 3 \, {\left (10 \, B c^{2} d^{4} e - 4 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} \]
1/6*(3*B*c^2*e^5*x^5 + 47*B*c^2*d^5 - 2*A*b^2*d^2*e^3 - 26*(2*B*b*c + A*c^ 2)*d^4*e + 11*(B*b^2 + 2*A*b*c)*d^3*e^2 - 3*(5*B*c^2*d*e^4 - 2*(2*B*b*c + A*c^2)*e^5)*x^4 - 9*(7*B*c^2*d^2*e^3 - 2*(2*B*b*c + A*c^2)*d*e^4)*x^3 - 3* (3*B*c^2*d^3*e^2 + 2*A*b^2*e^5 + 6*(2*B*b*c + A*c^2)*d^2*e^3 - 6*(B*b^2 + 2*A*b*c)*d*e^4)*x^2 + 3*(27*B*c^2*d^4*e - 2*A*b^2*d*e^4 - 18*(2*B*b*c + A* c^2)*d^3*e^2 + 9*(B*b^2 + 2*A*b*c)*d^2*e^3)*x + 6*(10*B*c^2*d^5 - 4*(2*B*b *c + A*c^2)*d^4*e + (B*b^2 + 2*A*b*c)*d^3*e^2 + (10*B*c^2*d^2*e^3 - 4*(2*B *b*c + A*c^2)*d*e^4 + (B*b^2 + 2*A*b*c)*e^5)*x^3 + 3*(10*B*c^2*d^3*e^2 - 4 *(2*B*b*c + A*c^2)*d^2*e^3 + (B*b^2 + 2*A*b*c)*d*e^4)*x^2 + 3*(10*B*c^2*d^ 4*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)*log(e*x + d))/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x + d^3*e^6)
Time = 8.02 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.57 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {B c^{2} x^{2}}{2 e^{4}} + x \left (\frac {A c^{2}}{e^{4}} + \frac {2 B b c}{e^{4}} - \frac {4 B c^{2} d}{e^{5}}\right ) + \frac {- 2 A b^{2} d^{2} e^{3} + 22 A b c d^{3} e^{2} - 26 A c^{2} d^{4} e + 11 B b^{2} d^{3} e^{2} - 52 B b c d^{4} e + 47 B c^{2} d^{5} + x^{2} \left (- 6 A b^{2} e^{5} + 36 A b c d e^{4} - 36 A c^{2} d^{2} e^{3} + 18 B b^{2} d e^{4} - 72 B b c d^{2} e^{3} + 60 B c^{2} d^{3} e^{2}\right ) + x \left (- 6 A b^{2} d e^{4} + 54 A b c d^{2} e^{3} - 60 A c^{2} d^{3} e^{2} + 27 B b^{2} d^{2} e^{3} - 120 B b c d^{3} e^{2} + 105 B c^{2} d^{4} e\right )}{6 d^{3} e^{6} + 18 d^{2} e^{7} x + 18 d e^{8} x^{2} + 6 e^{9} x^{3}} + \frac {\left (2 A b c e^{2} - 4 A c^{2} d e + B b^{2} e^{2} - 8 B b c d e + 10 B c^{2} d^{2}\right ) \log {\left (d + e x \right )}}{e^{6}} \]
B*c**2*x**2/(2*e**4) + x*(A*c**2/e**4 + 2*B*b*c/e**4 - 4*B*c**2*d/e**5) + (-2*A*b**2*d**2*e**3 + 22*A*b*c*d**3*e**2 - 26*A*c**2*d**4*e + 11*B*b**2*d **3*e**2 - 52*B*b*c*d**4*e + 47*B*c**2*d**5 + x**2*(-6*A*b**2*e**5 + 36*A* b*c*d*e**4 - 36*A*c**2*d**2*e**3 + 18*B*b**2*d*e**4 - 72*B*b*c*d**2*e**3 + 60*B*c**2*d**3*e**2) + x*(-6*A*b**2*d*e**4 + 54*A*b*c*d**2*e**3 - 60*A*c* *2*d**3*e**2 + 27*B*b**2*d**2*e**3 - 120*B*b*c*d**3*e**2 + 105*B*c**2*d**4 *e))/(6*d**3*e**6 + 18*d**2*e**7*x + 18*d*e**8*x**2 + 6*e**9*x**3) + (2*A* b*c*e**2 - 4*A*c**2*d*e + B*b**2*e**2 - 8*B*b*c*d*e + 10*B*c**2*d**2)*log( d + e*x)/e**6
Time = 0.20 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.31 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {47 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 26 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e + 11 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 6 \, {\left (10 \, B c^{2} d^{3} e^{2} - A b^{2} e^{5} - 6 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} + 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 3 \, {\left (35 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 20 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + 9 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x}{6 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} + \frac {B c^{2} e x^{2} - 2 \, {\left (4 \, B c^{2} d - {\left (2 \, B b c + A c^{2}\right )} e\right )} x}{2 \, e^{5}} + \frac {{\left (10 \, B c^{2} d^{2} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e + {\left (B b^{2} + 2 \, A b c\right )} e^{2}\right )} \log \left (e x + d\right )}{e^{6}} \]
1/6*(47*B*c^2*d^5 - 2*A*b^2*d^2*e^3 - 26*(2*B*b*c + A*c^2)*d^4*e + 11*(B*b ^2 + 2*A*b*c)*d^3*e^2 + 6*(10*B*c^2*d^3*e^2 - A*b^2*e^5 - 6*(2*B*b*c + A*c ^2)*d^2*e^3 + 3*(B*b^2 + 2*A*b*c)*d*e^4)*x^2 + 3*(35*B*c^2*d^4*e - 2*A*b^2 *d*e^4 - 20*(2*B*b*c + A*c^2)*d^3*e^2 + 9*(B*b^2 + 2*A*b*c)*d^2*e^3)*x)/(e ^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x + d^3*e^6) + 1/2*(B*c^2*e*x^2 - 2*(4*B* c^2*d - (2*B*b*c + A*c^2)*e)*x)/e^5 + (10*B*c^2*d^2 - 4*(2*B*b*c + A*c^2)* d*e + (B*b^2 + 2*A*b*c)*e^2)*log(e*x + d)/e^6
Time = 0.26 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.32 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {{\left (10 \, B c^{2} d^{2} - 8 \, B b c d e - 4 \, A c^{2} d e + B b^{2} e^{2} + 2 \, A b c e^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{6}} + \frac {B c^{2} e^{4} x^{2} - 8 \, B c^{2} d e^{3} x + 4 \, B b c e^{4} x + 2 \, A c^{2} e^{4} x}{2 \, e^{8}} + \frac {47 \, B c^{2} d^{5} - 52 \, B b c d^{4} e - 26 \, A c^{2} d^{4} e + 11 \, B b^{2} d^{3} e^{2} + 22 \, A b c d^{3} e^{2} - 2 \, A b^{2} d^{2} e^{3} + 6 \, {\left (10 \, B c^{2} d^{3} e^{2} - 12 \, B b c d^{2} e^{3} - 6 \, A c^{2} d^{2} e^{3} + 3 \, B b^{2} d e^{4} + 6 \, A b c d e^{4} - A b^{2} e^{5}\right )} x^{2} + 3 \, {\left (35 \, B c^{2} d^{4} e - 40 \, B b c d^{3} e^{2} - 20 \, A c^{2} d^{3} e^{2} + 9 \, B b^{2} d^{2} e^{3} + 18 \, A b c d^{2} e^{3} - 2 \, A b^{2} d e^{4}\right )} x}{6 \, {\left (e x + d\right )}^{3} e^{6}} \]
(10*B*c^2*d^2 - 8*B*b*c*d*e - 4*A*c^2*d*e + B*b^2*e^2 + 2*A*b*c*e^2)*log(a bs(e*x + d))/e^6 + 1/2*(B*c^2*e^4*x^2 - 8*B*c^2*d*e^3*x + 4*B*b*c*e^4*x + 2*A*c^2*e^4*x)/e^8 + 1/6*(47*B*c^2*d^5 - 52*B*b*c*d^4*e - 26*A*c^2*d^4*e + 11*B*b^2*d^3*e^2 + 22*A*b*c*d^3*e^2 - 2*A*b^2*d^2*e^3 + 6*(10*B*c^2*d^3*e ^2 - 12*B*b*c*d^2*e^3 - 6*A*c^2*d^2*e^3 + 3*B*b^2*d*e^4 + 6*A*b*c*d*e^4 - A*b^2*e^5)*x^2 + 3*(35*B*c^2*d^4*e - 40*B*b*c*d^3*e^2 - 20*A*c^2*d^3*e^2 + 9*B*b^2*d^2*e^3 + 18*A*b*c*d^2*e^3 - 2*A*b^2*d*e^4)*x)/((e*x + d)^3*e^6)
Time = 0.14 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=x\,\left (\frac {A\,c^2+2\,B\,b\,c}{e^4}-\frac {4\,B\,c^2\,d}{e^5}\right )+\frac {\frac {11\,B\,b^2\,d^3\,e^2-2\,A\,b^2\,d^2\,e^3-52\,B\,b\,c\,d^4\,e+22\,A\,b\,c\,d^3\,e^2+47\,B\,c^2\,d^5-26\,A\,c^2\,d^4\,e}{6\,e}-x^2\,\left (-3\,B\,b^2\,d\,e^3+A\,b^2\,e^4+12\,B\,b\,c\,d^2\,e^2-6\,A\,b\,c\,d\,e^3-10\,B\,c^2\,d^3\,e+6\,A\,c^2\,d^2\,e^2\right )+x\,\left (\frac {9\,B\,b^2\,d^2\,e^2}{2}-A\,b^2\,d\,e^3-20\,B\,b\,c\,d^3\,e+9\,A\,b\,c\,d^2\,e^2+\frac {35\,B\,c^2\,d^4}{2}-10\,A\,c^2\,d^3\,e\right )}{d^3\,e^5+3\,d^2\,e^6\,x+3\,d\,e^7\,x^2+e^8\,x^3}+\frac {\ln \left (d+e\,x\right )\,\left (B\,b^2\,e^2-8\,B\,b\,c\,d\,e+2\,A\,b\,c\,e^2+10\,B\,c^2\,d^2-4\,A\,c^2\,d\,e\right )}{e^6}+\frac {B\,c^2\,x^2}{2\,e^4} \]
x*((A*c^2 + 2*B*b*c)/e^4 - (4*B*c^2*d)/e^5) + ((47*B*c^2*d^5 - 26*A*c^2*d^ 4*e - 2*A*b^2*d^2*e^3 + 11*B*b^2*d^3*e^2 - 52*B*b*c*d^4*e + 22*A*b*c*d^3*e ^2)/(6*e) - x^2*(A*b^2*e^4 - 3*B*b^2*d*e^3 - 10*B*c^2*d^3*e + 6*A*c^2*d^2* e^2 - 6*A*b*c*d*e^3 + 12*B*b*c*d^2*e^2) + x*((35*B*c^2*d^4)/2 - A*b^2*d*e^ 3 - 10*A*c^2*d^3*e + (9*B*b^2*d^2*e^2)/2 - 20*B*b*c*d^3*e + 9*A*b*c*d^2*e^ 2))/(d^3*e^5 + e^8*x^3 + 3*d^2*e^6*x + 3*d*e^7*x^2) + (log(d + e*x)*(B*b^2 *e^2 + 10*B*c^2*d^2 + 2*A*b*c*e^2 - 4*A*c^2*d*e - 8*B*b*c*d*e))/e^6 + (B*c ^2*x^2)/(2*e^4)