Integrand size = 18, antiderivative size = 156 \[ \int \frac {x^2 (c+d x)^3}{(a+b x)^3} \, dx=\frac {3 d (b c-2 a d) (b c-a d) x}{b^5}+\frac {3 d^2 (b c-a d) x^2}{2 b^4}+\frac {d^3 x^3}{3 b^3}-\frac {a^2 (b c-a d)^3}{2 b^6 (a+b x)^2}+\frac {a (2 b c-5 a d) (b c-a d)^2}{b^6 (a+b x)}+\frac {(b c-a d) \left (b^2 c^2-8 a b c d+10 a^2 d^2\right ) \log (a+b x)}{b^6} \]
3*d*(-2*a*d+b*c)*(-a*d+b*c)*x/b^5+3/2*d^2*(-a*d+b*c)*x^2/b^4+1/3*d^3*x^3/b ^3-1/2*a^2*(-a*d+b*c)^3/b^6/(b*x+a)^2+a*(-5*a*d+2*b*c)*(-a*d+b*c)^2/b^6/(b *x+a)+(-a*d+b*c)*(10*a^2*d^2-8*a*b*c*d+b^2*c^2)*ln(b*x+a)/b^6
Time = 0.05 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03 \[ \int \frac {x^2 (c+d x)^3}{(a+b x)^3} \, dx=\frac {18 b d \left (b^2 c^2-3 a b c d+2 a^2 d^2\right ) x+9 b^2 d^2 (b c-a d) x^2+2 b^3 d^3 x^3+\frac {3 a^2 (-b c+a d)^3}{(a+b x)^2}-\frac {6 a (b c-a d)^2 (-2 b c+5 a d)}{a+b x}+6 \left (b^3 c^3-9 a b^2 c^2 d+18 a^2 b c d^2-10 a^3 d^3\right ) \log (a+b x)}{6 b^6} \]
(18*b*d*(b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x + 9*b^2*d^2*(b*c - a*d)*x^2 + 2*b^3*d^3*x^3 + (3*a^2*(-(b*c) + a*d)^3)/(a + b*x)^2 - (6*a*(b*c - a*d)^2* (-2*b*c + 5*a*d))/(a + b*x) + 6*(b^3*c^3 - 9*a*b^2*c^2*d + 18*a^2*b*c*d^2 - 10*a^3*d^3)*Log[a + b*x])/(6*b^6)
Time = 0.36 (sec) , antiderivative size = 156, 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.
| \(\displaystyle \int \frac {x^2 (c+d x)^3}{(a+b x)^3} \, dx\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \int \left (-\frac {a^2 (a d-b c)^3}{b^5 (a+b x)^3}+\frac {(b c-a d) \left (10 a^2 d^2-8 a b c d+b^2 c^2\right )}{b^5 (a+b x)}+\frac {a (a d-b c)^2 (5 a d-2 b c)}{b^5 (a+b x)^2}+\frac {3 d (b c-2 a d) (b c-a d)}{b^5}+\frac {3 d^2 x (b c-a d)}{b^4}+\frac {d^3 x^2}{b^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 (b c-a d)^3}{2 b^6 (a+b x)^2}+\frac {\left (10 a^2 d^2-8 a b c d+b^2 c^2\right ) (b c-a d) \log (a+b x)}{b^6}+\frac {a (2 b c-5 a d) (b c-a d)^2}{b^6 (a+b x)}+\frac {3 d x (b c-2 a d) (b c-a d)}{b^5}+\frac {3 d^2 x^2 (b c-a d)}{2 b^4}+\frac {d^3 x^3}{3 b^3}\) |
(3*d*(b*c - 2*a*d)*(b*c - a*d)*x)/b^5 + (3*d^2*(b*c - a*d)*x^2)/(2*b^4) + (d^3*x^3)/(3*b^3) - (a^2*(b*c - a*d)^3)/(2*b^6*(a + b*x)^2) + (a*(2*b*c - 5*a*d)*(b*c - a*d)^2)/(b^6*(a + b*x)) + ((b*c - a*d)*(b^2*c^2 - 8*a*b*c*d + 10*a^2*d^2)*Log[a + b*x])/b^6
3.4.2.3.1 Defintions of rubi rules used
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]))
Time = 0.48 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.35
| method | result | size |
| default | \(\frac {d \left (\frac {1}{3} d^{2} x^{3} b^{2}-\frac {3}{2} x^{2} a b \,d^{2}+\frac {3}{2} x^{2} b^{2} c d +6 a^{2} d^{2} x -9 a b c d x +3 b^{2} c^{2} x \right )}{b^{5}}+\frac {\left (-10 a^{3} d^{3}+18 a^{2} b c \,d^{2}-9 a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \ln \left (b x +a \right )}{b^{6}}+\frac {a^{2} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{2 b^{6} \left (b x +a \right )^{2}}-\frac {a \left (5 a^{3} d^{3}-12 a^{2} b c \,d^{2}+9 a \,b^{2} c^{2} d -2 b^{3} c^{3}\right )}{b^{6} \left (b x +a \right )}\) | \(211\) |
| norman | \(\frac {-\frac {a^{2} \left (30 a^{3} d^{3}-54 a^{2} b c \,d^{2}+27 a \,b^{2} c^{2} d -3 b^{3} c^{3}\right )}{2 b^{6}}+\frac {d^{3} x^{5}}{3 b}-\frac {2 a \left (10 a^{3} d^{3}-18 a^{2} b c \,d^{2}+9 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{b^{5}}+\frac {d \left (10 a^{2} d^{2}-18 a b c d +9 b^{2} c^{2}\right ) x^{3}}{3 b^{3}}-\frac {d^{2} \left (5 a d -9 b c \right ) x^{4}}{6 b^{2}}}{\left (b x +a \right )^{2}}-\frac {\left (10 a^{3} d^{3}-18 a^{2} b c \,d^{2}+9 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (b x +a \right )}{b^{6}}\) | \(211\) |
| risch | \(\frac {d^{3} x^{3}}{3 b^{3}}-\frac {3 d^{3} x^{2} a}{2 b^{4}}+\frac {3 d^{2} x^{2} c}{2 b^{3}}+\frac {6 d^{3} a^{2} x}{b^{5}}-\frac {9 d^{2} a c x}{b^{4}}+\frac {3 d \,c^{2} x}{b^{3}}+\frac {\left (-5 a^{4} d^{3}+12 a^{3} b c \,d^{2}-9 a^{2} b^{2} c^{2} d +2 a \,b^{3} c^{3}\right ) x -\frac {3 a^{2} \left (3 a^{3} d^{3}-7 a^{2} b c \,d^{2}+5 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{2 b}}{b^{5} \left (b x +a \right )^{2}}-\frac {10 \ln \left (b x +a \right ) a^{3} d^{3}}{b^{6}}+\frac {18 \ln \left (b x +a \right ) a^{2} c \,d^{2}}{b^{5}}-\frac {9 \ln \left (b x +a \right ) a \,c^{2} d}{b^{4}}+\frac {\ln \left (b x +a \right ) c^{3}}{b^{3}}\) | \(233\) |
| parallelrisch | \(-\frac {90 a^{5} d^{3}-20 x^{3} a^{2} b^{3} d^{3}-18 x^{3} b^{5} c^{2} d -6 \ln \left (b x +a \right ) a^{2} b^{3} c^{3}+120 x \,a^{4} b \,d^{3}-12 x a \,b^{4} c^{3}+5 x^{4} a \,b^{4} d^{3}-9 x^{4} b^{5} c \,d^{2}-162 a^{4} b c \,d^{2}+81 a^{3} b^{2} c^{2} d +54 \ln \left (b x +a \right ) x^{2} a \,b^{4} c^{2} d -108 \ln \left (b x +a \right ) x^{2} a^{2} b^{3} c \,d^{2}-216 \ln \left (b x +a \right ) x \,a^{3} b^{2} c \,d^{2}+108 \ln \left (b x +a \right ) x \,a^{2} b^{3} c^{2} d +120 \ln \left (b x +a \right ) x \,a^{4} b \,d^{3}-12 \ln \left (b x +a \right ) x a \,b^{4} c^{3}+108 x \,a^{2} b^{3} c^{2} d -9 a^{2} b^{3} c^{3}+60 \ln \left (b x +a \right ) a^{5} d^{3}+36 x^{3} a \,b^{4} c \,d^{2}-2 d^{3} x^{5} b^{5}-108 \ln \left (b x +a \right ) a^{4} b c \,d^{2}+54 \ln \left (b x +a \right ) a^{3} b^{2} c^{2} d -216 x \,a^{3} b^{2} c \,d^{2}-6 \ln \left (b x +a \right ) x^{2} b^{5} c^{3}+60 \ln \left (b x +a \right ) x^{2} a^{3} b^{2} d^{3}}{6 b^{6} \left (b x +a \right )^{2}}\) | \(387\) |
d/b^5*(1/3*d^2*x^3*b^2-3/2*x^2*a*b*d^2+3/2*x^2*b^2*c*d+6*a^2*d^2*x-9*a*b*c *d*x+3*b^2*c^2*x)+(-10*a^3*d^3+18*a^2*b*c*d^2-9*a*b^2*c^2*d+b^3*c^3)/b^6*l n(b*x+a)+1/2*a^2*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/b^6/(b*x+a) ^2-a/b^6*(5*a^3*d^3-12*a^2*b*c*d^2+9*a*b^2*c^2*d-2*b^3*c^3)/(b*x+a)
Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (150) = 300\).
Time = 0.22 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.31 \[ \int \frac {x^2 (c+d x)^3}{(a+b x)^3} \, dx=\frac {2 \, b^{5} d^{3} x^{5} + 9 \, a^{2} b^{3} c^{3} - 45 \, a^{3} b^{2} c^{2} d + 63 \, a^{4} b c d^{2} - 27 \, a^{5} d^{3} + {\left (9 \, b^{5} c d^{2} - 5 \, a b^{4} d^{3}\right )} x^{4} + 2 \, {\left (9 \, b^{5} c^{2} d - 18 \, a b^{4} c d^{2} + 10 \, a^{2} b^{3} d^{3}\right )} x^{3} + 9 \, {\left (4 \, a b^{4} c^{2} d - 11 \, a^{2} b^{3} c d^{2} + 7 \, a^{3} b^{2} d^{3}\right )} x^{2} + 6 \, {\left (2 \, a b^{4} c^{3} - 6 \, a^{2} b^{3} c^{2} d + 3 \, a^{3} b^{2} c d^{2} + a^{4} b d^{3}\right )} x + 6 \, {\left (a^{2} b^{3} c^{3} - 9 \, a^{3} b^{2} c^{2} d + 18 \, a^{4} b c d^{2} - 10 \, a^{5} d^{3} + {\left (b^{5} c^{3} - 9 \, a b^{4} c^{2} d + 18 \, a^{2} b^{3} c d^{2} - 10 \, a^{3} b^{2} d^{3}\right )} x^{2} + 2 \, {\left (a b^{4} c^{3} - 9 \, a^{2} b^{3} c^{2} d + 18 \, a^{3} b^{2} c d^{2} - 10 \, a^{4} b d^{3}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{8} x^{2} + 2 \, a b^{7} x + a^{2} b^{6}\right )}} \]
1/6*(2*b^5*d^3*x^5 + 9*a^2*b^3*c^3 - 45*a^3*b^2*c^2*d + 63*a^4*b*c*d^2 - 2 7*a^5*d^3 + (9*b^5*c*d^2 - 5*a*b^4*d^3)*x^4 + 2*(9*b^5*c^2*d - 18*a*b^4*c* d^2 + 10*a^2*b^3*d^3)*x^3 + 9*(4*a*b^4*c^2*d - 11*a^2*b^3*c*d^2 + 7*a^3*b^ 2*d^3)*x^2 + 6*(2*a*b^4*c^3 - 6*a^2*b^3*c^2*d + 3*a^3*b^2*c*d^2 + a^4*b*d^ 3)*x + 6*(a^2*b^3*c^3 - 9*a^3*b^2*c^2*d + 18*a^4*b*c*d^2 - 10*a^5*d^3 + (b ^5*c^3 - 9*a*b^4*c^2*d + 18*a^2*b^3*c*d^2 - 10*a^3*b^2*d^3)*x^2 + 2*(a*b^4 *c^3 - 9*a^2*b^3*c^2*d + 18*a^3*b^2*c*d^2 - 10*a^4*b*d^3)*x)*log(b*x + a)) /(b^8*x^2 + 2*a*b^7*x + a^2*b^6)
Time = 0.68 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.51 \[ \int \frac {x^2 (c+d x)^3}{(a+b x)^3} \, dx=x^{2} \left (- \frac {3 a d^{3}}{2 b^{4}} + \frac {3 c d^{2}}{2 b^{3}}\right ) + x \left (\frac {6 a^{2} d^{3}}{b^{5}} - \frac {9 a c d^{2}}{b^{4}} + \frac {3 c^{2} d}{b^{3}}\right ) + \frac {- 9 a^{5} d^{3} + 21 a^{4} b c d^{2} - 15 a^{3} b^{2} c^{2} d + 3 a^{2} b^{3} c^{3} + x \left (- 10 a^{4} b d^{3} + 24 a^{3} b^{2} c d^{2} - 18 a^{2} b^{3} c^{2} d + 4 a b^{4} c^{3}\right )}{2 a^{2} b^{6} + 4 a b^{7} x + 2 b^{8} x^{2}} + \frac {d^{3} x^{3}}{3 b^{3}} - \frac {\left (a d - b c\right ) \left (10 a^{2} d^{2} - 8 a b c d + b^{2} c^{2}\right ) \log {\left (a + b x \right )}}{b^{6}} \]
x**2*(-3*a*d**3/(2*b**4) + 3*c*d**2/(2*b**3)) + x*(6*a**2*d**3/b**5 - 9*a* c*d**2/b**4 + 3*c**2*d/b**3) + (-9*a**5*d**3 + 21*a**4*b*c*d**2 - 15*a**3* b**2*c**2*d + 3*a**2*b**3*c**3 + x*(-10*a**4*b*d**3 + 24*a**3*b**2*c*d**2 - 18*a**2*b**3*c**2*d + 4*a*b**4*c**3))/(2*a**2*b**6 + 4*a*b**7*x + 2*b**8 *x**2) + d**3*x**3/(3*b**3) - (a*d - b*c)*(10*a**2*d**2 - 8*a*b*c*d + b**2 *c**2)*log(a + b*x)/b**6
Time = 0.21 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.46 \[ \int \frac {x^2 (c+d x)^3}{(a+b x)^3} \, dx=\frac {3 \, a^{2} b^{3} c^{3} - 15 \, a^{3} b^{2} c^{2} d + 21 \, a^{4} b c d^{2} - 9 \, a^{5} d^{3} + 2 \, {\left (2 \, a b^{4} c^{3} - 9 \, a^{2} b^{3} c^{2} d + 12 \, a^{3} b^{2} c d^{2} - 5 \, a^{4} b d^{3}\right )} x}{2 \, {\left (b^{8} x^{2} + 2 \, a b^{7} x + a^{2} b^{6}\right )}} + \frac {2 \, b^{2} d^{3} x^{3} + 9 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2} + 18 \, {\left (b^{2} c^{2} d - 3 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x}{6 \, b^{5}} + \frac {{\left (b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 18 \, a^{2} b c d^{2} - 10 \, a^{3} d^{3}\right )} \log \left (b x + a\right )}{b^{6}} \]
1/2*(3*a^2*b^3*c^3 - 15*a^3*b^2*c^2*d + 21*a^4*b*c*d^2 - 9*a^5*d^3 + 2*(2* a*b^4*c^3 - 9*a^2*b^3*c^2*d + 12*a^3*b^2*c*d^2 - 5*a^4*b*d^3)*x)/(b^8*x^2 + 2*a*b^7*x + a^2*b^6) + 1/6*(2*b^2*d^3*x^3 + 9*(b^2*c*d^2 - a*b*d^3)*x^2 + 18*(b^2*c^2*d - 3*a*b*c*d^2 + 2*a^2*d^3)*x)/b^5 + (b^3*c^3 - 9*a*b^2*c^2 *d + 18*a^2*b*c*d^2 - 10*a^3*d^3)*log(b*x + a)/b^6
Time = 0.28 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.42 \[ \int \frac {x^2 (c+d x)^3}{(a+b x)^3} \, dx=\frac {{\left (b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 18 \, a^{2} b c d^{2} - 10 \, a^{3} d^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{6}} + \frac {3 \, a^{2} b^{3} c^{3} - 15 \, a^{3} b^{2} c^{2} d + 21 \, a^{4} b c d^{2} - 9 \, a^{5} d^{3} + 2 \, {\left (2 \, a b^{4} c^{3} - 9 \, a^{2} b^{3} c^{2} d + 12 \, a^{3} b^{2} c d^{2} - 5 \, a^{4} b d^{3}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{6}} + \frac {2 \, b^{6} d^{3} x^{3} + 9 \, b^{6} c d^{2} x^{2} - 9 \, a b^{5} d^{3} x^{2} + 18 \, b^{6} c^{2} d x - 54 \, a b^{5} c d^{2} x + 36 \, a^{2} b^{4} d^{3} x}{6 \, b^{9}} \]
(b^3*c^3 - 9*a*b^2*c^2*d + 18*a^2*b*c*d^2 - 10*a^3*d^3)*log(abs(b*x + a))/ b^6 + 1/2*(3*a^2*b^3*c^3 - 15*a^3*b^2*c^2*d + 21*a^4*b*c*d^2 - 9*a^5*d^3 + 2*(2*a*b^4*c^3 - 9*a^2*b^3*c^2*d + 12*a^3*b^2*c*d^2 - 5*a^4*b*d^3)*x)/((b *x + a)^2*b^6) + 1/6*(2*b^6*d^3*x^3 + 9*b^6*c*d^2*x^2 - 9*a*b^5*d^3*x^2 + 18*b^6*c^2*d*x - 54*a*b^5*c*d^2*x + 36*a^2*b^4*d^3*x)/b^9
Time = 0.48 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.59 \[ \int \frac {x^2 (c+d x)^3}{(a+b x)^3} \, dx=x\,\left (\frac {3\,c^2\,d}{b^3}+\frac {3\,a\,\left (\frac {3\,a\,d^3}{b^4}-\frac {3\,c\,d^2}{b^3}\right )}{b}-\frac {3\,a^2\,d^3}{b^5}\right )-x^2\,\left (\frac {3\,a\,d^3}{2\,b^4}-\frac {3\,c\,d^2}{2\,b^3}\right )-\frac {x\,\left (5\,a^4\,d^3-12\,a^3\,b\,c\,d^2+9\,a^2\,b^2\,c^2\,d-2\,a\,b^3\,c^3\right )+\frac {3\,\left (3\,a^5\,d^3-7\,a^4\,b\,c\,d^2+5\,a^3\,b^2\,c^2\,d-a^2\,b^3\,c^3\right )}{2\,b}}{a^2\,b^5+2\,a\,b^6\,x+b^7\,x^2}+\frac {d^3\,x^3}{3\,b^3}-\frac {\ln \left (a+b\,x\right )\,\left (10\,a^3\,d^3-18\,a^2\,b\,c\,d^2+9\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{b^6} \]
x*((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) - x^2*((3*a*d^3)/(2*b^4) - (3*c*d^2)/(2*b^3)) - (x*(5*a^4*d^3 - 2*a*b^ 3*c^3 + 9*a^2*b^2*c^2*d - 12*a^3*b*c*d^2) + (3*(3*a^5*d^3 - a^2*b^3*c^3 + 5*a^3*b^2*c^2*d - 7*a^4*b*c*d^2))/(2*b))/(a^2*b^5 + b^7*x^2 + 2*a*b^6*x) + (d^3*x^3)/(3*b^3) - (log(a + b*x)*(10*a^3*d^3 - b^3*c^3 + 9*a*b^2*c^2*d - 18*a^2*b*c*d^2))/b^6