Integrand size = 18, antiderivative size = 104 \[ \int \frac {x^2 (c+d x)^2}{(a+b x)^2} \, dx=\frac {(b c-3 a d) (b c-a d) x}{b^4}+\frac {d (b c-a d) x^2}{b^3}+\frac {d^2 x^3}{3 b^2}-\frac {a^2 (b c-a d)^2}{b^5 (a+b x)}-\frac {2 a (b c-2 a d) (b c-a d) \log (a+b x)}{b^5} \]
(-3*a*d+b*c)*(-a*d+b*c)*x/b^4+d*(-a*d+b*c)*x^2/b^3+1/3*d^2*x^3/b^2-a^2*(-a *d+b*c)^2/b^5/(b*x+a)-2*a*(-2*a*d+b*c)*(-a*d+b*c)*ln(b*x+a)/b^5
Time = 0.05 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.10 \[ \int \frac {x^2 (c+d x)^2}{(a+b x)^2} \, dx=\frac {3 b \left (b^2 c^2-4 a b c d+3 a^2 d^2\right ) x+3 b^2 d (b c-a d) x^2+b^3 d^2 x^3-\frac {3 a^2 (b c-a d)^2}{a+b x}-6 a \left (b^2 c^2-3 a b c d+2 a^2 d^2\right ) \log (a+b x)}{3 b^5} \]
(3*b*(b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x + 3*b^2*d*(b*c - a*d)*x^2 + b^3*d ^2*x^3 - (3*a^2*(b*c - a*d)^2)/(a + b*x) - 6*a*(b^2*c^2 - 3*a*b*c*d + 2*a^ 2*d^2)*Log[a + b*x])/(3*b^5)
Time = 0.28 (sec) , antiderivative size = 104, 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.
\(\displaystyle \int \frac {x^2 (c+d x)^2}{(a+b x)^2} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (\frac {a^2 (a d-b c)^2}{b^4 (a+b x)^2}+\frac {2 a (b c-2 a d) (a d-b c)}{b^4 (a+b x)}+\frac {(b c-3 a d) (b c-a d)}{b^4}+\frac {2 d x (b c-a d)}{b^3}+\frac {d^2 x^2}{b^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 (b c-a d)^2}{b^5 (a+b x)}-\frac {2 a (b c-2 a d) (b c-a d) \log (a+b x)}{b^5}+\frac {x (b c-3 a d) (b c-a d)}{b^4}+\frac {d x^2 (b c-a d)}{b^3}+\frac {d^2 x^3}{3 b^2}\) |
((b*c - 3*a*d)*(b*c - a*d)*x)/b^4 + (d*(b*c - a*d)*x^2)/b^3 + (d^2*x^3)/(3 *b^2) - (a^2*(b*c - a*d)^2)/(b^5*(a + b*x)) - (2*a*(b*c - 2*a*d)*(b*c - a* d)*Log[a + b*x])/b^5
3.3.61.3.1 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 = 0.46 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.26
method | result | size |
default | \(\frac {\frac {1}{3} d^{2} x^{3} b^{2}-x^{2} a b \,d^{2}+x^{2} b^{2} c d +3 a^{2} d^{2} x -4 a b c d x +b^{2} c^{2} x}{b^{4}}-\frac {2 a \left (2 a^{2} d^{2}-3 a b c d +b^{2} c^{2}\right ) \ln \left (b x +a \right )}{b^{5}}-\frac {a^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{b^{5} \left (b x +a \right )}\) | \(131\) |
norman | \(\frac {\frac {\left (2 a^{2} d^{2}-3 a b c d +b^{2} c^{2}\right ) x^{2}}{b^{3}}+\frac {\left (4 a^{4} d^{2}-6 a^{3} b c d +2 a^{2} b^{2} c^{2}\right ) x}{b^{4} a}+\frac {d^{2} x^{4}}{3 b}-\frac {d \left (2 a d -3 b c \right ) x^{3}}{3 b^{2}}}{b x +a}-\frac {2 a \left (2 a^{2} d^{2}-3 a b c d +b^{2} c^{2}\right ) \ln \left (b x +a \right )}{b^{5}}\) | \(139\) |
risch | \(\frac {d^{2} x^{3}}{3 b^{2}}-\frac {x^{2} a \,d^{2}}{b^{3}}+\frac {x^{2} c d}{b^{2}}+\frac {3 a^{2} d^{2} x}{b^{4}}-\frac {4 a c d x}{b^{3}}+\frac {c^{2} x}{b^{2}}-\frac {4 a^{3} \ln \left (b x +a \right ) d^{2}}{b^{5}}+\frac {6 a^{2} \ln \left (b x +a \right ) c d}{b^{4}}-\frac {2 a \ln \left (b x +a \right ) c^{2}}{b^{3}}-\frac {a^{4} d^{2}}{b^{5} \left (b x +a \right )}+\frac {2 a^{3} c d}{b^{4} \left (b x +a \right )}-\frac {a^{2} c^{2}}{b^{3} \left (b x +a \right )}\) | \(164\) |
parallelrisch | \(-\frac {-d^{2} x^{4} b^{4}+2 x^{3} a \,b^{3} d^{2}-3 x^{3} b^{4} c d +12 \ln \left (b x +a \right ) x \,a^{3} b \,d^{2}-18 \ln \left (b x +a \right ) x \,a^{2} b^{2} c d +6 \ln \left (b x +a \right ) x a \,b^{3} c^{2}-6 x^{2} a^{2} b^{2} d^{2}+9 x^{2} a \,b^{3} c d -3 x^{2} b^{4} c^{2}+12 \ln \left (b x +a \right ) a^{4} d^{2}-18 \ln \left (b x +a \right ) a^{3} b c d +6 \ln \left (b x +a \right ) a^{2} b^{2} c^{2}+12 a^{4} d^{2}-18 a^{3} b c d +6 a^{2} b^{2} c^{2}}{3 b^{5} \left (b x +a \right )}\) | \(204\) |
1/b^4*(1/3*d^2*x^3*b^2-x^2*a*b*d^2+x^2*b^2*c*d+3*a^2*d^2*x-4*a*b*c*d*x+b^2 *c^2*x)-2*a/b^5*(2*a^2*d^2-3*a*b*c*d+b^2*c^2)*ln(b*x+a)-a^2*(a^2*d^2-2*a*b *c*d+b^2*c^2)/b^5/(b*x+a)
Time = 0.23 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.94 \[ \int \frac {x^2 (c+d x)^2}{(a+b x)^2} \, dx=\frac {b^{4} d^{2} x^{4} - 3 \, a^{2} b^{2} c^{2} + 6 \, a^{3} b c d - 3 \, a^{4} d^{2} + {\left (3 \, b^{4} c d - 2 \, a b^{3} d^{2}\right )} x^{3} + 3 \, {\left (b^{4} c^{2} - 3 \, a b^{3} c d + 2 \, a^{2} b^{2} d^{2}\right )} x^{2} + 3 \, {\left (a b^{3} c^{2} - 4 \, a^{2} b^{2} c d + 3 \, a^{3} b d^{2}\right )} x - 6 \, {\left (a^{2} b^{2} c^{2} - 3 \, a^{3} b c d + 2 \, a^{4} d^{2} + {\left (a b^{3} c^{2} - 3 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x\right )} \log \left (b x + a\right )}{3 \, {\left (b^{6} x + a b^{5}\right )}} \]
1/3*(b^4*d^2*x^4 - 3*a^2*b^2*c^2 + 6*a^3*b*c*d - 3*a^4*d^2 + (3*b^4*c*d - 2*a*b^3*d^2)*x^3 + 3*(b^4*c^2 - 3*a*b^3*c*d + 2*a^2*b^2*d^2)*x^2 + 3*(a*b^ 3*c^2 - 4*a^2*b^2*c*d + 3*a^3*b*d^2)*x - 6*(a^2*b^2*c^2 - 3*a^3*b*c*d + 2* a^4*d^2 + (a*b^3*c^2 - 3*a^2*b^2*c*d + 2*a^3*b*d^2)*x)*log(b*x + a))/(b^6* x + a*b^5)
Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.21 \[ \int \frac {x^2 (c+d x)^2}{(a+b x)^2} \, dx=- \frac {2 a \left (a d - b c\right ) \left (2 a d - b c\right ) \log {\left (a + b x \right )}}{b^{5}} + x^{2} \left (- \frac {a d^{2}}{b^{3}} + \frac {c d}{b^{2}}\right ) + x \left (\frac {3 a^{2} d^{2}}{b^{4}} - \frac {4 a c d}{b^{3}} + \frac {c^{2}}{b^{2}}\right ) + \frac {- a^{4} d^{2} + 2 a^{3} b c d - a^{2} b^{2} c^{2}}{a b^{5} + b^{6} x} + \frac {d^{2} x^{3}}{3 b^{2}} \]
-2*a*(a*d - b*c)*(2*a*d - b*c)*log(a + b*x)/b**5 + x**2*(-a*d**2/b**3 + c* d/b**2) + x*(3*a**2*d**2/b**4 - 4*a*c*d/b**3 + c**2/b**2) + (-a**4*d**2 + 2*a**3*b*c*d - a**2*b**2*c**2)/(a*b**5 + b**6*x) + d**2*x**3/(3*b**2)
Time = 0.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.33 \[ \int \frac {x^2 (c+d x)^2}{(a+b x)^2} \, dx=-\frac {a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}}{b^{6} x + a b^{5}} + \frac {b^{2} d^{2} x^{3} + 3 \, {\left (b^{2} c d - a b d^{2}\right )} x^{2} + 3 \, {\left (b^{2} c^{2} - 4 \, a b c d + 3 \, a^{2} d^{2}\right )} x}{3 \, b^{4}} - \frac {2 \, {\left (a b^{2} c^{2} - 3 \, a^{2} b c d + 2 \, a^{3} d^{2}\right )} \log \left (b x + a\right )}{b^{5}} \]
-(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)/(b^6*x + a*b^5) + 1/3*(b^2*d^2*x^3 + 3*(b^2*c*d - a*b*d^2)*x^2 + 3*(b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x)/b^4 - 2*(a*b^2*c^2 - 3*a^2*b*c*d + 2*a^3*d^2)*log(b*x + a)/b^5
Time = 0.27 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.81 \[ \int \frac {x^2 (c+d x)^2}{(a+b x)^2} \, dx=\frac {{\left (d^{2} + \frac {3 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )}}{{\left (b x + a\right )} b} + \frac {3 \, {\left (b^{4} c^{2} - 6 \, a b^{3} c d + 6 \, a^{2} b^{2} d^{2}\right )}}{{\left (b x + a\right )}^{2} b^{2}}\right )} {\left (b x + a\right )}^{3}}{3 \, b^{5}} + \frac {2 \, {\left (a b^{2} c^{2} - 3 \, a^{2} b c d + 2 \, a^{3} d^{2}\right )} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{5}} - \frac {\frac {a^{2} b^{5} c^{2}}{b x + a} - \frac {2 \, a^{3} b^{4} c d}{b x + a} + \frac {a^{4} b^{3} d^{2}}{b x + a}}{b^{8}} \]
1/3*(d^2 + 3*(b^2*c*d - 2*a*b*d^2)/((b*x + a)*b) + 3*(b^4*c^2 - 6*a*b^3*c* d + 6*a^2*b^2*d^2)/((b*x + a)^2*b^2))*(b*x + a)^3/b^5 + 2*(a*b^2*c^2 - 3*a ^2*b*c*d + 2*a^3*d^2)*log(abs(b*x + a)/((b*x + a)^2*abs(b)))/b^5 - (a^2*b^ 5*c^2/(b*x + a) - 2*a^3*b^4*c*d/(b*x + a) + a^4*b^3*d^2/(b*x + a))/b^8
Time = 0.39 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.52 \[ \int \frac {x^2 (c+d x)^2}{(a+b x)^2} \, dx=x\,\left (\frac {c^2}{b^2}+\frac {2\,a\,\left (\frac {2\,a\,d^2}{b^3}-\frac {2\,c\,d}{b^2}\right )}{b}-\frac {a^2\,d^2}{b^4}\right )-x^2\,\left (\frac {a\,d^2}{b^3}-\frac {c\,d}{b^2}\right )-\frac {a^4\,d^2-2\,a^3\,b\,c\,d+a^2\,b^2\,c^2}{b\,\left (x\,b^5+a\,b^4\right )}-\frac {\ln \left (a+b\,x\right )\,\left (4\,a^3\,d^2-6\,a^2\,b\,c\,d+2\,a\,b^2\,c^2\right )}{b^5}+\frac {d^2\,x^3}{3\,b^2} \]