\(\int \frac {(A+B x) (d+e x)^3}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [321]

Optimal result

Integrand size = 31, antiderivative size = 144 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {B e^3 x}{b^4}-\frac {(A b-a B) (b d-a e)^3}{3 b^5 (a+b x)^3}-\frac {(b d-a e)^2 (b B d+3 A b e-4 a B e)}{2 b^5 (a+b x)^2}-\frac {3 e (b d-a e) (b B d+A b e-2 a B e)}{b^5 (a+b x)}+\frac {e^2 (3 b B d+A b e-4 a B e) \log (a+b x)}{b^5} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.51 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {A b (b d-a e) \left (11 a^2 e^2+a b e (5 d+27 e x)+b^2 \left (2 d^2+9 d e x+18 e^2 x^2\right )\right )+B \left (26 a^4 e^3+3 a^3 b e^2 (-11 d+18 e x)+3 a^2 b^2 e \left (2 d^2-27 d e x+6 e^2 x^2\right )+a b^3 \left (d^3+18 d^2 e x-54 d e^2 x^2-18 e^3 x^3\right )+3 b^4 x \left (d^3+6 d^2 e x-2 e^3 x^3\right )\right )-6 e^2 (3 b B d+A b e-4 a B e) (a+b x)^3 \log (a+b x)}{6 b^5 (a+b x)^3} \] Input:

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

Output:

-1/6*(A*b*(b*d - a*e)*(11*a^2*e^2 + a*b*e*(5*d + 27*e*x) + b^2*(2*d^2 + 9* 
d*e*x + 18*e^2*x^2)) + B*(26*a^4*e^3 + 3*a^3*b*e^2*(-11*d + 18*e*x) + 3*a^ 
2*b^2*e*(2*d^2 - 27*d*e*x + 6*e^2*x^2) + a*b^3*(d^3 + 18*d^2*e*x - 54*d*e^ 
2*x^2 - 18*e^3*x^3) + 3*b^4*x*(d^3 + 6*d^2*e*x - 2*e^3*x^3)) - 6*e^2*(3*b* 
B*d + A*b*e - 4*a*B*e)*(a + b*x)^3*Log[a + b*x])/(b^5*(a + b*x)^3)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1184, 27, 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.

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

\(\Big \downarrow \) 1184

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 1184

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p   Int[(d + e*x)^m*(f + g*x 
)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E 
qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

rule 2009

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

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.90

method	result	size
default	\(\frac {B \,e^{3} x}{b^{4}}-\frac {-A \,a^{3} b \,e^{3}+3 A \,a^{2} b^{2} d \,e^{2}-3 A a \,b^{3} d^{2} e +A \,d^{3} b^{4}+B \,e^{3} a^{4}-3 B \,a^{3} b d \,e^{2}+3 B \,a^{2} b^{2} d^{2} e -B a \,b^{3} d^{3}}{3 b^{5} \left (b x +a \right )^{3}}+\frac {3 e \left (A a b \,e^{2}-A \,b^{2} d e -2 B \,e^{2} a^{2}+3 B a b d e -B \,b^{2} d^{2}\right )}{b^{5} \left (b x +a \right )}+\frac {e^{2} \left (A b e -4 B a e +3 B b d \right ) \ln \left (b x +a \right )}{b^{5}}-\frac {3 A \,a^{2} b \,e^{3}-6 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e -4 B \,e^{3} a^{3}+9 B \,a^{2} b d \,e^{2}-6 B a \,b^{2} d^{2} e +B \,b^{3} d^{3}}{2 b^{5} \left (b x +a \right )^{2}}\)	\(273\)
norman	\(\frac {\frac {B \,e^{3} x^{4}}{b}+\frac {11 A \,a^{3} b \,e^{3}-6 A \,a^{2} b^{2} d \,e^{2}-3 A a \,b^{3} d^{2} e -2 A \,d^{3} b^{4}-44 B \,e^{3} a^{4}+33 B \,a^{3} b d \,e^{2}-6 B \,a^{2} b^{2} d^{2} e -B a \,b^{3} d^{3}}{6 b^{5}}+\frac {3 \left (A a b \,e^{3}-A \,b^{2} d \,e^{2}-4 B \,e^{3} a^{2}+3 B a b d \,e^{2}-B \,b^{2} d^{2} e \right ) x^{2}}{b^{3}}+\frac {\left (9 A \,a^{2} b \,e^{3}-6 A a \,b^{2} d \,e^{2}-3 A \,b^{3} d^{2} e -36 B \,e^{3} a^{3}+27 B \,a^{2} b d \,e^{2}-6 B a \,b^{2} d^{2} e -B \,b^{3} d^{3}\right ) x}{2 b^{4}}}{\left (b x +a \right )^{3}}+\frac {e^{2} \left (A b e -4 B a e +3 B b d \right ) \ln \left (b x +a \right )}{b^{5}}\)	\(274\)
risch	\(\frac {B \,e^{3} x}{b^{4}}+\frac {\left (3 A a \,b^{2} e^{3}-3 A \,b^{3} d \,e^{2}-6 B \,a^{2} b \,e^{3}+9 B a \,b^{2} d \,e^{2}-3 B \,b^{3} d^{2} e \right ) x^{2}+\left (\frac {9}{2} A \,a^{2} b \,e^{3}-3 A a \,b^{2} d \,e^{2}-\frac {3}{2} A \,b^{3} d^{2} e -10 B \,e^{3} a^{3}+\frac {27}{2} B \,a^{2} b d \,e^{2}-3 B a \,b^{2} d^{2} e -\frac {1}{2} B \,b^{3} d^{3}\right ) x +\frac {11 A \,a^{3} b \,e^{3}-6 A \,a^{2} b^{2} d \,e^{2}-3 A a \,b^{3} d^{2} e -2 A \,d^{3} b^{4}-26 B \,e^{3} a^{4}+33 B \,a^{3} b d \,e^{2}-6 B \,a^{2} b^{2} d^{2} e -B a \,b^{3} d^{3}}{6 b}}{b^{4} \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}+\frac {e^{3} \ln \left (b x +a \right ) A}{b^{4}}-\frac {4 e^{3} \ln \left (b x +a \right ) B a}{b^{5}}+\frac {3 e^{2} \ln \left (b x +a \right ) B d}{b^{4}}\)	\(309\)
parallelrisch	\(\frac {54 B \ln \left (b x +a \right ) x \,a^{2} b^{2} d \,e^{2}-9 A \,b^{4} d^{2} e x +54 B \ln \left (b x +a \right ) x^{2} a \,b^{3} d \,e^{2}-108 B \,a^{3} b \,e^{3} x -6 B \,a^{2} b^{2} d^{2} e +33 B \,a^{3} b d \,e^{2}-6 A \,a^{2} b^{2} d \,e^{2}-3 A a \,b^{3} d^{2} e +6 B \,x^{4} e^{3} b^{4}-24 B \ln \left (b x +a \right ) a^{4} e^{3}-3 B \,b^{4} d^{3} x +6 A \ln \left (b x +a \right ) a^{3} b \,e^{3}+6 A \ln \left (b x +a \right ) x^{3} b^{4} e^{3}+27 A x \,a^{2} b^{2} e^{3}-B a \,b^{3} d^{3}+11 A \,a^{3} b \,e^{3}+18 A \,x^{2} a \,b^{3} e^{3}-18 A \,x^{2} b^{4} d \,e^{2}-72 B \,x^{2} a^{2} b^{2} e^{3}-18 B \,x^{2} b^{4} d^{2} e +54 B \,x^{2} a \,b^{3} d \,e^{2}-18 A x a \,b^{3} d \,e^{2}+81 B x \,a^{2} b^{2} d \,e^{2}-18 B x a \,b^{3} d^{2} e -24 B \ln \left (b x +a \right ) x^{3} a \,b^{3} e^{3}+18 B \ln \left (b x +a \right ) x^{3} b^{4} d \,e^{2}-44 B \,e^{3} a^{4}+18 A \ln \left (b x +a \right ) x^{2} a \,b^{3} e^{3}-72 B \ln \left (b x +a \right ) x^{2} a^{2} b^{2} e^{3}+18 B \ln \left (b x +a \right ) a^{3} b d \,e^{2}+18 A \ln \left (b x +a \right ) x \,a^{2} b^{2} e^{3}-72 B \ln \left (b x +a \right ) x \,a^{3} b \,e^{3}-2 A \,d^{3} b^{4}}{6 b^{5} \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \left (b x +a \right )}\)	\(501\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (140) = 280\).

Time = 0.08 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.97 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {6 \, B b^{4} e^{3} x^{4} + 18 \, B a b^{3} e^{3} x^{3} - {\left (B a b^{3} + 2 \, A b^{4}\right )} d^{3} - 3 \, {\left (2 \, B a^{2} b^{2} + A a b^{3}\right )} d^{2} e + 3 \, {\left (11 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} d e^{2} - {\left (26 \, B a^{4} - 11 \, A a^{3} b\right )} e^{3} - 18 \, {\left (B b^{4} d^{2} e - {\left (3 \, B a b^{3} - A b^{4}\right )} d e^{2} + {\left (B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} - 3 \, {\left (B b^{4} d^{3} + 3 \, {\left (2 \, B a b^{3} + A b^{4}\right )} d^{2} e - 3 \, {\left (9 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} d e^{2} + 9 \, {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x + 6 \, {\left (3 \, B a^{3} b d e^{2} - {\left (4 \, B a^{4} - A a^{3} b\right )} e^{3} + {\left (3 \, B b^{4} d e^{2} - {\left (4 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + 3 \, {\left (3 \, B a b^{3} d e^{2} - {\left (4 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 3 \, {\left (3 \, B a^{2} b^{2} d e^{2} - {\left (4 \, B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}} \] Input:

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

Output:

1/6*(6*B*b^4*e^3*x^4 + 18*B*a*b^3*e^3*x^3 - (B*a*b^3 + 2*A*b^4)*d^3 - 3*(2 
*B*a^2*b^2 + A*a*b^3)*d^2*e + 3*(11*B*a^3*b - 2*A*a^2*b^2)*d*e^2 - (26*B*a 
^4 - 11*A*a^3*b)*e^3 - 18*(B*b^4*d^2*e - (3*B*a*b^3 - A*b^4)*d*e^2 + (B*a^ 
2*b^2 - A*a*b^3)*e^3)*x^2 - 3*(B*b^4*d^3 + 3*(2*B*a*b^3 + A*b^4)*d^2*e - 3 
*(9*B*a^2*b^2 - 2*A*a*b^3)*d*e^2 + 9*(2*B*a^3*b - A*a^2*b^2)*e^3)*x + 6*(3 
*B*a^3*b*d*e^2 - (4*B*a^4 - A*a^3*b)*e^3 + (3*B*b^4*d*e^2 - (4*B*a*b^3 - A 
*b^4)*e^3)*x^3 + 3*(3*B*a*b^3*d*e^2 - (4*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 3 
*(3*B*a^2*b^2*d*e^2 - (4*B*a^3*b - A*a^2*b^2)*e^3)*x)*log(b*x + a))/(b^8*x 
^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (144) = 288\).

Time = 5.86 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.34 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {B e^{3} x}{b^{4}} + \frac {11 A a^{3} b e^{3} - 6 A a^{2} b^{2} d e^{2} - 3 A a b^{3} d^{2} e - 2 A b^{4} d^{3} - 26 B a^{4} e^{3} + 33 B a^{3} b d e^{2} - 6 B a^{2} b^{2} d^{2} e - B a b^{3} d^{3} + x^{2} \cdot \left (18 A a b^{3} e^{3} - 18 A b^{4} d e^{2} - 36 B a^{2} b^{2} e^{3} + 54 B a b^{3} d e^{2} - 18 B b^{4} d^{2} e\right ) + x \left (27 A a^{2} b^{2} e^{3} - 18 A a b^{3} d e^{2} - 9 A b^{4} d^{2} e - 60 B a^{3} b e^{3} + 81 B a^{2} b^{2} d e^{2} - 18 B a b^{3} d^{2} e - 3 B b^{4} d^{3}\right )}{6 a^{3} b^{5} + 18 a^{2} b^{6} x + 18 a b^{7} x^{2} + 6 b^{8} x^{3}} - \frac {e^{2} \left (- A b e + 4 B a e - 3 B b d\right ) \log {\left (a + b x \right )}}{b^{5}} \] Input:

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

Output:

B*e**3*x/b**4 + (11*A*a**3*b*e**3 - 6*A*a**2*b**2*d*e**2 - 3*A*a*b**3*d**2 
*e - 2*A*b**4*d**3 - 26*B*a**4*e**3 + 33*B*a**3*b*d*e**2 - 6*B*a**2*b**2*d 
**2*e - B*a*b**3*d**3 + x**2*(18*A*a*b**3*e**3 - 18*A*b**4*d*e**2 - 36*B*a 
**2*b**2*e**3 + 54*B*a*b**3*d*e**2 - 18*B*b**4*d**2*e) + x*(27*A*a**2*b**2 
*e**3 - 18*A*a*b**3*d*e**2 - 9*A*b**4*d**2*e - 60*B*a**3*b*e**3 + 81*B*a** 
2*b**2*d*e**2 - 18*B*a*b**3*d**2*e - 3*B*b**4*d**3))/(6*a**3*b**5 + 18*a** 
2*b**6*x + 18*a*b**7*x**2 + 6*b**8*x**3) - e**2*(-A*b*e + 4*B*a*e - 3*B*b* 
d)*log(a + b*x)/b**5
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (140) = 280\).

Time = 0.07 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.03 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {B e^{3} x}{b^{4}} - \frac {{\left (B a b^{3} + 2 \, A b^{4}\right )} d^{3} + 3 \, {\left (2 \, B a^{2} b^{2} + A a b^{3}\right )} d^{2} e - 3 \, {\left (11 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} d e^{2} + {\left (26 \, B a^{4} - 11 \, A a^{3} b\right )} e^{3} + 18 \, {\left (B b^{4} d^{2} e - {\left (3 \, B a b^{3} - A b^{4}\right )} d e^{2} + {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 3 \, {\left (B b^{4} d^{3} + 3 \, {\left (2 \, B a b^{3} + A b^{4}\right )} d^{2} e - 3 \, {\left (9 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} d e^{2} + {\left (20 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} e^{3}\right )} x}{6 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}} + \frac {{\left (3 \, B b d e^{2} - {\left (4 \, B a - A b\right )} e^{3}\right )} \log \left (b x + a\right )}{b^{5}} \] Input:

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

Output:

B*e^3*x/b^4 - 1/6*((B*a*b^3 + 2*A*b^4)*d^3 + 3*(2*B*a^2*b^2 + A*a*b^3)*d^2 
*e - 3*(11*B*a^3*b - 2*A*a^2*b^2)*d*e^2 + (26*B*a^4 - 11*A*a^3*b)*e^3 + 18 
*(B*b^4*d^2*e - (3*B*a*b^3 - A*b^4)*d*e^2 + (2*B*a^2*b^2 - A*a*b^3)*e^3)*x 
^2 + 3*(B*b^4*d^3 + 3*(2*B*a*b^3 + A*b^4)*d^2*e - 3*(9*B*a^2*b^2 - 2*A*a*b 
^3)*d*e^2 + (20*B*a^3*b - 9*A*a^2*b^2)*e^3)*x)/(b^8*x^3 + 3*a*b^7*x^2 + 3* 
a^2*b^6*x + a^3*b^5) + (3*B*b*d*e^2 - (4*B*a - A*b)*e^3)*log(b*x + a)/b^5
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.92 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {B e^{3} x}{b^{4}} + \frac {{\left (3 \, B b d e^{2} - 4 \, B a e^{3} + A b e^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} - \frac {B a b^{3} d^{3} + 2 \, A b^{4} d^{3} + 6 \, B a^{2} b^{2} d^{2} e + 3 \, A a b^{3} d^{2} e - 33 \, B a^{3} b d e^{2} + 6 \, A a^{2} b^{2} d e^{2} + 26 \, B a^{4} e^{3} - 11 \, A a^{3} b e^{3} + 18 \, {\left (B b^{4} d^{2} e - 3 \, B a b^{3} d e^{2} + A b^{4} d e^{2} + 2 \, B a^{2} b^{2} e^{3} - A a b^{3} e^{3}\right )} x^{2} + 3 \, {\left (B b^{4} d^{3} + 6 \, B a b^{3} d^{2} e + 3 \, A b^{4} d^{2} e - 27 \, B a^{2} b^{2} d e^{2} + 6 \, A a b^{3} d e^{2} + 20 \, B a^{3} b e^{3} - 9 \, A a^{2} b^{2} e^{3}\right )} x}{6 \, {\left (b x + a\right )}^{3} b^{5}} \] Input:

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

Output:

B*e^3*x/b^4 + (3*B*b*d*e^2 - 4*B*a*e^3 + A*b*e^3)*log(abs(b*x + a))/b^5 - 
1/6*(B*a*b^3*d^3 + 2*A*b^4*d^3 + 6*B*a^2*b^2*d^2*e + 3*A*a*b^3*d^2*e - 33* 
B*a^3*b*d*e^2 + 6*A*a^2*b^2*d*e^2 + 26*B*a^4*e^3 - 11*A*a^3*b*e^3 + 18*(B* 
b^4*d^2*e - 3*B*a*b^3*d*e^2 + A*b^4*d*e^2 + 2*B*a^2*b^2*e^3 - A*a*b^3*e^3) 
*x^2 + 3*(B*b^4*d^3 + 6*B*a*b^3*d^2*e + 3*A*b^4*d^2*e - 27*B*a^2*b^2*d*e^2 
 + 6*A*a*b^3*d*e^2 + 20*B*a^3*b*e^3 - 9*A*a^2*b^2*e^3)*x)/((b*x + a)^3*b^5 
)
 

Mupad [B] (verification not implemented)

Time = 11.80 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.09 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\ln \left (a+b\,x\right )\,\left (A\,b\,e^3-4\,B\,a\,e^3+3\,B\,b\,d\,e^2\right )}{b^5}-\frac {\frac {26\,B\,a^4\,e^3-33\,B\,a^3\,b\,d\,e^2-11\,A\,a^3\,b\,e^3+6\,B\,a^2\,b^2\,d^2\,e+6\,A\,a^2\,b^2\,d\,e^2+B\,a\,b^3\,d^3+3\,A\,a\,b^3\,d^2\,e+2\,A\,b^4\,d^3}{6\,b}+x\,\left (10\,B\,a^3\,e^3-\frac {27\,B\,a^2\,b\,d\,e^2}{2}-\frac {9\,A\,a^2\,b\,e^3}{2}+3\,B\,a\,b^2\,d^2\,e+3\,A\,a\,b^2\,d\,e^2+\frac {B\,b^3\,d^3}{2}+\frac {3\,A\,b^3\,d^2\,e}{2}\right )+x^2\,\left (6\,B\,a^2\,b\,e^3-9\,B\,a\,b^2\,d\,e^2-3\,A\,a\,b^2\,e^3+3\,B\,b^3\,d^2\,e+3\,A\,b^3\,d\,e^2\right )}{a^3\,b^4+3\,a^2\,b^5\,x+3\,a\,b^6\,x^2+b^7\,x^3}+\frac {B\,e^3\,x}{b^4} \] Input:

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

Output:

(log(a + b*x)*(A*b*e^3 - 4*B*a*e^3 + 3*B*b*d*e^2))/b^5 - ((2*A*b^4*d^3 + 2 
6*B*a^4*e^3 - 11*A*a^3*b*e^3 + B*a*b^3*d^3 + 6*A*a^2*b^2*d*e^2 + 6*B*a^2*b 
^2*d^2*e + 3*A*a*b^3*d^2*e - 33*B*a^3*b*d*e^2)/(6*b) + x*(10*B*a^3*e^3 + ( 
B*b^3*d^3)/2 - (9*A*a^2*b*e^3)/2 + (3*A*b^3*d^2*e)/2 + 3*A*a*b^2*d*e^2 + 3 
*B*a*b^2*d^2*e - (27*B*a^2*b*d*e^2)/2) + x^2*(6*B*a^2*b*e^3 - 3*A*a*b^2*e^ 
3 + 3*A*b^3*d*e^2 + 3*B*b^3*d^2*e - 9*B*a*b^2*d*e^2))/(a^3*b^4 + b^7*x^3 + 
 3*a^2*b^5*x + 3*a*b^6*x^2) + (B*e^3*x)/b^4
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.45 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {-6 \,\mathrm {log}\left (b x +a \right ) a^{4} e^{3}+6 \,\mathrm {log}\left (b x +a \right ) a^{3} b d \,e^{2}-12 \,\mathrm {log}\left (b x +a \right ) a^{3} b \,e^{3} x +12 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} d \,e^{2} x -6 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} e^{3} x^{2}+6 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} d \,e^{2} x^{2}-3 a^{4} e^{3}+3 a^{3} b d \,e^{2}+6 a^{2} b^{2} e^{3} x^{2}-a \,b^{3} d^{3}-6 a \,b^{3} d \,e^{2} x^{2}+2 a \,b^{3} e^{3} x^{3}+3 b^{4} d^{2} e \,x^{2}}{2 a \,b^{4} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:

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

Output:

( - 6*log(a + b*x)*a**4*e**3 + 6*log(a + b*x)*a**3*b*d*e**2 - 12*log(a + b 
*x)*a**3*b*e**3*x + 12*log(a + b*x)*a**2*b**2*d*e**2*x - 6*log(a + b*x)*a* 
*2*b**2*e**3*x**2 + 6*log(a + b*x)*a*b**3*d*e**2*x**2 - 3*a**4*e**3 + 3*a* 
*3*b*d*e**2 + 6*a**2*b**2*e**3*x**2 - a*b**3*d**3 - 6*a*b**3*d*e**2*x**2 + 
 2*a*b**3*e**3*x**3 + 3*b**4*d**2*e*x**2)/(2*a*b**4*(a**2 + 2*a*b*x + b**2 
*x**2))