Integrand size = 24, antiderivative size = 248 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {d^2 (B d-A e) (c d-b e)^2}{5 e^6 (d+e x)^5}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{4 e^6 (d+e x)^4}-\frac {A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )}{3 e^6 (d+e x)^3}+\frac {2 A c e (2 c d-b e)-B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )}{2 e^6 (d+e x)^2}+\frac {c (5 B c d-2 b B e-A c e)}{e^6 (d+e x)}+\frac {B c^2 \log (d+e x)}{e^6} \] Output:
1/5*d^2*(-A*e+B*d)*(-b*e+c*d)^2/e^6/(e*x+d)^5-1/4*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)^4-1/3*(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)^3+1/2*(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)^2+c*(-A*c*e-2*B *b*e+5*B*c*d)/e^6/(e*x+d)+B*c^2*ln(e*x+d)/e^6
Time = 0.15 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {-2 A e \left (b^2 e^2 \left (d^2+5 d e x+10 e^2 x^2\right )+3 b c e \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )+6 c^2 \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )\right )+B \left (-3 b^2 e^2 \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )-24 b c e \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )+c^2 d \left (137 d^4+625 d^3 e x+1100 d^2 e^2 x^2+900 d e^3 x^3+300 e^4 x^4\right )\right )+60 B c^2 (d+e x)^5 \log (d+e x)}{60 e^6 (d+e x)^5} \] Input:
Integrate[((A + B*x)*(b*x + c*x^2)^2)/(d + e*x)^6,x]
Output:
(-2*A*e*(b^2*e^2*(d^2 + 5*d*e*x + 10*e^2*x^2) + 3*b*c*e*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3) + 6*c^2*(d^4 + 5*d^3*e*x + 10*d^2*e^2*x^2 + 10 *d*e^3*x^3 + 5*e^4*x^4)) + B*(-3*b^2*e^2*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3) - 24*b*c*e*(d^4 + 5*d^3*e*x + 10*d^2*e^2*x^2 + 10*d*e^3*x^3 + 5*e^4*x^4) + c^2*d*(137*d^4 + 625*d^3*e*x + 1100*d^2*e^2*x^2 + 900*d*e^3* x^3 + 300*e^4*x^4)) + 60*B*c^2*(d + e*x)^5*Log[d + e*x])/(60*e^6*(d + e*x) ^5)
Time = 0.83 (sec) , antiderivative size = 248, 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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \int \left (\frac {B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )-2 A c e (2 c d-b e)}{e^5 (d+e x)^3}+\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{e^5 (d+e x)^4}-\frac {d^2 (B d-A e) (c d-b e)^2}{e^5 (d+e x)^6}+\frac {c (A c e+2 b B e-5 B c d)}{e^5 (d+e x)^2}+\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^5 (d+e x)^5}+\frac {B c^2}{e^5 (d+e x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 A c e (2 c d-b e)-B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )}{2 e^6 (d+e x)^2}-\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{3 e^6 (d+e x)^3}+\frac {d^2 (B d-A e) (c d-b e)^2}{5 e^6 (d+e x)^5}+\frac {c (-A c e-2 b B e+5 B c d)}{e^6 (d+e x)}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{4 e^6 (d+e x)^4}+\frac {B c^2 \log (d+e x)}{e^6}\) |
Input:
Int[((A + B*x)*(b*x + c*x^2)^2)/(d + e*x)^6,x]
Output:
(d^2*(B*d - A*e)*(c*d - b*e)^2)/(5*e^6*(d + e*x)^5) - (d*(c*d - b*e)*(B*d* (5*c*d - 3*b*e) - 2*A*e*(2*c*d - b*e)))/(4*e^6*(d + e*x)^4) - (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))/(3 *e^6*(d + e*x)^3) + (2*A*c*e*(2*c*d - b*e) - B*(10*c^2*d^2 - 8*b*c*d*e + b ^2*e^2))/(2*e^6*(d + e*x)^2) + (c*(5*B*c*d - 2*b*B*e - A*c*e))/(e^6*(d + e *x)) + (B*c^2*Log[d + e*x])/e^6
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 = 0.84 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.18
| method | result | size |
| risch | \(\frac {-\frac {c \left (A c e +2 B b e -5 B c d \right ) x^{4}}{e^{2}}-\frac {\left (2 A b c \,e^{2}+4 A \,c^{2} d e +B \,e^{2} b^{2}+8 B b c d e -30 B \,c^{2} d^{2}\right ) x^{3}}{2 e^{3}}-\frac {\left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -110 B \,c^{2} d^{3}\right ) x^{2}}{6 e^{4}}-\frac {d \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -125 B \,c^{2} d^{3}\right ) x}{12 e^{5}}-\frac {d^{2} \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -137 B \,c^{2} d^{3}\right )}{60 e^{6}}}{\left (e x +d \right )^{5}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}\) | \(292\) |
| norman | \(\frac {-\frac {d^{2} \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -137 B \,c^{2} d^{3}\right )}{60 e^{6}}-\frac {\left (A \,c^{2} e +2 B e b c -5 B \,c^{2} d \right ) x^{4}}{e^{2}}-\frac {\left (2 A b c \,e^{2}+4 A \,c^{2} d e +B \,e^{2} b^{2}+8 B b c d e -30 B \,c^{2} d^{2}\right ) x^{3}}{2 e^{3}}-\frac {\left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -110 B \,c^{2} d^{3}\right ) x^{2}}{6 e^{4}}-\frac {d \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -125 B \,c^{2} d^{3}\right ) x}{12 e^{5}}}{\left (e x +d \right )^{5}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}\) | \(296\) |
| default | \(-\frac {A \,b^{2} e^{3}-6 A b c d \,e^{2}+6 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+12 B b c \,d^{2} e -10 B \,c^{2} d^{3}}{3 e^{6} \left (e x +d \right )^{3}}+\frac {d \left (2 A \,b^{2} e^{3}-6 A b c d \,e^{2}+4 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+8 B b c \,d^{2} e -5 B \,c^{2} d^{3}\right )}{4 e^{6} \left (e x +d \right )^{4}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}-\frac {d^{2} \left (A \,b^{2} e^{3}-2 A b c d \,e^{2}+A \,c^{2} d^{2} e -B \,b^{2} d \,e^{2}+2 B b c \,d^{2} e -B \,c^{2} d^{3}\right )}{5 e^{6} \left (e x +d \right )^{5}}-\frac {c \left (A c e +2 B b e -5 B c d \right )}{e^{6} \left (e x +d \right )}-\frac {2 A b c \,e^{2}-4 A \,c^{2} d e +B \,e^{2} b^{2}-8 B b c d e +10 B \,c^{2} d^{2}}{2 e^{6} \left (e x +d \right )^{2}}\) | \(305\) |
| parallelrisch | \(-\frac {-1100 B \,x^{2} c^{2} d^{3} e^{2}+10 A x \,b^{2} d \,e^{4}+60 A x \,c^{2} d^{3} e^{2}+6 A b c \,d^{3} e^{2}+24 B b c \,d^{4} e -600 B \ln \left (e x +d \right ) x^{3} c^{2} d^{2} e^{3}+15 B x \,b^{2} d^{2} e^{3}-625 B x \,c^{2} d^{4} e +120 B \,x^{4} b c \,e^{5}+120 A \,x^{3} c^{2} d \,e^{4}-900 B \,x^{3} c^{2} d^{2} e^{3}+120 A \,x^{2} c^{2} d^{2} e^{3}+30 B \,x^{2} b^{2} d \,e^{4}+12 A \,c^{2} d^{4} e +60 A \,x^{3} b c \,e^{5}+30 A x b c \,d^{2} e^{3}+120 B x b c \,d^{3} e^{2}+60 A \,x^{2} b c d \,e^{4}+240 B \,x^{2} b c \,d^{2} e^{3}-60 B \ln \left (e x +d \right ) x^{5} c^{2} e^{5}+240 B \,x^{3} b c d \,e^{4}+2 A \,b^{2} d^{2} e^{3}+3 B \,b^{2} d^{3} e^{2}+60 A \,x^{4} c^{2} e^{5}+30 B \,x^{3} b^{2} e^{5}+20 A \,x^{2} b^{2} e^{5}-60 B \ln \left (e x +d \right ) c^{2} d^{5}-300 B \ln \left (e x +d \right ) x \,c^{2} d^{4} e -300 B \,x^{4} c^{2} d \,e^{4}-300 B \ln \left (e x +d \right ) x^{4} c^{2} d \,e^{4}-600 B \ln \left (e x +d \right ) x^{2} c^{2} d^{3} e^{2}-137 B \,c^{2} d^{5}}{60 e^{6} \left (e x +d \right )^{5}}\) | \(440\) |
Input:
int((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x,method=_RETURNVERBOSE)
Output:
(-c*(A*c*e+2*B*b*e-5*B*c*d)/e^2*x^4-1/2*(2*A*b*c*e^2+4*A*c^2*d*e+B*b^2*e^2 +8*B*b*c*d*e-30*B*c^2*d^2)/e^3*x^3-1/6*(2*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-110*B*c^2*d^3)/e^4*x^2-1/12*d*(2*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-125*B*c^2*d ^3)/e^5*x-1/60*d^2*(2*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-137*B*c^2*d^3)/e^6)/(e*x+d)^5+B*c^2*ln(e*x+d)/e^6
Time = 0.08 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.64 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {137 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 60 \, {\left (5 \, B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} - {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e^{2} - 2 \, A b^{2} e^{5} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x + 60 \, {\left (B c^{2} e^{5} x^{5} + 5 \, B c^{2} d e^{4} x^{4} + 10 \, B c^{2} d^{2} e^{3} x^{3} + 10 \, B c^{2} d^{3} e^{2} x^{2} + 5 \, B c^{2} d^{4} e x + B c^{2} d^{5}\right )} \log \left (e x + d\right )}{60 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="fricas")
Output:
1/60*(137*B*c^2*d^5 - 2*A*b^2*d^2*e^3 - 12*(2*B*b*c + A*c^2)*d^4*e - 3*(B* b^2 + 2*A*b*c)*d^3*e^2 + 60*(5*B*c^2*d*e^4 - (2*B*b*c + A*c^2)*e^5)*x^4 + 30*(30*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 + 10*(110*B*c^2*d^3*e^2 - 2*A*b^2*e^5 - 12*(2*B*b*c + A*c^2)*d^2*e^3 - 3*(B*b^2 + 2*A*b*c)*d*e^4)*x^2 + 5*(125*B*c^2*d^4*e - 2*A*b^2*d*e^4 - 12* (2*B*b*c + A*c^2)*d^3*e^2 - 3*(B*b^2 + 2*A*b*c)*d^2*e^3)*x + 60*(B*c^2*e^5 *x^5 + 5*B*c^2*d*e^4*x^4 + 10*B*c^2*d^2*e^3*x^3 + 10*B*c^2*d^3*e^2*x^2 + 5 *B*c^2*d^4*e*x + B*c^2*d^5)*log(e*x + d))/(e^11*x^5 + 5*d*e^10*x^4 + 10*d^ 2*e^9*x^3 + 10*d^3*e^8*x^2 + 5*d^4*e^7*x + d^5*e^6)
Time = 55.97 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {B c^{2} \log {\left (d + e x \right )}}{e^{6}} + \frac {- 2 A b^{2} d^{2} e^{3} - 6 A b c d^{3} e^{2} - 12 A c^{2} d^{4} e - 3 B b^{2} d^{3} e^{2} - 24 B b c d^{4} e + 137 B c^{2} d^{5} + x^{4} \left (- 60 A c^{2} e^{5} - 120 B b c e^{5} + 300 B c^{2} d e^{4}\right ) + x^{3} \left (- 60 A b c e^{5} - 120 A c^{2} d e^{4} - 30 B b^{2} e^{5} - 240 B b c d e^{4} + 900 B c^{2} d^{2} e^{3}\right ) + x^{2} \left (- 20 A b^{2} e^{5} - 60 A b c d e^{4} - 120 A c^{2} d^{2} e^{3} - 30 B b^{2} d e^{4} - 240 B b c d^{2} e^{3} + 1100 B c^{2} d^{3} e^{2}\right ) + x \left (- 10 A b^{2} d e^{4} - 30 A b c d^{2} e^{3} - 60 A c^{2} d^{3} e^{2} - 15 B b^{2} d^{2} e^{3} - 120 B b c d^{3} e^{2} + 625 B c^{2} d^{4} e\right )}{60 d^{5} e^{6} + 300 d^{4} e^{7} x + 600 d^{3} e^{8} x^{2} + 600 d^{2} e^{9} x^{3} + 300 d e^{10} x^{4} + 60 e^{11} x^{5}} \] Input:
integrate((B*x+A)*(c*x**2+b*x)**2/(e*x+d)**6,x)
Output:
B*c**2*log(d + e*x)/e**6 + (-2*A*b**2*d**2*e**3 - 6*A*b*c*d**3*e**2 - 12*A *c**2*d**4*e - 3*B*b**2*d**3*e**2 - 24*B*b*c*d**4*e + 137*B*c**2*d**5 + x* *4*(-60*A*c**2*e**5 - 120*B*b*c*e**5 + 300*B*c**2*d*e**4) + x**3*(-60*A*b* c*e**5 - 120*A*c**2*d*e**4 - 30*B*b**2*e**5 - 240*B*b*c*d*e**4 + 900*B*c** 2*d**2*e**3) + x**2*(-20*A*b**2*e**5 - 60*A*b*c*d*e**4 - 120*A*c**2*d**2*e **3 - 30*B*b**2*d*e**4 - 240*B*b*c*d**2*e**3 + 1100*B*c**2*d**3*e**2) + x* (-10*A*b**2*d*e**4 - 30*A*b*c*d**2*e**3 - 60*A*c**2*d**3*e**2 - 15*B*b**2* d**2*e**3 - 120*B*b*c*d**3*e**2 + 625*B*c**2*d**4*e))/(60*d**5*e**6 + 300* d**4*e**7*x + 600*d**3*e**8*x**2 + 600*d**2*e**9*x**3 + 300*d*e**10*x**4 + 60*e**11*x**5)
Time = 0.04 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {137 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 60 \, {\left (5 \, B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} - {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e^{2} - 2 \, A b^{2} e^{5} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x}{60 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} + \frac {B c^{2} \log \left (e x + d\right )}{e^{6}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="maxima")
Output:
1/60*(137*B*c^2*d^5 - 2*A*b^2*d^2*e^3 - 12*(2*B*b*c + A*c^2)*d^4*e - 3*(B* b^2 + 2*A*b*c)*d^3*e^2 + 60*(5*B*c^2*d*e^4 - (2*B*b*c + A*c^2)*e^5)*x^4 + 30*(30*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 + 10*(110*B*c^2*d^3*e^2 - 2*A*b^2*e^5 - 12*(2*B*b*c + A*c^2)*d^2*e^3 - 3*(B*b^2 + 2*A*b*c)*d*e^4)*x^2 + 5*(125*B*c^2*d^4*e - 2*A*b^2*d*e^4 - 12* (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^11*x^5 + 5* d*e^10*x^4 + 10*d^2*e^9*x^3 + 10*d^3*e^8*x^2 + 5*d^4*e^7*x + d^5*e^6) + B* c^2*log(e*x + d)/e^6
Time = 0.26 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.27 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {B c^{2} \log \left ({\left | e x + d \right |}\right )}{e^{6}} + \frac {60 \, {\left (5 \, B c^{2} d e^{3} - 2 \, B b c e^{4} - A c^{2} e^{4}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{2} - 8 \, B b c d e^{3} - 4 \, A c^{2} d e^{3} - B b^{2} e^{4} - 2 \, A b c e^{4}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e - 24 \, B b c d^{2} e^{2} - 12 \, A c^{2} d^{2} e^{2} - 3 \, B b^{2} d e^{3} - 6 \, A b c d e^{3} - 2 \, A b^{2} e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} - 24 \, B b c d^{3} e - 12 \, A c^{2} d^{3} e - 3 \, B b^{2} d^{2} e^{2} - 6 \, A b c d^{2} e^{2} - 2 \, A b^{2} d e^{3}\right )} x + \frac {137 \, B c^{2} d^{5} - 24 \, B b c d^{4} e - 12 \, A c^{2} d^{4} e - 3 \, B b^{2} d^{3} e^{2} - 6 \, A b c d^{3} e^{2} - 2 \, A b^{2} d^{2} e^{3}}{e}}{60 \, {\left (e x + d\right )}^{5} e^{5}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="giac")
Output:
B*c^2*log(abs(e*x + d))/e^6 + 1/60*(60*(5*B*c^2*d*e^3 - 2*B*b*c*e^4 - A*c^ 2*e^4)*x^4 + 30*(30*B*c^2*d^2*e^2 - 8*B*b*c*d*e^3 - 4*A*c^2*d*e^3 - B*b^2* e^4 - 2*A*b*c*e^4)*x^3 + 10*(110*B*c^2*d^3*e - 24*B*b*c*d^2*e^2 - 12*A*c^2 *d^2*e^2 - 3*B*b^2*d*e^3 - 6*A*b*c*d*e^3 - 2*A*b^2*e^4)*x^2 + 5*(125*B*c^2 *d^4 - 24*B*b*c*d^3*e - 12*A*c^2*d^3*e - 3*B*b^2*d^2*e^2 - 6*A*b*c*d^2*e^2 - 2*A*b^2*d*e^3)*x + (137*B*c^2*d^5 - 24*B*b*c*d^4*e - 12*A*c^2*d^4*e - 3 *B*b^2*d^3*e^2 - 6*A*b*c*d^3*e^2 - 2*A*b^2*d^2*e^3)/e)/((e*x + d)^5*e^5)
Time = 10.88 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {B\,c^2\,\ln \left (d+e\,x\right )}{e^6}-\frac {\frac {3\,B\,b^2\,d^3\,e^2+2\,A\,b^2\,d^2\,e^3+24\,B\,b\,c\,d^4\,e+6\,A\,b\,c\,d^3\,e^2-137\,B\,c^2\,d^5+12\,A\,c^2\,d^4\,e}{60\,e^6}+\frac {x^3\,\left (B\,b^2\,e^2+8\,B\,b\,c\,d\,e+2\,A\,b\,c\,e^2-30\,B\,c^2\,d^2+4\,A\,c^2\,d\,e\right )}{2\,e^3}+\frac {x\,\left (3\,B\,b^2\,d^2\,e^2+2\,A\,b^2\,d\,e^3+24\,B\,b\,c\,d^3\,e+6\,A\,b\,c\,d^2\,e^2-125\,B\,c^2\,d^4+12\,A\,c^2\,d^3\,e\right )}{12\,e^5}+\frac {x^2\,\left (3\,B\,b^2\,d\,e^2+2\,A\,b^2\,e^3+24\,B\,b\,c\,d^2\,e+6\,A\,b\,c\,d\,e^2-110\,B\,c^2\,d^3+12\,A\,c^2\,d^2\,e\right )}{6\,e^4}+\frac {c\,x^4\,\left (A\,c\,e+2\,B\,b\,e-5\,B\,c\,d\right )}{e^2}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \] Input:
int(((b*x + c*x^2)^2*(A + B*x))/(d + e*x)^6,x)
Output:
(B*c^2*log(d + e*x))/e^6 - ((12*A*c^2*d^4*e - 137*B*c^2*d^5 + 2*A*b^2*d^2* e^3 + 3*B*b^2*d^3*e^2 + 24*B*b*c*d^4*e + 6*A*b*c*d^3*e^2)/(60*e^6) + (x^3* (B*b^2*e^2 - 30*B*c^2*d^2 + 2*A*b*c*e^2 + 4*A*c^2*d*e + 8*B*b*c*d*e))/(2*e ^3) + (x*(2*A*b^2*d*e^3 - 125*B*c^2*d^4 + 12*A*c^2*d^3*e + 3*B*b^2*d^2*e^2 + 24*B*b*c*d^3*e + 6*A*b*c*d^2*e^2))/(12*e^5) + (x^2*(2*A*b^2*e^3 - 110*B *c^2*d^3 + 12*A*c^2*d^2*e + 3*B*b^2*d*e^2 + 6*A*b*c*d*e^2 + 24*B*b*c*d^2*e ))/(6*e^4) + (c*x^4*(A*c*e + 2*B*b*e - 5*B*c*d))/e^2)/(d^5 + e^5*x^5 + 5*d *e^4*x^4 + 10*d^3*e^2*x^2 + 10*d^2*e^3*x^3 + 5*d^4*e*x)
Time = 0.25 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.60 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {60 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{6}+300 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{5} e x +600 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{4} e^{2} x^{2}+600 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{3} e^{3} x^{3}+300 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{2} e^{4} x^{4}+60 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d \,e^{5} x^{5}-2 a \,b^{2} d^{3} e^{3}-10 a \,b^{2} d^{2} e^{4} x -20 a \,b^{2} d \,e^{5} x^{2}-6 a b c \,d^{4} e^{2}-30 a b c \,d^{3} e^{3} x -60 a b c \,d^{2} e^{4} x^{2}-60 a b c d \,e^{5} x^{3}+12 a \,c^{2} e^{6} x^{5}-3 b^{3} d^{4} e^{2}-15 b^{3} d^{3} e^{3} x -30 b^{3} d^{2} e^{4} x^{2}-30 b^{3} d \,e^{5} x^{3}+24 b^{2} c \,e^{6} x^{5}+77 b \,c^{2} d^{6}+325 b \,c^{2} d^{5} e x +500 b \,c^{2} d^{4} e^{2} x^{2}+300 b \,c^{2} d^{3} e^{3} x^{3}-60 b \,c^{2} d \,e^{5} x^{5}}{60 d \,e^{6} \left (e^{5} x^{5}+5 d \,e^{4} x^{4}+10 d^{2} e^{3} x^{3}+10 d^{3} e^{2} x^{2}+5 d^{4} e x +d^{5}\right )} \] Input:
int((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x)
Output:
(60*log(d + e*x)*b*c**2*d**6 + 300*log(d + e*x)*b*c**2*d**5*e*x + 600*log( d + e*x)*b*c**2*d**4*e**2*x**2 + 600*log(d + e*x)*b*c**2*d**3*e**3*x**3 + 300*log(d + e*x)*b*c**2*d**2*e**4*x**4 + 60*log(d + e*x)*b*c**2*d*e**5*x** 5 - 2*a*b**2*d**3*e**3 - 10*a*b**2*d**2*e**4*x - 20*a*b**2*d*e**5*x**2 - 6 *a*b*c*d**4*e**2 - 30*a*b*c*d**3*e**3*x - 60*a*b*c*d**2*e**4*x**2 - 60*a*b *c*d*e**5*x**3 + 12*a*c**2*e**6*x**5 - 3*b**3*d**4*e**2 - 15*b**3*d**3*e** 3*x - 30*b**3*d**2*e**4*x**2 - 30*b**3*d*e**5*x**3 + 24*b**2*c*e**6*x**5 + 77*b*c**2*d**6 + 325*b*c**2*d**5*e*x + 500*b*c**2*d**4*e**2*x**2 + 300*b* c**2*d**3*e**3*x**3 - 60*b*c**2*d*e**5*x**5)/(60*d*e**6*(d**5 + 5*d**4*e*x + 10*d**3*e**2*x**2 + 10*d**2*e**3*x**3 + 5*d*e**4*x**4 + e**5*x**5))