Integrand size = 18, antiderivative size = 117 \[ \int \frac {x^3 (c+d x)^2}{a+b x} \, dx=\frac {a^2 (b c-a d)^2 x}{b^5}-\frac {a (b c-a d)^2 x^2}{2 b^4}+\frac {(b c-a d)^2 x^3}{3 b^3}+\frac {d (2 b c-a d) x^4}{4 b^2}+\frac {d^2 x^5}{5 b}-\frac {a^3 (b c-a d)^2 \log (a+b x)}{b^6} \]
a^2*(-a*d+b*c)^2*x/b^5-1/2*a*(-a*d+b*c)^2*x^2/b^4+1/3*(-a*d+b*c)^2*x^3/b^3 +1/4*d*(-a*d+2*b*c)*x^4/b^2+1/5*d^2*x^5/b-a^3*(-a*d+b*c)^2*ln(b*x+a)/b^6
Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 (c+d x)^2}{a+b x} \, dx=\frac {60 a^2 b (b c-a d)^2 x-30 a b^2 (b c-a d)^2 x^2+20 b^3 (b c-a d)^2 x^3+15 b^4 d (2 b c-a d) x^4+12 b^5 d^2 x^5-60 a^3 (b c-a d)^2 \log (a+b x)}{60 b^6} \]
(60*a^2*b*(b*c - a*d)^2*x - 30*a*b^2*(b*c - a*d)^2*x^2 + 20*b^3*(b*c - a*d )^2*x^3 + 15*b^4*d*(2*b*c - a*d)*x^4 + 12*b^5*d^2*x^5 - 60*a^3*(b*c - a*d) ^2*Log[a + b*x])/(60*b^6)
Time = 0.30 (sec) , antiderivative size = 117, 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.
\(\displaystyle \int \frac {x^3 (c+d x)^2}{a+b x} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {a^3 (a d-b c)^2}{b^5 (a+b x)}+\frac {a^2 (a d-b c)^2}{b^5}-\frac {a x (a d-b c)^2}{b^4}+\frac {x^2 (b c-a d)^2}{b^3}+\frac {d x^3 (2 b c-a d)}{b^2}+\frac {d^2 x^4}{b}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 (b c-a d)^2 \log (a+b x)}{b^6}+\frac {a^2 x (b c-a d)^2}{b^5}-\frac {a x^2 (b c-a d)^2}{2 b^4}+\frac {x^3 (b c-a d)^2}{3 b^3}+\frac {d x^4 (2 b c-a d)}{4 b^2}+\frac {d^2 x^5}{5 b}\) |
(a^2*(b*c - a*d)^2*x)/b^5 - (a*(b*c - a*d)^2*x^2)/(2*b^4) + ((b*c - a*d)^2 *x^3)/(3*b^3) + (d*(2*b*c - a*d)*x^4)/(4*b^2) + (d^2*x^5)/(5*b) - (a^3*(b* c - a*d)^2*Log[a + b*x])/b^6
3.3.13.3.1 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 = 0.44 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.31
method | result | size |
norman | \(\frac {a^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x}{b^{5}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{3}}{3 b^{3}}+\frac {d^{2} x^{5}}{5 b}-\frac {a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{2}}{2 b^{4}}-\frac {d \left (a d -2 b c \right ) x^{4}}{4 b^{2}}-\frac {a^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b x +a \right )}{b^{6}}\) | \(153\) |
default | \(\frac {\frac {1}{5} d^{2} x^{5} b^{4}-\frac {1}{4} a \,b^{3} d^{2} x^{4}+\frac {1}{2} b^{4} c d \,x^{4}+\frac {1}{3} a^{2} b^{2} d^{2} x^{3}-\frac {2}{3} a \,b^{3} c d \,x^{3}+\frac {1}{3} b^{4} c^{2} x^{3}-\frac {1}{2} a^{3} b \,d^{2} x^{2}+a^{2} b^{2} c d \,x^{2}-\frac {1}{2} a \,b^{3} c^{2} x^{2}+a^{4} d^{2} x -2 a^{3} b c d x +a^{2} b^{2} c^{2} x}{b^{5}}-\frac {a^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b x +a \right )}{b^{6}}\) | \(175\) |
risch | \(\frac {d^{2} x^{5}}{5 b}-\frac {a \,d^{2} x^{4}}{4 b^{2}}+\frac {c d \,x^{4}}{2 b}+\frac {a^{2} d^{2} x^{3}}{3 b^{3}}-\frac {2 a c d \,x^{3}}{3 b^{2}}+\frac {c^{2} x^{3}}{3 b}-\frac {a^{3} d^{2} x^{2}}{2 b^{4}}+\frac {a^{2} c d \,x^{2}}{b^{3}}-\frac {a \,c^{2} x^{2}}{2 b^{2}}+\frac {a^{4} d^{2} x}{b^{5}}-\frac {2 a^{3} c d x}{b^{4}}+\frac {a^{2} c^{2} x}{b^{3}}-\frac {a^{5} \ln \left (b x +a \right ) d^{2}}{b^{6}}+\frac {2 a^{4} \ln \left (b x +a \right ) c d}{b^{5}}-\frac {a^{3} \ln \left (b x +a \right ) c^{2}}{b^{4}}\) | \(192\) |
parallelrisch | \(-\frac {-12 d^{2} x^{5} b^{5}+15 x^{4} a \,b^{4} d^{2}-30 x^{4} b^{5} c d -20 x^{3} a^{2} b^{3} d^{2}+40 x^{3} a \,b^{4} c d -20 x^{3} b^{5} c^{2}+30 x^{2} a^{3} b^{2} d^{2}-60 x^{2} a^{2} b^{3} c d +30 x^{2} a \,b^{4} c^{2}+60 \ln \left (b x +a \right ) a^{5} d^{2}-120 \ln \left (b x +a \right ) a^{4} b c d +60 \ln \left (b x +a \right ) a^{3} b^{2} c^{2}-60 x \,a^{4} b \,d^{2}+120 x \,a^{3} b^{2} c d -60 x \,a^{2} b^{3} c^{2}}{60 b^{6}}\) | \(193\) |
a^2*(a^2*d^2-2*a*b*c*d+b^2*c^2)/b^5*x+1/3/b^3*(a^2*d^2-2*a*b*c*d+b^2*c^2)* x^3+1/5*d^2*x^5/b-1/2*a/b^4*(a^2*d^2-2*a*b*c*d+b^2*c^2)*x^2-1/4/b^2*d*(a*d -2*b*c)*x^4-a^3*(a^2*d^2-2*a*b*c*d+b^2*c^2)/b^6*ln(b*x+a)
Time = 0.22 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.45 \[ \int \frac {x^3 (c+d x)^2}{a+b x} \, dx=\frac {12 \, b^{5} d^{2} x^{5} + 15 \, {\left (2 \, b^{5} c d - a b^{4} d^{2}\right )} x^{4} + 20 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{3} - 30 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} x^{2} + 60 \, {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} x - 60 \, {\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )} \log \left (b x + a\right )}{60 \, b^{6}} \]
1/60*(12*b^5*d^2*x^5 + 15*(2*b^5*c*d - a*b^4*d^2)*x^4 + 20*(b^5*c^2 - 2*a* b^4*c*d + a^2*b^3*d^2)*x^3 - 30*(a*b^4*c^2 - 2*a^2*b^3*c*d + a^3*b^2*d^2)* x^2 + 60*(a^2*b^3*c^2 - 2*a^3*b^2*c*d + a^4*b*d^2)*x - 60*(a^3*b^2*c^2 - 2 *a^4*b*c*d + a^5*d^2)*log(b*x + a))/b^6
Time = 0.19 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.32 \[ \int \frac {x^3 (c+d x)^2}{a+b x} \, dx=- \frac {a^{3} \left (a d - b c\right )^{2} \log {\left (a + b x \right )}}{b^{6}} + x^{4} \left (- \frac {a d^{2}}{4 b^{2}} + \frac {c d}{2 b}\right ) + x^{3} \left (\frac {a^{2} d^{2}}{3 b^{3}} - \frac {2 a c d}{3 b^{2}} + \frac {c^{2}}{3 b}\right ) + x^{2} \left (- \frac {a^{3} d^{2}}{2 b^{4}} + \frac {a^{2} c d}{b^{3}} - \frac {a c^{2}}{2 b^{2}}\right ) + x \left (\frac {a^{4} d^{2}}{b^{5}} - \frac {2 a^{3} c d}{b^{4}} + \frac {a^{2} c^{2}}{b^{3}}\right ) + \frac {d^{2} x^{5}}{5 b} \]
-a**3*(a*d - b*c)**2*log(a + b*x)/b**6 + x**4*(-a*d**2/(4*b**2) + c*d/(2*b )) + x**3*(a**2*d**2/(3*b**3) - 2*a*c*d/(3*b**2) + c**2/(3*b)) + x**2*(-a* *3*d**2/(2*b**4) + a**2*c*d/b**3 - a*c**2/(2*b**2)) + x*(a**4*d**2/b**5 - 2*a**3*c*d/b**4 + a**2*c**2/b**3) + d**2*x**5/(5*b)
Time = 0.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.44 \[ \int \frac {x^3 (c+d x)^2}{a+b x} \, dx=\frac {12 \, b^{4} d^{2} x^{5} + 15 \, {\left (2 \, b^{4} c d - a b^{3} d^{2}\right )} x^{4} + 20 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{3} - 30 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2} + 60 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} x}{60 \, b^{5}} - \frac {{\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )} \log \left (b x + a\right )}{b^{6}} \]
1/60*(12*b^4*d^2*x^5 + 15*(2*b^4*c*d - a*b^3*d^2)*x^4 + 20*(b^4*c^2 - 2*a* b^3*c*d + a^2*b^2*d^2)*x^3 - 30*(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^ 2 + 60*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*x)/b^5 - (a^3*b^2*c^2 - 2*a^4 *b*c*d + a^5*d^2)*log(b*x + a)/b^6
Time = 0.29 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.55 \[ \int \frac {x^3 (c+d x)^2}{a+b x} \, dx=\frac {12 \, b^{4} d^{2} x^{5} + 30 \, b^{4} c d x^{4} - 15 \, a b^{3} d^{2} x^{4} + 20 \, b^{4} c^{2} x^{3} - 40 \, a b^{3} c d x^{3} + 20 \, a^{2} b^{2} d^{2} x^{3} - 30 \, a b^{3} c^{2} x^{2} + 60 \, a^{2} b^{2} c d x^{2} - 30 \, a^{3} b d^{2} x^{2} + 60 \, a^{2} b^{2} c^{2} x - 120 \, a^{3} b c d x + 60 \, a^{4} d^{2} x}{60 \, b^{5}} - \frac {{\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{6}} \]
1/60*(12*b^4*d^2*x^5 + 30*b^4*c*d*x^4 - 15*a*b^3*d^2*x^4 + 20*b^4*c^2*x^3 - 40*a*b^3*c*d*x^3 + 20*a^2*b^2*d^2*x^3 - 30*a*b^3*c^2*x^2 + 60*a^2*b^2*c* d*x^2 - 30*a^3*b*d^2*x^2 + 60*a^2*b^2*c^2*x - 120*a^3*b*c*d*x + 60*a^4*d^2 *x)/b^5 - (a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d^2)*log(abs(b*x + a))/b^6
Time = 0.05 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.55 \[ \int \frac {x^3 (c+d x)^2}{a+b x} \, dx=x^3\,\left (\frac {c^2}{3\,b}+\frac {a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{3\,b}\right )-x^4\,\left (\frac {a\,d^2}{4\,b^2}-\frac {c\,d}{2\,b}\right )-\frac {\ln \left (a+b\,x\right )\,\left (a^5\,d^2-2\,a^4\,b\,c\,d+a^3\,b^2\,c^2\right )}{b^6}+\frac {d^2\,x^5}{5\,b}-\frac {a\,x^2\,\left (\frac {c^2}{b}+\frac {a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}\right )}{2\,b}+\frac {a^2\,x\,\left (\frac {c^2}{b}+\frac {a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}\right )}{b^2} \]