\(\int \frac {(A+B x) (b x+c x^2)^2}{(d+e x)^6} \, dx\) [1124]

Optimal result

Integrand size = 24, antiderivative size = 248 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {d^2 (B d-A e) (c d-b e)^2}{5 e^6 (d+e x)^5}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{4 e^6 (d+e x)^4}-\frac {A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )}{3 e^6 (d+e x)^3}+\frac {2 A c e (2 c d-b e)-B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )}{2 e^6 (d+e x)^2}+\frac {c (5 B c d-2 b B e-A c e)}{e^6 (d+e x)}+\frac {B c^2 \log (d+e x)}{e^6} \]

[Out]

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*l
n(e*x+d)/e^6

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {785} \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {2 A c e (2 c d-b e)-B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )}{2 e^6 (d+e x)^2}-\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{3 e^6 (d+e x)^3}+\frac {d^2 (B d-A e) (c d-b e)^2}{5 e^6 (d+e x)^5}+\frac {c (-A c e-2 b B e+5 B c d)}{e^6 (d+e x)}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{4 e^6 (d+e x)^4}+\frac {B c^2 \log (d+e x)}{e^6} \]

[In]

Int[((A + B*x)*(b*x + c*x^2)^2)/(d + e*x)^6,x]

[Out]

(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

Rule 785

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {d^2 (B d-A e) (c d-b e)^2}{e^5 (d+e x)^6}+\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 {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^5 (d+e x)^4}+\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^5 (d+e x)^3}+\frac {c (-5 B c d+2 b B e+A c e)}{e^5 (d+e x)^2}+\frac {B c^2}{e^5 (d+e x)}\right ) \, 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {-2 A e \left (b^2 e^2 \left (d^2+5 d e x+10 e^2 x^2\right )+3 b c e \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )+6 c^2 \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )\right )+B \left (-3 b^2 e^2 \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )-24 b c e \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )+c^2 d \left (137 d^4+625 d^3 e x+1100 d^2 e^2 x^2+900 d e^3 x^3+300 e^4 x^4\right )\right )+60 B c^2 (d+e x)^5 \log (d+e x)}{60 e^6 (d+e x)^5} \]

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^2)/(d + e*x)^6,x]

[Out]

(-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)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.18

method	result	size
risch	\(\frac {-\frac {c \left (A c e +2 B b e -5 B c d \right ) x^{4}}{e^{2}}-\frac {\left (2 A b c \,e^{2}+4 A \,c^{2} d e +B \,b^{2} e^{2}+8 B b c d e -30 B \,c^{2} d^{2}\right ) x^{3}}{2 e^{3}}-\frac {\left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -110 B \,c^{2} d^{3}\right ) x^{2}}{6 e^{4}}-\frac {d \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -125 B \,c^{2} d^{3}\right ) x}{12 e^{5}}-\frac {d^{2} \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -137 B \,c^{2} d^{3}\right )}{60 e^{6}}}{\left (e x +d \right )^{5}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}\)	\(292\)
norman	\(\frac {-\frac {d^{2} \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -137 B \,c^{2} d^{3}\right )}{60 e^{6}}-\frac {\left (A \,c^{2} e +2 B e b c -5 B \,c^{2} d \right ) x^{4}}{e^{2}}-\frac {\left (2 A b c \,e^{2}+4 A \,c^{2} d e +B \,b^{2} e^{2}+8 B b c d e -30 B \,c^{2} d^{2}\right ) x^{3}}{2 e^{3}}-\frac {\left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -110 B \,c^{2} d^{3}\right ) x^{2}}{6 e^{4}}-\frac {d \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -125 B \,c^{2} d^{3}\right ) x}{12 e^{5}}}{\left (e x +d \right )^{5}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}\)	\(296\)
default	\(-\frac {d^{2} \left (A \,b^{2} e^{3}-2 A b c d \,e^{2}+A \,c^{2} d^{2} e -B \,b^{2} d \,e^{2}+2 B b c \,d^{2} e -B \,c^{2} d^{3}\right )}{5 e^{6} \left (e x +d \right )^{5}}-\frac {A \,b^{2} e^{3}-6 A b c d \,e^{2}+6 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+12 B b c \,d^{2} e -10 B \,c^{2} d^{3}}{3 e^{6} \left (e x +d \right )^{3}}-\frac {c \left (A c e +2 B b e -5 B c d \right )}{e^{6} \left (e x +d \right )}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}+\frac {d \left (2 A \,b^{2} e^{3}-6 A b c d \,e^{2}+4 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+8 B b c \,d^{2} e -5 B \,c^{2} d^{3}\right )}{4 e^{6} \left (e x +d \right )^{4}}-\frac {2 A b c \,e^{2}-4 A \,c^{2} d e +B \,b^{2} e^{2}-8 B b c d e +10 B \,c^{2} d^{2}}{2 e^{6} \left (e x +d \right )^{2}}\)	\(305\)
parallelrisch	\(-\frac {-137 B \,c^{2} d^{5}-300 B \ln \left (e x +d \right ) x \,c^{2} d^{4} e -600 B \ln \left (e x +d \right ) x^{3} c^{2} d^{2} e^{3}-300 B \ln \left (e x +d \right ) x^{4} c^{2} d \,e^{4}-600 B \ln \left (e x +d \right ) x^{2} c^{2} d^{3} e^{2}-60 B \ln \left (e x +d \right ) c^{2} d^{5}+60 A \,x^{4} c^{2} e^{5}+30 B \,x^{3} b^{2} e^{5}+20 A \,x^{2} b^{2} e^{5}+120 B x b c \,d^{3} e^{2}+30 A x b c \,d^{2} e^{3}+240 B \,x^{2} b c \,d^{2} e^{3}+60 A \,x^{2} b c d \,e^{4}+240 B \,x^{3} b c d \,e^{4}+2 A \,b^{2} d^{2} e^{3}+12 A \,c^{2} d^{4} e +3 B \,b^{2} d^{3} e^{2}+6 A b c \,d^{3} e^{2}+24 B b c \,d^{4} e -60 B \ln \left (e x +d \right ) x^{5} c^{2} e^{5}+60 A \,x^{3} b c \,e^{5}+120 A \,x^{3} c^{2} d \,e^{4}-900 B \,x^{3} c^{2} d^{2} e^{3}+120 A \,x^{2} c^{2} d^{2} e^{3}+30 B \,x^{2} b^{2} d \,e^{4}-1100 B \,x^{2} c^{2} d^{3} e^{2}+10 A x \,b^{2} d \,e^{4}+60 A x \,c^{2} d^{3} e^{2}+15 B x \,b^{2} d^{2} e^{3}-625 B x \,c^{2} d^{4} e +120 B \,x^{4} b c \,e^{5}-300 B \,x^{4} c^{2} d \,e^{4}}{60 e^{6} \left (e x +d \right )^{5}}\)	\(440\)

[In]

int((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x,method=_RETURNVERBOSE)

[Out]

(-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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.64 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {137 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 60 \, {\left (5 \, B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} - {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e^{2} - 2 \, A b^{2} e^{5} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x + 60 \, {\left (B c^{2} e^{5} x^{5} + 5 \, B c^{2} d e^{4} x^{4} + 10 \, B c^{2} d^{2} e^{3} x^{3} + 10 \, B c^{2} d^{3} e^{2} x^{2} + 5 \, B c^{2} d^{4} e x + B c^{2} d^{5}\right )} \log \left (e x + d\right )}{60 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="fricas")

[Out]

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)

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\text {Timed out} \]

[In]

integrate((B*x+A)*(c*x**2+b*x)**2/(e*x+d)**6,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {137 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 60 \, {\left (5 \, B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} - {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e^{2} - 2 \, A b^{2} e^{5} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x}{60 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} + \frac {B c^{2} \log \left (e x + d\right )}{e^{6}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.27 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {B c^{2} \log \left ({\left | e x + d \right |}\right )}{e^{6}} + \frac {60 \, {\left (5 \, B c^{2} d e^{3} - 2 \, B b c e^{4} - A c^{2} e^{4}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{2} - 8 \, B b c d e^{3} - 4 \, A c^{2} d e^{3} - B b^{2} e^{4} - 2 \, A b c e^{4}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e - 24 \, B b c d^{2} e^{2} - 12 \, A c^{2} d^{2} e^{2} - 3 \, B b^{2} d e^{3} - 6 \, A b c d e^{3} - 2 \, A b^{2} e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} - 24 \, B b c d^{3} e - 12 \, A c^{2} d^{3} e - 3 \, B b^{2} d^{2} e^{2} - 6 \, A b c d^{2} e^{2} - 2 \, A b^{2} d e^{3}\right )} x + \frac {137 \, B c^{2} d^{5} - 24 \, B b c d^{4} e - 12 \, A c^{2} d^{4} e - 3 \, B b^{2} d^{3} e^{2} - 6 \, A b c d^{3} e^{2} - 2 \, A b^{2} d^{2} e^{3}}{e}}{60 \, {\left (e x + d\right )}^{5} e^{5}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 10.87 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {B\,c^2\,\ln \left (d+e\,x\right )}{e^6}-\frac {\frac {3\,B\,b^2\,d^3\,e^2+2\,A\,b^2\,d^2\,e^3+24\,B\,b\,c\,d^4\,e+6\,A\,b\,c\,d^3\,e^2-137\,B\,c^2\,d^5+12\,A\,c^2\,d^4\,e}{60\,e^6}+\frac {x^3\,\left (B\,b^2\,e^2+8\,B\,b\,c\,d\,e+2\,A\,b\,c\,e^2-30\,B\,c^2\,d^2+4\,A\,c^2\,d\,e\right )}{2\,e^3}+\frac {x\,\left (3\,B\,b^2\,d^2\,e^2+2\,A\,b^2\,d\,e^3+24\,B\,b\,c\,d^3\,e+6\,A\,b\,c\,d^2\,e^2-125\,B\,c^2\,d^4+12\,A\,c^2\,d^3\,e\right )}{12\,e^5}+\frac {x^2\,\left (3\,B\,b^2\,d\,e^2+2\,A\,b^2\,e^3+24\,B\,b\,c\,d^2\,e+6\,A\,b\,c\,d\,e^2-110\,B\,c^2\,d^3+12\,A\,c^2\,d^2\,e\right )}{6\,e^4}+\frac {c\,x^4\,\left (A\,c\,e+2\,B\,b\,e-5\,B\,c\,d\right )}{e^2}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \]

[In]

int(((b*x + c*x^2)^2*(A + B*x))/(d + e*x)^6,x)

[Out]

(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)