Integrand size = 24, antiderivative size = 240 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {B c^2 x}{e^5}+\frac {d^2 (B d-A e) (c d-b e)^2}{4 e^6 (d+e x)^4}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{3 e^6 (d+e x)^3}-\frac {A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )}{2 e^6 (d+e x)^2}+\frac {2 A c e (2 c d-b e)-B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )}{e^6 (d+e x)}-\frac {c (5 B c d-2 b B e-A c e) \log (d+e x)}{e^6} \]
B*c^2*x/e^5+1/4*d^2*(-A*e+B*d)*(-b*e+c*d)^2/e^6/(e*x+d)^4-1/3*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)^3+1/2*(-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+ (2*A*c*e*(-b*e+2*c*d)-B*(b^2*e^2-8*b*c*d*e+10*c^2*d^2))/e^6/(e*x+d)-c*(-A* c*e-2*B*b*e+5*B*c*d)*ln(e*x+d)/e^6
Time = 0.10 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.15 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {A e \left (b^2 e^2 \left (d^2+4 d e x+6 e^2 x^2\right )+6 b c e \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )-c^2 d \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )\right )+B \left (3 b^2 e^2 \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )-2 b c d e \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )+c^2 \left (77 d^5+248 d^4 e x+252 d^3 e^2 x^2+48 d^2 e^3 x^3-48 d e^4 x^4-12 e^5 x^5\right )\right )+12 c (5 B c d-2 b B e-A c e) (d+e x)^4 \log (d+e x)}{12 e^6 (d+e x)^4} \]
-1/12*(A*e*(b^2*e^2*(d^2 + 4*d*e*x + 6*e^2*x^2) + 6*b*c*e*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3) - c^2*d*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^3*x^3)) + B*(3*b^2*e^2*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3) - 2*b*c*d*e*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^3*x^3) + c^2*(77*d^ 5 + 248*d^4*e*x + 252*d^3*e^2*x^2 + 48*d^2*e^3*x^3 - 48*d*e^4*x^4 - 12*e^5 *x^5)) + 12*c*(5*B*c*d - 2*b*B*e - A*c*e)*(d + e*x)^4*Log[d + e*x])/(e^6*( d + e*x)^4)
Time = 0.48 (sec) , antiderivative size = 240, 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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^5} \, dx\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \int \left (\frac {B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )-2 A c e (2 c d-b e)}{e^5 (d+e x)^2}+\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{e^5 (d+e x)^3}-\frac {d^2 (B d-A e) (c d-b e)^2}{e^5 (d+e x)^5}+\frac {c (A c e+2 b B e-5 B c d)}{e^5 (d+e x)}+\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^5 (d+e x)^4}+\frac {B c^2}{e^5}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 A c e (2 c d-b e)-B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )}{e^6 (d+e x)}-\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{2 e^6 (d+e x)^2}+\frac {d^2 (B d-A e) (c d-b e)^2}{4 e^6 (d+e x)^4}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{3 e^6 (d+e x)^3}-\frac {c \log (d+e x) (-A c e-2 b B e+5 B c d)}{e^6}+\frac {B c^2 x}{e^5}\) |
(B*c^2*x)/e^5 + (d^2*(B*d - A*e)*(c*d - b*e)^2)/(4*e^6*(d + e*x)^4) - (d*( c*d - b*e)*(B*d*(5*c*d - 3*b*e) - 2*A*e*(2*c*d - b*e)))/(3*e^6*(d + e*x)^3 ) - (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))/(2*e^6*(d + e*x)^2) + (2*A*c*e*(2*c*d - b*e) - B*(10*c^2*d^2 - 8*b*c*d*e + b^2*e^2))/(e^6*(d + e*x)) - (c*(5*B*c*d - 2*b*B*e - A*c*e)* Log[d + e*x])/e^6
3.12.23.3.1 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 = 0.24 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.20
| method | result | size |
| norman | \(\frac {\frac {B \,c^{2} x^{5}}{e}-\frac {d^{2} \left (A \,b^{2} e^{3}+6 A b c d \,e^{2}-25 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}-50 B b c \,d^{2} e +125 B \,c^{2} d^{3}\right )}{12 e^{6}}-\frac {\left (2 A b c \,e^{2}-4 A \,c^{2} d e +B \,b^{2} e^{2}-8 B b c d e +20 B \,c^{2} d^{2}\right ) x^{3}}{e^{3}}-\frac {\left (A \,b^{2} e^{3}+6 A b c d \,e^{2}-18 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}-36 B b c \,d^{2} e +90 B \,c^{2} d^{3}\right ) x^{2}}{2 e^{4}}-\frac {d \left (A \,b^{2} e^{3}+6 A b c d \,e^{2}-22 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}-44 B b c \,d^{2} e +110 B \,c^{2} d^{3}\right ) x}{3 e^{5}}}{\left (e x +d \right )^{4}}+\frac {c \left (A c e +2 B b e -5 B c d \right ) \ln \left (e x +d \right )}{e^{6}}\) | \(288\) |
| default | \(\frac {B \,c^{2} x}{e^{5}}+\frac {d \left (2 A \,b^{2} e^{3}-6 A b c d \,e^{2}+4 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+8 B b c \,d^{2} e -5 B \,c^{2} d^{3}\right )}{3 e^{6} \left (e x +d \right )^{3}}-\frac {2 A b c \,e^{2}-4 A \,c^{2} d e +B \,b^{2} e^{2}-8 B b c d e +10 B \,c^{2} d^{2}}{e^{6} \left (e x +d \right )}+\frac {c \left (A c e +2 B b e -5 B c d \right ) \ln \left (e x +d \right )}{e^{6}}-\frac {d^{2} \left (A \,b^{2} e^{3}-2 A b c d \,e^{2}+A \,c^{2} d^{2} e -B \,b^{2} d \,e^{2}+2 B b c \,d^{2} e -B \,c^{2} d^{3}\right )}{4 e^{6} \left (e x +d \right )^{4}}-\frac {A \,b^{2} e^{3}-6 A b c d \,e^{2}+6 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+12 B b c \,d^{2} e -10 B \,c^{2} d^{3}}{2 e^{6} \left (e x +d \right )^{2}}\) | \(298\) |
| risch | \(\frac {B \,c^{2} x}{e^{5}}+\frac {\left (-2 A b c \,e^{4}+4 A \,c^{2} d \,e^{3}-B \,b^{2} e^{4}+8 B b c d \,e^{3}-10 B \,c^{2} d^{2} e^{2}\right ) x^{3}-\frac {e \left (A \,b^{2} e^{3}+6 A b c d \,e^{2}-18 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}-36 B b c \,d^{2} e +50 B \,c^{2} d^{3}\right ) x^{2}}{2}-\frac {d \left (A \,b^{2} e^{3}+6 A b c d \,e^{2}-22 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}-44 B b c \,d^{2} e +65 B \,c^{2} d^{3}\right ) x}{3}-\frac {d^{2} \left (A \,b^{2} e^{3}+6 A b c d \,e^{2}-25 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}-50 B b c \,d^{2} e +77 B \,c^{2} d^{3}\right )}{12 e}}{e^{5} \left (e x +d \right )^{4}}+\frac {c^{2} \ln \left (e x +d \right ) A}{e^{5}}+\frac {2 c \ln \left (e x +d \right ) B b}{e^{5}}-\frac {5 c^{2} \ln \left (e x +d \right ) B d}{e^{6}}\) | \(306\) |
| parallelrisch | \(\frac {-125 B \,c^{2} d^{5}-240 B \ln \left (e x +d \right ) x \,c^{2} d^{4} e +48 A \ln \left (e x +d \right ) x \,c^{2} d^{3} e^{2}+48 A \ln \left (e x +d \right ) x^{3} c^{2} d \,e^{4}-240 B \ln \left (e x +d \right ) x^{3} c^{2} d^{2} e^{3}+24 B \ln \left (e x +d \right ) x^{4} b c \,e^{5}-60 B \ln \left (e x +d \right ) x^{4} c^{2} d \,e^{4}+72 A \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{3}-360 B \ln \left (e x +d \right ) x^{2} c^{2} d^{3} e^{2}-60 B \ln \left (e x +d \right ) c^{2} d^{5}-12 B \,x^{3} b^{2} e^{5}-6 A \,x^{2} b^{2} e^{5}+12 B \,x^{5} c^{2} e^{5}+24 B \ln \left (e x +d \right ) b c \,d^{4} e +176 B x b c \,d^{3} e^{2}-24 A x b c \,d^{2} e^{3}+216 B \,x^{2} b c \,d^{2} e^{3}-36 A \,x^{2} b c d \,e^{4}+96 B \,x^{3} b c d \,e^{4}-A \,b^{2} d^{2} e^{3}+25 A \,c^{2} d^{4} e -3 B \,b^{2} d^{3} e^{2}+96 B \ln \left (e x +d \right ) x^{3} b c d \,e^{4}-6 A b c \,d^{3} e^{2}+50 B b c \,d^{4} e +12 A \ln \left (e x +d \right ) x^{4} c^{2} e^{5}+144 B \ln \left (e x +d \right ) x^{2} b c \,d^{2} e^{3}+96 B \ln \left (e x +d \right ) x b c \,d^{3} e^{2}-24 A \,x^{3} b c \,e^{5}+48 A \,x^{3} c^{2} d \,e^{4}-240 B \,x^{3} c^{2} d^{2} e^{3}+108 A \,x^{2} c^{2} d^{2} e^{3}-18 B \,x^{2} b^{2} d \,e^{4}-540 B \,x^{2} c^{2} d^{3} e^{2}-4 A x \,b^{2} d \,e^{4}+88 A x \,c^{2} d^{3} e^{2}-12 B x \,b^{2} d^{2} e^{3}-440 B x \,c^{2} d^{4} e +12 A \ln \left (e x +d \right ) c^{2} d^{4} e}{12 e^{6} \left (e x +d \right )^{4}}\) | \(579\) |
(B*c^2*x^5/e-1/12*d^2*(A*b^2*e^3+6*A*b*c*d*e^2-25*A*c^2*d^2*e+3*B*b^2*d*e^ 2-50*B*b*c*d^2*e+125*B*c^2*d^3)/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+20*B*c^2*d^2)/e^3*x^3-1/2*(A*b^2*e^3+6*A*b*c*d*e^2-18*A*c^2*d^2*e +3*B*b^2*d*e^2-36*B*b*c*d^2*e+90*B*c^2*d^3)/e^4*x^2-1/3*d*(A*b^2*e^3+6*A*b *c*d*e^2-22*A*c^2*d^2*e+3*B*b^2*d*e^2-44*B*b*c*d^2*e+110*B*c^2*d^3)/e^5*x) /(e*x+d)^4+c/e^6*(A*c*e+2*B*b*e-5*B*c*d)*ln(e*x+d)
Leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (234) = 468\).
Time = 0.31 (sec) , antiderivative size = 475, normalized size of antiderivative = 1.98 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {12 \, B c^{2} e^{5} x^{5} + 48 \, B c^{2} d e^{4} x^{4} - 77 \, B c^{2} d^{5} - A b^{2} d^{2} e^{3} + 25 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} - 12 \, {\left (4 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} + {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} - 6 \, {\left (42 \, B c^{2} d^{3} e^{2} + A b^{2} e^{5} - 18 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} + 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} - 4 \, {\left (62 \, B c^{2} d^{4} e + A b^{2} d e^{4} - 22 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x - 12 \, {\left (5 \, B c^{2} d^{5} - {\left (2 \, B b c + A c^{2}\right )} d^{4} e + {\left (5 \, B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 4 \, {\left (5 \, B c^{2} d^{2} e^{3} - {\left (2 \, B b c + A c^{2}\right )} d e^{4}\right )} x^{3} + 6 \, {\left (5 \, B c^{2} d^{3} e^{2} - {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3}\right )} x^{2} + 4 \, {\left (5 \, B c^{2} d^{4} e - {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2}\right )} x\right )} \log \left (e x + d\right )}{12 \, {\left (e^{10} x^{4} + 4 \, d e^{9} x^{3} + 6 \, d^{2} e^{8} x^{2} + 4 \, d^{3} e^{7} x + d^{4} e^{6}\right )}} \]
1/12*(12*B*c^2*e^5*x^5 + 48*B*c^2*d*e^4*x^4 - 77*B*c^2*d^5 - A*b^2*d^2*e^3 + 25*(2*B*b*c + A*c^2)*d^4*e - 3*(B*b^2 + 2*A*b*c)*d^3*e^2 - 12*(4*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 - 6*(42*B *c^2*d^3*e^2 + A*b^2*e^5 - 18*(2*B*b*c + A*c^2)*d^2*e^3 + 3*(B*b^2 + 2*A*b *c)*d*e^4)*x^2 - 4*(62*B*c^2*d^4*e + A*b^2*d*e^4 - 22*(2*B*b*c + A*c^2)*d^ 3*e^2 + 3*(B*b^2 + 2*A*b*c)*d^2*e^3)*x - 12*(5*B*c^2*d^5 - (2*B*b*c + A*c^ 2)*d^4*e + (5*B*c^2*d*e^4 - (2*B*b*c + A*c^2)*e^5)*x^4 + 4*(5*B*c^2*d^2*e^ 3 - (2*B*b*c + A*c^2)*d*e^4)*x^3 + 6*(5*B*c^2*d^3*e^2 - (2*B*b*c + A*c^2)* d^2*e^3)*x^2 + 4*(5*B*c^2*d^4*e - (2*B*b*c + A*c^2)*d^3*e^2)*x)*log(e*x + d))/(e^10*x^4 + 4*d*e^9*x^3 + 6*d^2*e^8*x^2 + 4*d^3*e^7*x + d^4*e^6)
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.34 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {77 \, B c^{2} d^{5} + A b^{2} d^{2} e^{3} - 25 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e + 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 12 \, {\left (10 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} + {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} + 6 \, {\left (50 \, B c^{2} d^{3} e^{2} + A b^{2} e^{5} - 18 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} + 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 4 \, {\left (65 \, B c^{2} d^{4} e + A b^{2} d e^{4} - 22 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x}{12 \, {\left (e^{10} x^{4} + 4 \, d e^{9} x^{3} + 6 \, d^{2} e^{8} x^{2} + 4 \, d^{3} e^{7} x + d^{4} e^{6}\right )}} + \frac {B c^{2} x}{e^{5}} - \frac {{\left (5 \, B c^{2} d - {\left (2 \, B b c + A c^{2}\right )} e\right )} \log \left (e x + d\right )}{e^{6}} \]
-1/12*(77*B*c^2*d^5 + A*b^2*d^2*e^3 - 25*(2*B*b*c + A*c^2)*d^4*e + 3*(B*b^ 2 + 2*A*b*c)*d^3*e^2 + 12*(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 + 6*(50*B*c^2*d^3*e^2 + A*b^2*e^5 - 18*(2*B*b*c + A*c^2)*d^2*e^3 + 3*(B*b^2 + 2*A*b*c)*d*e^4)*x^2 + 4*(65*B*c^2*d^4*e + A *b^2*d*e^4 - 22*(2*B*b*c + A*c^2)*d^3*e^2 + 3*(B*b^2 + 2*A*b*c)*d^2*e^3)*x )/(e^10*x^4 + 4*d*e^9*x^3 + 6*d^2*e^8*x^2 + 4*d^3*e^7*x + d^4*e^6) + B*c^2 *x/e^5 - (5*B*c^2*d - (2*B*b*c + A*c^2)*e)*log(e*x + d)/e^6
Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (234) = 468\).
Time = 0.27 (sec) , antiderivative size = 474, normalized size of antiderivative = 1.98 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {{\left (e x + d\right )} B c^{2}}{e^{6}} + \frac {{\left (5 \, B c^{2} d - 2 \, B b c e - A c^{2} e\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{6}} - \frac {\frac {120 \, B c^{2} d^{2} e^{22}}{e x + d} - \frac {60 \, B c^{2} d^{3} e^{22}}{{\left (e x + d\right )}^{2}} + \frac {20 \, B c^{2} d^{4} e^{22}}{{\left (e x + d\right )}^{3}} - \frac {3 \, B c^{2} d^{5} e^{22}}{{\left (e x + d\right )}^{4}} - \frac {96 \, B b c d e^{23}}{e x + d} - \frac {48 \, A c^{2} d e^{23}}{e x + d} + \frac {72 \, B b c d^{2} e^{23}}{{\left (e x + d\right )}^{2}} + \frac {36 \, A c^{2} d^{2} e^{23}}{{\left (e x + d\right )}^{2}} - \frac {32 \, B b c d^{3} e^{23}}{{\left (e x + d\right )}^{3}} - \frac {16 \, A c^{2} d^{3} e^{23}}{{\left (e x + d\right )}^{3}} + \frac {6 \, B b c d^{4} e^{23}}{{\left (e x + d\right )}^{4}} + \frac {3 \, A c^{2} d^{4} e^{23}}{{\left (e x + d\right )}^{4}} + \frac {12 \, B b^{2} e^{24}}{e x + d} + \frac {24 \, A b c e^{24}}{e x + d} - \frac {18 \, B b^{2} d e^{24}}{{\left (e x + d\right )}^{2}} - \frac {36 \, A b c d e^{24}}{{\left (e x + d\right )}^{2}} + \frac {12 \, B b^{2} d^{2} e^{24}}{{\left (e x + d\right )}^{3}} + \frac {24 \, A b c d^{2} e^{24}}{{\left (e x + d\right )}^{3}} - \frac {3 \, B b^{2} d^{3} e^{24}}{{\left (e x + d\right )}^{4}} - \frac {6 \, A b c d^{3} e^{24}}{{\left (e x + d\right )}^{4}} + \frac {6 \, A b^{2} e^{25}}{{\left (e x + d\right )}^{2}} - \frac {8 \, A b^{2} d e^{25}}{{\left (e x + d\right )}^{3}} + \frac {3 \, A b^{2} d^{2} e^{25}}{{\left (e x + d\right )}^{4}}}{12 \, e^{28}} \]
(e*x + d)*B*c^2/e^6 + (5*B*c^2*d - 2*B*b*c*e - A*c^2*e)*log(abs(e*x + d)/( (e*x + d)^2*abs(e)))/e^6 - 1/12*(120*B*c^2*d^2*e^22/(e*x + d) - 60*B*c^2*d ^3*e^22/(e*x + d)^2 + 20*B*c^2*d^4*e^22/(e*x + d)^3 - 3*B*c^2*d^5*e^22/(e* x + d)^4 - 96*B*b*c*d*e^23/(e*x + d) - 48*A*c^2*d*e^23/(e*x + d) + 72*B*b* c*d^2*e^23/(e*x + d)^2 + 36*A*c^2*d^2*e^23/(e*x + d)^2 - 32*B*b*c*d^3*e^23 /(e*x + d)^3 - 16*A*c^2*d^3*e^23/(e*x + d)^3 + 6*B*b*c*d^4*e^23/(e*x + d)^ 4 + 3*A*c^2*d^4*e^23/(e*x + d)^4 + 12*B*b^2*e^24/(e*x + d) + 24*A*b*c*e^24 /(e*x + d) - 18*B*b^2*d*e^24/(e*x + d)^2 - 36*A*b*c*d*e^24/(e*x + d)^2 + 1 2*B*b^2*d^2*e^24/(e*x + d)^3 + 24*A*b*c*d^2*e^24/(e*x + d)^3 - 3*B*b^2*d^3 *e^24/(e*x + d)^4 - 6*A*b*c*d^3*e^24/(e*x + d)^4 + 6*A*b^2*e^25/(e*x + d)^ 2 - 8*A*b^2*d*e^25/(e*x + d)^3 + 3*A*b^2*d^2*e^25/(e*x + d)^4)/e^28
Time = 10.85 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.41 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (A\,c^2\,e-5\,B\,c^2\,d+2\,B\,b\,c\,e\right )}{e^6}-\frac {x^2\,\left (\frac {3\,B\,b^2\,d\,e^3}{2}+\frac {A\,b^2\,e^4}{2}-18\,B\,b\,c\,d^2\,e^2+3\,A\,b\,c\,d\,e^3+25\,B\,c^2\,d^3\,e-9\,A\,c^2\,d^2\,e^2\right )+\frac {3\,B\,b^2\,d^3\,e^2+A\,b^2\,d^2\,e^3-50\,B\,b\,c\,d^4\,e+6\,A\,b\,c\,d^3\,e^2+77\,B\,c^2\,d^5-25\,A\,c^2\,d^4\,e}{12\,e}+x\,\left (B\,b^2\,d^2\,e^2+\frac {A\,b^2\,d\,e^3}{3}-\frac {44\,B\,b\,c\,d^3\,e}{3}+2\,A\,b\,c\,d^2\,e^2+\frac {65\,B\,c^2\,d^4}{3}-\frac {22\,A\,c^2\,d^3\,e}{3}\right )+x^3\,\left (B\,b^2\,e^4-8\,B\,b\,c\,d\,e^3+2\,A\,b\,c\,e^4+10\,B\,c^2\,d^2\,e^2-4\,A\,c^2\,d\,e^3\right )}{d^4\,e^5+4\,d^3\,e^6\,x+6\,d^2\,e^7\,x^2+4\,d\,e^8\,x^3+e^9\,x^4}+\frac {B\,c^2\,x}{e^5} \]
(log(d + e*x)*(A*c^2*e - 5*B*c^2*d + 2*B*b*c*e))/e^6 - (x^2*((A*b^2*e^4)/2 + (3*B*b^2*d*e^3)/2 + 25*B*c^2*d^3*e - 9*A*c^2*d^2*e^2 + 3*A*b*c*d*e^3 - 18*B*b*c*d^2*e^2) + (77*B*c^2*d^5 - 25*A*c^2*d^4*e + A*b^2*d^2*e^3 + 3*B*b ^2*d^3*e^2 - 50*B*b*c*d^4*e + 6*A*b*c*d^3*e^2)/(12*e) + x*((65*B*c^2*d^4)/ 3 + (A*b^2*d*e^3)/3 - (22*A*c^2*d^3*e)/3 + B*b^2*d^2*e^2 - (44*B*b*c*d^3*e )/3 + 2*A*b*c*d^2*e^2) + x^3*(B*b^2*e^4 + 2*A*b*c*e^4 - 4*A*c^2*d*e^3 + 10 *B*c^2*d^2*e^2 - 8*B*b*c*d*e^3))/(d^4*e^5 + e^9*x^4 + 4*d^3*e^6*x + 4*d*e^ 8*x^3 + 6*d^2*e^7*x^2) + (B*c^2*x)/e^5