3.4.1 \(\int \frac {x^3 (c+d x)^3}{(a+b x)^3} \, dx\) [301]

3.4.1.1 Optimal result
3.4.1.2 Mathematica [A] (verified)
3.4.1.3 Rubi [A] (verified)
3.4.1.4 Maple [A] (verified)
3.4.1.5 Fricas [B] (verification not implemented)
3.4.1.6 Sympy [A] (verification not implemented)
3.4.1.7 Maxima [A] (verification not implemented)
3.4.1.8 Giac [A] (verification not implemented)
3.4.1.9 Mupad [B] (verification not implemented)

3.4.1.1 Optimal result

Integrand size = 18, antiderivative size = 196 \[ \int \frac {x^3 (c+d x)^3}{(a+b x)^3} \, dx=\frac {(b c-a d) \left (b^2 c^2-8 a b c d+10 a^2 d^2\right ) x}{b^6}+\frac {3 d (b c-2 a d) (b c-a d) x^2}{2 b^5}+\frac {d^2 (b c-a d) x^3}{b^4}+\frac {d^3 x^4}{4 b^3}+\frac {a^3 (b c-a d)^3}{2 b^7 (a+b x)^2}-\frac {3 a^2 (b c-2 a d) (b c-a d)^2}{b^7 (a+b x)}-\frac {3 a (b c-a d) \left (b^2 c^2-5 a b c d+5 a^2 d^2\right ) \log (a+b x)}{b^7} \]

output
(-a*d+b*c)*(10*a^2*d^2-8*a*b*c*d+b^2*c^2)*x/b^6+3/2*d*(-2*a*d+b*c)*(-a*d+b 
*c)*x^2/b^5+d^2*(-a*d+b*c)*x^3/b^4+1/4*d^3*x^4/b^3+1/2*a^3*(-a*d+b*c)^3/b^ 
7/(b*x+a)^2-3*a^2*(-2*a*d+b*c)*(-a*d+b*c)^2/b^7/(b*x+a)-3*a*(-a*d+b*c)*(5* 
a^2*d^2-5*a*b*c*d+b^2*c^2)*ln(b*x+a)/b^7
 
3.4.1.2 Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 (c+d x)^3}{(a+b x)^3} \, dx=\frac {4 b \left (b^3 c^3-9 a b^2 c^2 d+18 a^2 b c d^2-10 a^3 d^3\right ) x+6 b^2 d \left (b^2 c^2-3 a b c d+2 a^2 d^2\right ) x^2+4 b^3 d^2 (b c-a d) x^3+b^4 d^3 x^4+\frac {2 a^3 (b c-a d)^3}{(a+b x)^2}+\frac {12 a^2 (b c-a d)^2 (-b c+2 a d)}{a+b x}+12 a \left (-b^3 c^3+6 a b^2 c^2 d-10 a^2 b c d^2+5 a^3 d^3\right ) \log (a+b x)}{4 b^7} \]

input
Integrate[(x^3*(c + d*x)^3)/(a + b*x)^3,x]
 
output
(4*b*(b^3*c^3 - 9*a*b^2*c^2*d + 18*a^2*b*c*d^2 - 10*a^3*d^3)*x + 6*b^2*d*( 
b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^2 + 4*b^3*d^2*(b*c - a*d)*x^3 + b^4*d^3 
*x^4 + (2*a^3*(b*c - a*d)^3)/(a + b*x)^2 + (12*a^2*(b*c - a*d)^2*(-(b*c) + 
 2*a*d))/(a + b*x) + 12*a*(-(b^3*c^3) + 6*a*b^2*c^2*d - 10*a^2*b*c*d^2 + 5 
*a^3*d^3)*Log[a + b*x])/(4*b^7)
 
3.4.1.3 Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {99, 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.

\(\displaystyle \int \frac {x^3 (c+d x)^3}{(a+b x)^3} \, dx\)

\(\Big \downarrow \) 99

\(\displaystyle \int \left (\frac {a^3 (a d-b c)^3}{b^6 (a+b x)^3}-\frac {3 a^2 (2 a d-b c) (a d-b c)^2}{b^6 (a+b x)^2}+\frac {3 a (b c-a d) \left (-5 a^2 d^2+5 a b c d-b^2 c^2\right )}{b^6 (a+b x)}+\frac {(b c-a d) \left (10 a^2 d^2-8 a b c d+b^2 c^2\right )}{b^6}+\frac {3 d x (b c-2 a d) (b c-a d)}{b^5}+\frac {3 d^2 x^2 (b c-a d)}{b^4}+\frac {d^3 x^3}{b^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 (b c-a d)^3}{2 b^7 (a+b x)^2}-\frac {3 a^2 (b c-2 a d) (b c-a d)^2}{b^7 (a+b x)}-\frac {3 a (b c-a d) \left (5 a^2 d^2-5 a b c d+b^2 c^2\right ) \log (a+b x)}{b^7}+\frac {x (b c-a d) \left (10 a^2 d^2-8 a b c d+b^2 c^2\right )}{b^6}+\frac {3 d x^2 (b c-2 a d) (b c-a d)}{2 b^5}+\frac {d^2 x^3 (b c-a d)}{b^4}+\frac {d^3 x^4}{4 b^3}\)

input
Int[(x^3*(c + d*x)^3)/(a + b*x)^3,x]
 
output
((b*c - a*d)*(b^2*c^2 - 8*a*b*c*d + 10*a^2*d^2)*x)/b^6 + (3*d*(b*c - 2*a*d 
)*(b*c - a*d)*x^2)/(2*b^5) + (d^2*(b*c - a*d)*x^3)/b^4 + (d^3*x^4)/(4*b^3) 
 + (a^3*(b*c - a*d)^3)/(2*b^7*(a + b*x)^2) - (3*a^2*(b*c - 2*a*d)*(b*c - a 
*d)^2)/(b^7*(a + b*x)) - (3*a*(b*c - a*d)*(b^2*c^2 - 5*a*b*c*d + 5*a^2*d^2 
)*Log[a + b*x])/b^7
 

3.4.1.3.1 Defintions of rubi rules used

rule 99
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
3.4.1.4 Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.34

method result size
norman \(\frac {\frac {a^{2} \left (45 a^{4} d^{3}-90 a^{3} b c \,d^{2}+54 a^{2} b^{2} c^{2} d -9 a \,b^{3} c^{3}\right )}{2 b^{7}}-\frac {\left (5 a^{3} d^{3}-10 a^{2} b c \,d^{2}+6 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x^{3}}{b^{4}}+\frac {d^{3} x^{6}}{4 b}+\frac {2 a \left (15 a^{4} d^{3}-30 a^{3} b c \,d^{2}+18 a^{2} b^{2} c^{2} d -3 a \,b^{3} c^{3}\right ) x}{b^{6}}+\frac {d \left (5 a^{2} d^{2}-10 a b c d +6 b^{2} c^{2}\right ) x^{4}}{4 b^{3}}-\frac {d^{2} \left (a d -2 b c \right ) x^{5}}{2 b^{2}}}{\left (b x +a \right )^{2}}+\frac {3 a \left (5 a^{3} d^{3}-10 a^{2} b c \,d^{2}+6 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (b x +a \right )}{b^{7}}\) \(262\)
default \(-\frac {-\frac {1}{4} d^{3} x^{4} b^{3}+x^{3} a \,b^{2} d^{3}-x^{3} b^{3} c \,d^{2}-3 x^{2} a^{2} b \,d^{3}+\frac {9}{2} x^{2} a \,b^{2} c \,d^{2}-\frac {3}{2} x^{2} b^{3} c^{2} d +10 a^{3} d^{3} x -18 a^{2} b c \,d^{2} x +9 a \,b^{2} c^{2} d x -b^{3} c^{3} x}{b^{6}}+\frac {3 a \left (5 a^{3} d^{3}-10 a^{2} b c \,d^{2}+6 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (b x +a \right )}{b^{7}}-\frac {a^{3} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{2 b^{7} \left (b x +a \right )^{2}}+\frac {3 a^{2} \left (2 a^{3} d^{3}-5 a^{2} b c \,d^{2}+4 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{b^{7} \left (b x +a \right )}\) \(271\)
risch \(\frac {d^{3} x^{4}}{4 b^{3}}-\frac {x^{3} a \,d^{3}}{b^{4}}+\frac {x^{3} c \,d^{2}}{b^{3}}+\frac {3 x^{2} a^{2} d^{3}}{b^{5}}-\frac {9 x^{2} a c \,d^{2}}{2 b^{4}}+\frac {3 x^{2} c^{2} d}{2 b^{3}}-\frac {10 a^{3} d^{3} x}{b^{6}}+\frac {18 a^{2} c \,d^{2} x}{b^{5}}-\frac {9 a \,c^{2} d x}{b^{4}}+\frac {c^{3} x}{b^{3}}+\frac {\left (6 a^{5} d^{3}-15 a^{4} b c \,d^{2}+12 a^{3} b^{2} c^{2} d -3 a^{2} b^{3} c^{3}\right ) x +\frac {a^{3} \left (11 a^{3} d^{3}-27 a^{2} b c \,d^{2}+21 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right )}{2 b}}{b^{6} \left (b x +a \right )^{2}}+\frac {15 a^{4} \ln \left (b x +a \right ) d^{3}}{b^{7}}-\frac {30 a^{3} \ln \left (b x +a \right ) c \,d^{2}}{b^{6}}+\frac {18 a^{2} \ln \left (b x +a \right ) c^{2} d}{b^{5}}-\frac {3 a \ln \left (b x +a \right ) c^{3}}{b^{4}}\) \(288\)
parallelrisch \(\frac {60 \ln \left (b x +a \right ) x^{2} a^{4} b^{2} d^{3}-12 \ln \left (b x +a \right ) x^{2} a \,b^{5} c^{3}+90 a^{6} d^{3}-180 a^{5} b c \,d^{2}+108 a^{4} b^{2} c^{2} d +72 \ln \left (b x +a \right ) x^{2} a^{2} b^{4} c^{2} d -120 \ln \left (b x +a \right ) x^{2} a^{3} b^{3} c \,d^{2}+144 \ln \left (b x +a \right ) x \,a^{3} b^{3} c^{2} d -240 \ln \left (b x +a \right ) x \,a^{4} b^{2} c \,d^{2}-10 x^{4} a \,b^{5} c \,d^{2}+40 x^{3} a^{2} b^{4} c \,d^{2}-24 x^{3} a \,b^{5} c^{2} d -120 \ln \left (b x +a \right ) a^{5} b c \,d^{2}+72 \ln \left (b x +a \right ) a^{4} b^{2} c^{2} d -240 x \,a^{4} b^{2} c \,d^{2}+144 x \,a^{3} b^{3} c^{2} d -18 a^{3} b^{3} c^{3}+120 \ln \left (b x +a \right ) x \,a^{5} b \,d^{3}-24 \ln \left (b x +a \right ) x \,a^{2} b^{4} c^{3}+4 x^{3} b^{6} c^{3}+60 \ln \left (b x +a \right ) a^{6} d^{3}+x^{6} d^{3} b^{6}-2 x^{5} a \,b^{5} d^{3}+4 x^{5} b^{6} c \,d^{2}+5 x^{4} a^{2} b^{4} d^{3}+6 x^{4} b^{6} c^{2} d -20 x^{3} a^{3} b^{3} d^{3}-12 \ln \left (b x +a \right ) a^{3} b^{3} c^{3}+120 x \,a^{5} b \,d^{3}-24 x \,a^{2} b^{4} c^{3}}{4 b^{7} \left (b x +a \right )^{2}}\) \(446\)

input
int(x^3*(d*x+c)^3/(b*x+a)^3,x,method=_RETURNVERBOSE)
 
output
(1/2*a^2*(45*a^4*d^3-90*a^3*b*c*d^2+54*a^2*b^2*c^2*d-9*a*b^3*c^3)/b^7-1/b^ 
4*(5*a^3*d^3-10*a^2*b*c*d^2+6*a*b^2*c^2*d-b^3*c^3)*x^3+1/4*d^3*x^6/b+2*a*( 
15*a^4*d^3-30*a^3*b*c*d^2+18*a^2*b^2*c^2*d-3*a*b^3*c^3)/b^6*x+1/4*d*(5*a^2 
*d^2-10*a*b*c*d+6*b^2*c^2)/b^3*x^4-1/2*d^2*(a*d-2*b*c)/b^2*x^5)/(b*x+a)^2+ 
3*a/b^7*(5*a^3*d^3-10*a^2*b*c*d^2+6*a*b^2*c^2*d-b^3*c^3)*ln(b*x+a)
 
3.4.1.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (190) = 380\).

Time = 0.23 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.17 \[ \int \frac {x^3 (c+d x)^3}{(a+b x)^3} \, dx=\frac {b^{6} d^{3} x^{6} - 10 \, a^{3} b^{3} c^{3} + 42 \, a^{4} b^{2} c^{2} d - 54 \, a^{5} b c d^{2} + 22 \, a^{6} d^{3} + 2 \, {\left (2 \, b^{6} c d^{2} - a b^{5} d^{3}\right )} x^{5} + {\left (6 \, b^{6} c^{2} d - 10 \, a b^{5} c d^{2} + 5 \, a^{2} b^{4} d^{3}\right )} x^{4} + 4 \, {\left (b^{6} c^{3} - 6 \, a b^{5} c^{2} d + 10 \, a^{2} b^{4} c d^{2} - 5 \, a^{3} b^{3} d^{3}\right )} x^{3} + 2 \, {\left (4 \, a b^{5} c^{3} - 33 \, a^{2} b^{4} c^{2} d + 63 \, a^{3} b^{3} c d^{2} - 34 \, a^{4} b^{2} d^{3}\right )} x^{2} - 4 \, {\left (2 \, a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d - 3 \, a^{4} b^{2} c d^{2} + 4 \, a^{5} b d^{3}\right )} x - 12 \, {\left (a^{3} b^{3} c^{3} - 6 \, a^{4} b^{2} c^{2} d + 10 \, a^{5} b c d^{2} - 5 \, a^{6} d^{3} + {\left (a b^{5} c^{3} - 6 \, a^{2} b^{4} c^{2} d + 10 \, a^{3} b^{3} c d^{2} - 5 \, a^{4} b^{2} d^{3}\right )} x^{2} + 2 \, {\left (a^{2} b^{4} c^{3} - 6 \, a^{3} b^{3} c^{2} d + 10 \, a^{4} b^{2} c d^{2} - 5 \, a^{5} b d^{3}\right )} x\right )} \log \left (b x + a\right )}{4 \, {\left (b^{9} x^{2} + 2 \, a b^{8} x + a^{2} b^{7}\right )}} \]

input
integrate(x^3*(d*x+c)^3/(b*x+a)^3,x, algorithm="fricas")
 
output
1/4*(b^6*d^3*x^6 - 10*a^3*b^3*c^3 + 42*a^4*b^2*c^2*d - 54*a^5*b*c*d^2 + 22 
*a^6*d^3 + 2*(2*b^6*c*d^2 - a*b^5*d^3)*x^5 + (6*b^6*c^2*d - 10*a*b^5*c*d^2 
 + 5*a^2*b^4*d^3)*x^4 + 4*(b^6*c^3 - 6*a*b^5*c^2*d + 10*a^2*b^4*c*d^2 - 5* 
a^3*b^3*d^3)*x^3 + 2*(4*a*b^5*c^3 - 33*a^2*b^4*c^2*d + 63*a^3*b^3*c*d^2 - 
34*a^4*b^2*d^3)*x^2 - 4*(2*a^2*b^4*c^3 - 3*a^3*b^3*c^2*d - 3*a^4*b^2*c*d^2 
 + 4*a^5*b*d^3)*x - 12*(a^3*b^3*c^3 - 6*a^4*b^2*c^2*d + 10*a^5*b*c*d^2 - 5 
*a^6*d^3 + (a*b^5*c^3 - 6*a^2*b^4*c^2*d + 10*a^3*b^3*c*d^2 - 5*a^4*b^2*d^3 
)*x^2 + 2*(a^2*b^4*c^3 - 6*a^3*b^3*c^2*d + 10*a^4*b^2*c*d^2 - 5*a^5*b*d^3) 
*x)*log(b*x + a))/(b^9*x^2 + 2*a*b^8*x + a^2*b^7)
 
3.4.1.6 Sympy [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.45 \[ \int \frac {x^3 (c+d x)^3}{(a+b x)^3} \, dx=\frac {3 a \left (a d - b c\right ) \left (5 a^{2} d^{2} - 5 a b c d + b^{2} c^{2}\right ) \log {\left (a + b x \right )}}{b^{7}} + x^{3} \left (- \frac {a d^{3}}{b^{4}} + \frac {c d^{2}}{b^{3}}\right ) + x^{2} \cdot \left (\frac {3 a^{2} d^{3}}{b^{5}} - \frac {9 a c d^{2}}{2 b^{4}} + \frac {3 c^{2} d}{2 b^{3}}\right ) + x \left (- \frac {10 a^{3} d^{3}}{b^{6}} + \frac {18 a^{2} c d^{2}}{b^{5}} - \frac {9 a c^{2} d}{b^{4}} + \frac {c^{3}}{b^{3}}\right ) + \frac {11 a^{6} d^{3} - 27 a^{5} b c d^{2} + 21 a^{4} b^{2} c^{2} d - 5 a^{3} b^{3} c^{3} + x \left (12 a^{5} b d^{3} - 30 a^{4} b^{2} c d^{2} + 24 a^{3} b^{3} c^{2} d - 6 a^{2} b^{4} c^{3}\right )}{2 a^{2} b^{7} + 4 a b^{8} x + 2 b^{9} x^{2}} + \frac {d^{3} x^{4}}{4 b^{3}} \]

input
integrate(x**3*(d*x+c)**3/(b*x+a)**3,x)
 
output
3*a*(a*d - b*c)*(5*a**2*d**2 - 5*a*b*c*d + b**2*c**2)*log(a + b*x)/b**7 + 
x**3*(-a*d**3/b**4 + c*d**2/b**3) + x**2*(3*a**2*d**3/b**5 - 9*a*c*d**2/(2 
*b**4) + 3*c**2*d/(2*b**3)) + x*(-10*a**3*d**3/b**6 + 18*a**2*c*d**2/b**5 
- 9*a*c**2*d/b**4 + c**3/b**3) + (11*a**6*d**3 - 27*a**5*b*c*d**2 + 21*a** 
4*b**2*c**2*d - 5*a**3*b**3*c**3 + x*(12*a**5*b*d**3 - 30*a**4*b**2*c*d**2 
 + 24*a**3*b**3*c**2*d - 6*a**2*b**4*c**3))/(2*a**2*b**7 + 4*a*b**8*x + 2* 
b**9*x**2) + d**3*x**4/(4*b**3)
 
3.4.1.7 Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.41 \[ \int \frac {x^3 (c+d x)^3}{(a+b x)^3} \, dx=-\frac {5 \, a^{3} b^{3} c^{3} - 21 \, a^{4} b^{2} c^{2} d + 27 \, a^{5} b c d^{2} - 11 \, a^{6} d^{3} + 6 \, {\left (a^{2} b^{4} c^{3} - 4 \, a^{3} b^{3} c^{2} d + 5 \, a^{4} b^{2} c d^{2} - 2 \, a^{5} b d^{3}\right )} x}{2 \, {\left (b^{9} x^{2} + 2 \, a b^{8} x + a^{2} b^{7}\right )}} + \frac {b^{3} d^{3} x^{4} + 4 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{3} + 6 \, {\left (b^{3} c^{2} d - 3 \, a b^{2} c d^{2} + 2 \, a^{2} b d^{3}\right )} x^{2} + 4 \, {\left (b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 18 \, a^{2} b c d^{2} - 10 \, a^{3} d^{3}\right )} x}{4 \, b^{6}} - \frac {3 \, {\left (a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d + 10 \, a^{3} b c d^{2} - 5 \, a^{4} d^{3}\right )} \log \left (b x + a\right )}{b^{7}} \]

input
integrate(x^3*(d*x+c)^3/(b*x+a)^3,x, algorithm="maxima")
 
output
-1/2*(5*a^3*b^3*c^3 - 21*a^4*b^2*c^2*d + 27*a^5*b*c*d^2 - 11*a^6*d^3 + 6*( 
a^2*b^4*c^3 - 4*a^3*b^3*c^2*d + 5*a^4*b^2*c*d^2 - 2*a^5*b*d^3)*x)/(b^9*x^2 
 + 2*a*b^8*x + a^2*b^7) + 1/4*(b^3*d^3*x^4 + 4*(b^3*c*d^2 - a*b^2*d^3)*x^3 
 + 6*(b^3*c^2*d - 3*a*b^2*c*d^2 + 2*a^2*b*d^3)*x^2 + 4*(b^3*c^3 - 9*a*b^2* 
c^2*d + 18*a^2*b*c*d^2 - 10*a^3*d^3)*x)/b^6 - 3*(a*b^3*c^3 - 6*a^2*b^2*c^2 
*d + 10*a^3*b*c*d^2 - 5*a^4*d^3)*log(b*x + a)/b^7
 
3.4.1.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.41 \[ \int \frac {x^3 (c+d x)^3}{(a+b x)^3} \, dx=-\frac {3 \, {\left (a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d + 10 \, a^{3} b c d^{2} - 5 \, a^{4} d^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{7}} - \frac {5 \, a^{3} b^{3} c^{3} - 21 \, a^{4} b^{2} c^{2} d + 27 \, a^{5} b c d^{2} - 11 \, a^{6} d^{3} + 6 \, {\left (a^{2} b^{4} c^{3} - 4 \, a^{3} b^{3} c^{2} d + 5 \, a^{4} b^{2} c d^{2} - 2 \, a^{5} b d^{3}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{7}} + \frac {b^{9} d^{3} x^{4} + 4 \, b^{9} c d^{2} x^{3} - 4 \, a b^{8} d^{3} x^{3} + 6 \, b^{9} c^{2} d x^{2} - 18 \, a b^{8} c d^{2} x^{2} + 12 \, a^{2} b^{7} d^{3} x^{2} + 4 \, b^{9} c^{3} x - 36 \, a b^{8} c^{2} d x + 72 \, a^{2} b^{7} c d^{2} x - 40 \, a^{3} b^{6} d^{3} x}{4 \, b^{12}} \]

input
integrate(x^3*(d*x+c)^3/(b*x+a)^3,x, algorithm="giac")
 
output
-3*(a*b^3*c^3 - 6*a^2*b^2*c^2*d + 10*a^3*b*c*d^2 - 5*a^4*d^3)*log(abs(b*x 
+ a))/b^7 - 1/2*(5*a^3*b^3*c^3 - 21*a^4*b^2*c^2*d + 27*a^5*b*c*d^2 - 11*a^ 
6*d^3 + 6*(a^2*b^4*c^3 - 4*a^3*b^3*c^2*d + 5*a^4*b^2*c*d^2 - 2*a^5*b*d^3)* 
x)/((b*x + a)^2*b^7) + 1/4*(b^9*d^3*x^4 + 4*b^9*c*d^2*x^3 - 4*a*b^8*d^3*x^ 
3 + 6*b^9*c^2*d*x^2 - 18*a*b^8*c*d^2*x^2 + 12*a^2*b^7*d^3*x^2 + 4*b^9*c^3* 
x - 36*a*b^8*c^2*d*x + 72*a^2*b^7*c*d^2*x - 40*a^3*b^6*d^3*x)/b^12
 
3.4.1.9 Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.80 \[ \int \frac {x^3 (c+d x)^3}{(a+b x)^3} \, dx=\frac {x\,\left (6\,a^5\,d^3-15\,a^4\,b\,c\,d^2+12\,a^3\,b^2\,c^2\,d-3\,a^2\,b^3\,c^3\right )+\frac {11\,a^6\,d^3-27\,a^5\,b\,c\,d^2+21\,a^4\,b^2\,c^2\,d-5\,a^3\,b^3\,c^3}{2\,b}}{a^2\,b^6+2\,a\,b^7\,x+b^8\,x^2}-x^3\,\left (\frac {a\,d^3}{b^4}-\frac {c\,d^2}{b^3}\right )+x\,\left (\frac {c^3}{b^3}-\frac {3\,a\,\left (\frac {3\,c^2\,d}{b^3}+\frac {3\,a\,\left (\frac {3\,a\,d^3}{b^4}-\frac {3\,c\,d^2}{b^3}\right )}{b}-\frac {3\,a^2\,d^3}{b^5}\right )}{b}-\frac {a^3\,d^3}{b^6}+\frac {3\,a^2\,\left (\frac {3\,a\,d^3}{b^4}-\frac {3\,c\,d^2}{b^3}\right )}{b^2}\right )+x^2\,\left (\frac {3\,c^2\,d}{2\,b^3}+\frac {3\,a\,\left (\frac {3\,a\,d^3}{b^4}-\frac {3\,c\,d^2}{b^3}\right )}{2\,b}-\frac {3\,a^2\,d^3}{2\,b^5}\right )+\frac {d^3\,x^4}{4\,b^3}+\frac {\ln \left (a+b\,x\right )\,\left (15\,a^4\,d^3-30\,a^3\,b\,c\,d^2+18\,a^2\,b^2\,c^2\,d-3\,a\,b^3\,c^3\right )}{b^7} \]

input
int((x^3*(c + d*x)^3)/(a + b*x)^3,x)
 
output
(x*(6*a^5*d^3 - 3*a^2*b^3*c^3 + 12*a^3*b^2*c^2*d - 15*a^4*b*c*d^2) + (11*a 
^6*d^3 - 5*a^3*b^3*c^3 + 21*a^4*b^2*c^2*d - 27*a^5*b*c*d^2)/(2*b))/(a^2*b^ 
6 + b^8*x^2 + 2*a*b^7*x) - x^3*((a*d^3)/b^4 - (c*d^2)/b^3) + x*(c^3/b^3 - 
(3*a*((3*c^2*d)/b^3 + (3*a*((3*a*d^3)/b^4 - (3*c*d^2)/b^3))/b - (3*a^2*d^3 
)/b^5))/b - (a^3*d^3)/b^6 + (3*a^2*((3*a*d^3)/b^4 - (3*c*d^2)/b^3))/b^2) + 
 x^2*((3*c^2*d)/(2*b^3) + (3*a*((3*a*d^3)/b^4 - (3*c*d^2)/b^3))/(2*b) - (3 
*a^2*d^3)/(2*b^5)) + (d^3*x^4)/(4*b^3) + (log(a + b*x)*(15*a^4*d^3 - 3*a*b 
^3*c^3 + 18*a^2*b^2*c^2*d - 30*a^3*b*c*d^2))/b^7