\(\int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx\) [32]

Optimal result

Integrand size = 20, antiderivative size = 150 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx=\frac {3 b (b d-a e) (2 b B d-A b e-a B e) x}{e^4}-\frac {(b d-a e)^3 (B d-A e)}{e^5 (d+e x)}-\frac {b^2 (4 b B d-A b e-3 a B e) (d+e x)^2}{2 e^5}+\frac {b^3 B (d+e x)^3}{3 e^5}-\frac {(b d-a e)^2 (4 b B d-3 A b e-a B e) \log (d+e x)}{e^5} \] Output:

3*b*(-a*e+b*d)*(-A*b*e-B*a*e+2*B*b*d)*x/e^4-(-a*e+b*d)^3*(-A*e+B*d)/e^5/(e 
*x+d)-1/2*b^2*(-A*b*e-3*B*a*e+4*B*b*d)*(e*x+d)^2/e^5+1/3*b^3*B*(e*x+d)^3/e 
^5-(-a*e+b*d)^2*(-3*A*b*e-B*a*e+4*B*b*d)*ln(e*x+d)/e^5
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.63 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx=\frac {6 a^3 e^3 (B d-A e)+18 a^2 b e^2 \left (A d e+B \left (-d^2+d e x+e^2 x^2\right )\right )+9 a b^2 e \left (2 A e \left (-d^2+d e x+e^2 x^2\right )+B \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )\right )+b^3 \left (3 A e \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )+2 B \left (-3 d^4+9 d^3 e x+6 d^2 e^2 x^2-2 d e^3 x^3+e^4 x^4\right )\right )-6 (b d-a e)^2 (4 b B d-3 A b e-a B e) (d+e x) \log (d+e x)}{6 e^5 (d+e x)} \] Input:

Integrate[((a + b*x)^3*(A + B*x))/(d + e*x)^2,x]
 

Output:

(6*a^3*e^3*(B*d - A*e) + 18*a^2*b*e^2*(A*d*e + B*(-d^2 + d*e*x + e^2*x^2)) 
 + 9*a*b^2*e*(2*A*e*(-d^2 + d*e*x + e^2*x^2) + B*(2*d^3 - 4*d^2*e*x - 3*d* 
e^2*x^2 + e^3*x^3)) + b^3*(3*A*e*(2*d^3 - 4*d^2*e*x - 3*d*e^2*x^2 + e^3*x^ 
3) + 2*B*(-3*d^4 + 9*d^3*e*x + 6*d^2*e^2*x^2 - 2*d*e^3*x^3 + e^4*x^4)) - 6 
*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e)*(d + e*x)*Log[d + e*x])/(6*e^5* 
(d + e*x))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {86, 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.

\(\displaystyle \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx\)

\(\Big \downarrow \) 86

\(\displaystyle \int \left (\frac {b^2 (d+e x) (3 a B e+A b e-4 b B d)}{e^4}+\frac {(a e-b d)^2 (a B e+3 A b e-4 b B d)}{e^4 (d+e x)}+\frac {(a e-b d)^3 (A e-B d)}{e^4 (d+e x)^2}-\frac {3 b (b d-a e) (a B e+A b e-2 b B d)}{e^4}+\frac {b^3 B (d+e x)^2}{e^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {b^2 (d+e x)^2 (-3 a B e-A b e+4 b B d)}{2 e^5}-\frac {(b d-a e)^3 (B d-A e)}{e^5 (d+e x)}-\frac {(b d-a e)^2 \log (d+e x) (-a B e-3 A b e+4 b B d)}{e^5}+\frac {3 b x (b d-a e) (-a B e-A b e+2 b B d)}{e^4}+\frac {b^3 B (d+e x)^3}{3 e^5}\)

Input:

Int[((a + b*x)^3*(A + B*x))/(d + e*x)^2,x]
 

Output:

(3*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e)*x)/e^4 - ((b*d - a*e)^3*(B*d - 
A*e))/(e^5*(d + e*x)) - (b^2*(4*b*B*d - A*b*e - 3*a*B*e)*(d + e*x)^2)/(2*e 
^5) + (b^3*B*(d + e*x)^3)/(3*e^5) - ((b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a* 
B*e)*Log[d + e*x])/e^5
 

Defintions of rubi rules used

rule 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.84

method	result	size
norman	\(\frac {\frac {\left (a^{3} A \,e^{4}-3 A \,a^{2} b d \,e^{3}+6 A a \,b^{2} d^{2} e^{2}-3 A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+6 B \,a^{2} b \,d^{2} e^{2}-9 B a \,b^{2} d^{3} e +4 b^{3} B \,d^{4}\right ) x}{e^{4} d}+\frac {b \left (6 A a b \,e^{2}-3 A \,b^{2} d e +6 B \,a^{2} e^{2}-9 B a b d e +4 b^{2} B \,d^{2}\right ) x^{2}}{2 e^{3}}+\frac {b^{2} \left (3 A b e +9 B a e -4 B b d \right ) x^{3}}{6 e^{2}}+\frac {b^{3} B \,x^{4}}{3 e}}{e x +d}+\frac {\left (3 A \,a^{2} b \,e^{3}-6 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e +B \,a^{3} e^{3}-6 B \,a^{2} b d \,e^{2}+9 B a \,b^{2} d^{2} e -4 b^{3} B \,d^{3}\right ) \ln \left (e x +d \right )}{e^{5}}\)	\(276\)
default	\(\frac {b \left (\frac {1}{3} b^{2} B \,x^{3} e^{2}+\frac {1}{2} A \,b^{2} e^{2} x^{2}+\frac {3}{2} B a b \,e^{2} x^{2}-B \,b^{2} d e \,x^{2}+3 A a b \,e^{2} x -2 A \,b^{2} d e x +3 B \,a^{2} e^{2} x -6 B a b d e x +3 b^{2} B \,d^{2} x \right )}{e^{4}}-\frac {a^{3} A \,e^{4}-3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}-A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}-3 B a \,b^{2} d^{3} e +b^{3} B \,d^{4}}{e^{5} \left (e x +d \right )}+\frac {\left (3 A \,a^{2} b \,e^{3}-6 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e +B \,a^{3} e^{3}-6 B \,a^{2} b d \,e^{2}+9 B a \,b^{2} d^{2} e -4 b^{3} B \,d^{3}\right ) \ln \left (e x +d \right )}{e^{5}}\)	\(278\)
risch	\(\frac {b^{3} B \,x^{3}}{3 e^{2}}+\frac {b^{3} A \,x^{2}}{2 e^{2}}+\frac {3 b^{2} B a \,x^{2}}{2 e^{2}}-\frac {b^{3} B d \,x^{2}}{e^{3}}+\frac {3 b^{2} A a x}{e^{2}}-\frac {2 b^{3} A d x}{e^{3}}+\frac {3 b B \,a^{2} x}{e^{2}}-\frac {6 b^{2} B a d x}{e^{3}}+\frac {3 b^{3} B \,d^{2} x}{e^{4}}-\frac {a^{3} A}{e \left (e x +d \right )}+\frac {3 A \,a^{2} b d}{e^{2} \left (e x +d \right )}-\frac {3 A a \,b^{2} d^{2}}{e^{3} \left (e x +d \right )}+\frac {A \,b^{3} d^{3}}{e^{4} \left (e x +d \right )}+\frac {B \,a^{3} d}{e^{2} \left (e x +d \right )}-\frac {3 B \,a^{2} b \,d^{2}}{e^{3} \left (e x +d \right )}+\frac {3 B a \,b^{2} d^{3}}{e^{4} \left (e x +d \right )}-\frac {b^{3} B \,d^{4}}{e^{5} \left (e x +d \right )}+\frac {3 \ln \left (e x +d \right ) A \,a^{2} b}{e^{2}}-\frac {6 \ln \left (e x +d \right ) A a \,b^{2} d}{e^{3}}+\frac {3 \ln \left (e x +d \right ) A \,b^{3} d^{2}}{e^{4}}+\frac {\ln \left (e x +d \right ) B \,a^{3}}{e^{2}}-\frac {6 \ln \left (e x +d \right ) B \,a^{2} b d}{e^{3}}+\frac {9 \ln \left (e x +d \right ) B a \,b^{2} d^{2}}{e^{4}}-\frac {4 \ln \left (e x +d \right ) b^{3} B \,d^{3}}{e^{5}}\)	\(376\)
parallelrisch	\(\frac {-24 b^{3} B \,d^{4}+6 B \,a^{3} d \,e^{3}+18 A \,b^{3} d^{3} e +54 B a \,b^{2} d^{3} e -36 A a \,b^{2} d^{2} e^{2}-36 B \,a^{2} b \,d^{2} e^{2}+18 A \,a^{2} b d \,e^{3}+9 B \,x^{3} a \,b^{2} e^{4}-4 B \,x^{3} b^{3} d \,e^{3}+18 A \,x^{2} a \,b^{2} e^{4}-9 A \,x^{2} b^{3} d \,e^{3}+18 A \ln \left (e x +d \right ) x \,b^{3} d^{2} e^{2}-24 B \ln \left (e x +d \right ) x \,b^{3} d^{3} e -6 a^{3} A \,e^{4}+6 B \ln \left (e x +d \right ) x \,a^{3} e^{4}+2 B \,x^{4} b^{3} e^{4}-24 B \ln \left (e x +d \right ) b^{3} d^{4}+3 A \,x^{3} b^{3} e^{4}-27 B \,x^{2} a \,b^{2} d \,e^{3}+18 A \ln \left (e x +d \right ) a^{2} b d \,e^{3}-36 A \ln \left (e x +d \right ) a \,b^{2} d^{2} e^{2}-36 B \ln \left (e x +d \right ) a^{2} b \,d^{2} e^{2}+54 B \ln \left (e x +d \right ) a \,b^{2} d^{3} e +18 A \ln \left (e x +d \right ) x \,a^{2} b \,e^{4}-36 A \ln \left (e x +d \right ) x a \,b^{2} d \,e^{3}-36 B \ln \left (e x +d \right ) x \,a^{2} b d \,e^{3}+54 B \ln \left (e x +d \right ) x a \,b^{2} d^{2} e^{2}+18 A \ln \left (e x +d \right ) b^{3} d^{3} e +6 B \ln \left (e x +d \right ) a^{3} d \,e^{3}+18 B \,x^{2} a^{2} b \,e^{4}+12 B \,x^{2} b^{3} d^{2} e^{2}}{6 e^{5} \left (e x +d \right )}\)	\(462\)

Input:

int((b*x+a)^3*(B*x+A)/(e*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

((A*a^3*e^4-3*A*a^2*b*d*e^3+6*A*a*b^2*d^2*e^2-3*A*b^3*d^3*e-B*a^3*d*e^3+6* 
B*a^2*b*d^2*e^2-9*B*a*b^2*d^3*e+4*B*b^3*d^4)/e^4/d*x+1/2*b*(6*A*a*b*e^2-3* 
A*b^2*d*e+6*B*a^2*e^2-9*B*a*b*d*e+4*B*b^2*d^2)/e^3*x^2+1/6*b^2*(3*A*b*e+9* 
B*a*e-4*B*b*d)/e^2*x^3+1/3*b^3*B/e*x^4)/(e*x+d)+1/e^5*(3*A*a^2*b*e^3-6*A*a 
*b^2*d*e^2+3*A*b^3*d^2*e+B*a^3*e^3-6*B*a^2*b*d*e^2+9*B*a*b^2*d^2*e-4*B*b^3 
*d^3)*ln(e*x+d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (146) = 292\).

Time = 0.07 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.64 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx=\frac {2 \, B b^{3} e^{4} x^{4} - 6 \, B b^{3} d^{4} - 6 \, A a^{3} e^{4} + 6 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e - 18 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - {\left (4 \, B b^{3} d e^{3} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 3 \, {\left (4 \, B b^{3} d^{2} e^{2} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 6 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 6 \, {\left (3 \, B b^{3} d^{3} e - 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3}\right )} x - 6 \, {\left (4 \, B b^{3} d^{4} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 6 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + {\left (4 \, B b^{3} d^{3} e - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 6 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{6} x + d e^{5}\right )}} \] Input:

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^2,x, algorithm="fricas")
 

Output:

1/6*(2*B*b^3*e^4*x^4 - 6*B*b^3*d^4 - 6*A*a^3*e^4 + 6*(3*B*a*b^2 + A*b^3)*d 
^3*e - 18*(B*a^2*b + A*a*b^2)*d^2*e^2 + 6*(B*a^3 + 3*A*a^2*b)*d*e^3 - (4*B 
*b^3*d*e^3 - 3*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 3*(4*B*b^3*d^2*e^2 - 3*(3*B* 
a*b^2 + A*b^3)*d*e^3 + 6*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 6*(3*B*b^3*d^3*e - 
 2*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 3*(B*a^2*b + A*a*b^2)*d*e^3)*x - 6*(4*B*b 
^3*d^4 - 3*(3*B*a*b^2 + A*b^3)*d^3*e + 6*(B*a^2*b + A*a*b^2)*d^2*e^2 - (B* 
a^3 + 3*A*a^2*b)*d*e^3 + (4*B*b^3*d^3*e - 3*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 
6*(B*a^2*b + A*a*b^2)*d*e^3 - (B*a^3 + 3*A*a^2*b)*e^4)*x)*log(e*x + d))/(e 
^6*x + d*e^5)
 

Sympy [A] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.71 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx=\frac {B b^{3} x^{3}}{3 e^{2}} + x^{2} \left (\frac {A b^{3}}{2 e^{2}} + \frac {3 B a b^{2}}{2 e^{2}} - \frac {B b^{3} d}{e^{3}}\right ) + x \left (\frac {3 A a b^{2}}{e^{2}} - \frac {2 A b^{3} d}{e^{3}} + \frac {3 B a^{2} b}{e^{2}} - \frac {6 B a b^{2} d}{e^{3}} + \frac {3 B b^{3} d^{2}}{e^{4}}\right ) + \frac {- A a^{3} e^{4} + 3 A a^{2} b d e^{3} - 3 A a b^{2} d^{2} e^{2} + A b^{3} d^{3} e + B a^{3} d e^{3} - 3 B a^{2} b d^{2} e^{2} + 3 B a b^{2} d^{3} e - B b^{3} d^{4}}{d e^{5} + e^{6} x} + \frac {\left (a e - b d\right )^{2} \cdot \left (3 A b e + B a e - 4 B b d\right ) \log {\left (d + e x \right )}}{e^{5}} \] Input:

integrate((b*x+a)**3*(B*x+A)/(e*x+d)**2,x)
 

Output:

B*b**3*x**3/(3*e**2) + x**2*(A*b**3/(2*e**2) + 3*B*a*b**2/(2*e**2) - B*b** 
3*d/e**3) + x*(3*A*a*b**2/e**2 - 2*A*b**3*d/e**3 + 3*B*a**2*b/e**2 - 6*B*a 
*b**2*d/e**3 + 3*B*b**3*d**2/e**4) + (-A*a**3*e**4 + 3*A*a**2*b*d*e**3 - 3 
*A*a*b**2*d**2*e**2 + A*b**3*d**3*e + B*a**3*d*e**3 - 3*B*a**2*b*d**2*e**2 
 + 3*B*a*b**2*d**3*e - B*b**3*d**4)/(d*e**5 + e**6*x) + (a*e - b*d)**2*(3* 
A*b*e + B*a*e - 4*B*b*d)*log(d + e*x)/e**5
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.78 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx=-\frac {B b^{3} d^{4} + A a^{3} e^{4} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3}}{e^{6} x + d e^{5}} + \frac {2 \, B b^{3} e^{2} x^{3} - 3 \, {\left (2 \, B b^{3} d e - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{2}\right )} x^{2} + 6 \, {\left (3 \, B b^{3} d^{2} - 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e^{2}\right )} x}{6 \, e^{4}} - \frac {{\left (4 \, B b^{3} d^{3} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e + 6 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{3}\right )} \log \left (e x + d\right )}{e^{5}} \] Input:

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^2,x, algorithm="maxima")
 

Output:

-(B*b^3*d^4 + A*a^3*e^4 - (3*B*a*b^2 + A*b^3)*d^3*e + 3*(B*a^2*b + A*a*b^2 
)*d^2*e^2 - (B*a^3 + 3*A*a^2*b)*d*e^3)/(e^6*x + d*e^5) + 1/6*(2*B*b^3*e^2* 
x^3 - 3*(2*B*b^3*d*e - (3*B*a*b^2 + A*b^3)*e^2)*x^2 + 6*(3*B*b^3*d^2 - 2*( 
3*B*a*b^2 + A*b^3)*d*e + 3*(B*a^2*b + A*a*b^2)*e^2)*x)/e^4 - (4*B*b^3*d^3 
- 3*(3*B*a*b^2 + A*b^3)*d^2*e + 6*(B*a^2*b + A*a*b^2)*d*e^2 - (B*a^3 + 3*A 
*a^2*b)*e^3)*log(e*x + d)/e^5
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (146) = 292\).

Time = 0.12 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.48 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx=\frac {{\left (2 \, B b^{3} - \frac {3 \, {\left (4 \, B b^{3} d e - 3 \, B a b^{2} e^{2} - A b^{3} e^{2}\right )}}{{\left (e x + d\right )} e} + \frac {18 \, {\left (2 \, B b^{3} d^{2} e^{2} - 3 \, B a b^{2} d e^{3} - A b^{3} d e^{3} + B a^{2} b e^{4} + A a b^{2} e^{4}\right )}}{{\left (e x + d\right )}^{2} e^{2}}\right )} {\left (e x + d\right )}^{3}}{6 \, e^{5}} + \frac {{\left (4 \, B b^{3} d^{3} - 9 \, B a b^{2} d^{2} e - 3 \, A b^{3} d^{2} e + 6 \, B a^{2} b d e^{2} + 6 \, A a b^{2} d e^{2} - B a^{3} e^{3} - 3 \, A a^{2} b e^{3}\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{5}} - \frac {\frac {B b^{3} d^{4} e^{3}}{e x + d} - \frac {3 \, B a b^{2} d^{3} e^{4}}{e x + d} - \frac {A b^{3} d^{3} e^{4}}{e x + d} + \frac {3 \, B a^{2} b d^{2} e^{5}}{e x + d} + \frac {3 \, A a b^{2} d^{2} e^{5}}{e x + d} - \frac {B a^{3} d e^{6}}{e x + d} - \frac {3 \, A a^{2} b d e^{6}}{e x + d} + \frac {A a^{3} e^{7}}{e x + d}}{e^{8}} \] Input:

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^2,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

1/6*(2*B*b^3 - 3*(4*B*b^3*d*e - 3*B*a*b^2*e^2 - A*b^3*e^2)/((e*x + d)*e) + 
 18*(2*B*b^3*d^2*e^2 - 3*B*a*b^2*d*e^3 - A*b^3*d*e^3 + B*a^2*b*e^4 + A*a*b 
^2*e^4)/((e*x + d)^2*e^2))*(e*x + d)^3/e^5 + (4*B*b^3*d^3 - 9*B*a*b^2*d^2* 
e - 3*A*b^3*d^2*e + 6*B*a^2*b*d*e^2 + 6*A*a*b^2*d*e^2 - B*a^3*e^3 - 3*A*a^ 
2*b*e^3)*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^5 - (B*b^3*d^4*e^3/(e*x 
+ d) - 3*B*a*b^2*d^3*e^4/(e*x + d) - A*b^3*d^3*e^4/(e*x + d) + 3*B*a^2*b*d 
^2*e^5/(e*x + d) + 3*A*a*b^2*d^2*e^5/(e*x + d) - B*a^3*d*e^6/(e*x + d) - 3 
*A*a^2*b*d*e^6/(e*x + d) + A*a^3*e^7/(e*x + d))/e^8
 

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.95 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx=x^2\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{2\,e^2}-\frac {B\,b^3\,d}{e^3}\right )-x\,\left (\frac {2\,d\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{e^2}-\frac {2\,B\,b^3\,d}{e^3}\right )}{e}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )}{e^2}+\frac {B\,b^3\,d^2}{e^4}\right )+\frac {\ln \left (d+e\,x\right )\,\left (B\,a^3\,e^3-6\,B\,a^2\,b\,d\,e^2+3\,A\,a^2\,b\,e^3+9\,B\,a\,b^2\,d^2\,e-6\,A\,a\,b^2\,d\,e^2-4\,B\,b^3\,d^3+3\,A\,b^3\,d^2\,e\right )}{e^5}-\frac {-B\,a^3\,d\,e^3+A\,a^3\,e^4+3\,B\,a^2\,b\,d^2\,e^2-3\,A\,a^2\,b\,d\,e^3-3\,B\,a\,b^2\,d^3\,e+3\,A\,a\,b^2\,d^2\,e^2+B\,b^3\,d^4-A\,b^3\,d^3\,e}{e\,\left (x\,e^5+d\,e^4\right )}+\frac {B\,b^3\,x^3}{3\,e^2} \] Input:

int(((A + B*x)*(a + b*x)^3)/(d + e*x)^2,x)
 

Output:

x^2*((A*b^3 + 3*B*a*b^2)/(2*e^2) - (B*b^3*d)/e^3) - x*((2*d*((A*b^3 + 3*B* 
a*b^2)/e^2 - (2*B*b^3*d)/e^3))/e - (3*a*b*(A*b + B*a))/e^2 + (B*b^3*d^2)/e 
^4) + (log(d + e*x)*(B*a^3*e^3 - 4*B*b^3*d^3 + 3*A*a^2*b*e^3 + 3*A*b^3*d^2 
*e - 6*A*a*b^2*d*e^2 + 9*B*a*b^2*d^2*e - 6*B*a^2*b*d*e^2))/e^5 - (A*a^3*e^ 
4 + B*b^3*d^4 - A*b^3*d^3*e - B*a^3*d*e^3 + 3*A*a*b^2*d^2*e^2 + 3*B*a^2*b* 
d^2*e^2 - 3*A*a^2*b*d*e^3 - 3*B*a*b^2*d^3*e)/(e*(d*e^4 + e^5*x)) + (B*b^3* 
x^3)/(3*e^2)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.98 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx=\frac {12 \,\mathrm {log}\left (e x +d \right ) a^{3} b \,d^{2} e^{3}+12 \,\mathrm {log}\left (e x +d \right ) a^{3} b d \,e^{4} x -36 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{2} d^{3} e^{2}-36 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{2} d^{2} e^{3} x +36 \,\mathrm {log}\left (e x +d \right ) a \,b^{3} d^{4} e +36 \,\mathrm {log}\left (e x +d \right ) a \,b^{3} d^{3} e^{2} x -12 \,\mathrm {log}\left (e x +d \right ) b^{4} d^{5}-12 \,\mathrm {log}\left (e x +d \right ) b^{4} d^{4} e x +3 a^{4} e^{5} x -12 a^{3} b d \,e^{4} x +36 a^{2} b^{2} d^{2} e^{3} x +18 a^{2} b^{2} d \,e^{4} x^{2}-36 a \,b^{3} d^{3} e^{2} x -18 a \,b^{3} d^{2} e^{3} x^{2}+6 a \,b^{3} d \,e^{4} x^{3}+12 b^{4} d^{4} e x +6 b^{4} d^{3} e^{2} x^{2}-2 b^{4} d^{2} e^{3} x^{3}+b^{4} d \,e^{4} x^{4}}{3 d \,e^{5} \left (e x +d \right )} \] Input:

int((b*x+a)^3*(B*x+A)/(e*x+d)^2,x)
 

Output:

(12*log(d + e*x)*a**3*b*d**2*e**3 + 12*log(d + e*x)*a**3*b*d*e**4*x - 36*l 
og(d + e*x)*a**2*b**2*d**3*e**2 - 36*log(d + e*x)*a**2*b**2*d**2*e**3*x + 
36*log(d + e*x)*a*b**3*d**4*e + 36*log(d + e*x)*a*b**3*d**3*e**2*x - 12*lo 
g(d + e*x)*b**4*d**5 - 12*log(d + e*x)*b**4*d**4*e*x + 3*a**4*e**5*x - 12* 
a**3*b*d*e**4*x + 36*a**2*b**2*d**2*e**3*x + 18*a**2*b**2*d*e**4*x**2 - 36 
*a*b**3*d**3*e**2*x - 18*a*b**3*d**2*e**3*x**2 + 6*a*b**3*d*e**4*x**3 + 12 
*b**4*d**4*e*x + 6*b**4*d**3*e**2*x**2 - 2*b**4*d**2*e**3*x**3 + b**4*d*e* 
*4*x**4)/(3*d*e**5*(d + e*x))