Integrand size = 16, antiderivative size = 108 \[ \int \frac {(a+b x)^5 (A+B x)}{x^3} \, dx=-\frac {a^5 A}{2 x^2}-\frac {a^4 (5 A b+a B)}{x}+10 a^2 b^2 (A b+a B) x+\frac {5}{2} a b^3 (A b+2 a B) x^2+\frac {1}{3} b^4 (A b+5 a B) x^3+\frac {1}{4} b^5 B x^4+5 a^3 b (2 A b+a B) \log (x) \]
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Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {77} \[ \int \frac {(a+b x)^5 (A+B x)}{x^3} \, dx=-\frac {a^5 A}{2 x^2}-\frac {a^4 (a B+5 A b)}{x}+5 a^3 b \log (x) (a B+2 A b)+10 a^2 b^2 x (a B+A b)+\frac {1}{3} b^4 x^3 (5 a B+A b)+\frac {5}{2} a b^3 x^2 (2 a B+A b)+\frac {1}{4} b^5 B x^4 \]
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Rule 77
Rubi steps \begin{align*} \text {integral}& = \int \left (10 a^2 b^2 (A b+a B)+\frac {a^5 A}{x^3}+\frac {a^4 (5 A b+a B)}{x^2}+\frac {5 a^3 b (2 A b+a B)}{x}+5 a b^3 (A b+2 a B) x+b^4 (A b+5 a B) x^2+b^5 B x^3\right ) \, dx \\ & = -\frac {a^5 A}{2 x^2}-\frac {a^4 (5 A b+a B)}{x}+10 a^2 b^2 (A b+a B) x+\frac {5}{2} a b^3 (A b+2 a B) x^2+\frac {1}{3} b^4 (A b+5 a B) x^3+\frac {1}{4} b^5 B x^4+5 a^3 b (2 A b+a B) \log (x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^5 (A+B x)}{x^3} \, dx=-\frac {5 a^4 A b}{x}+10 a^3 b^2 B x+5 a^2 b^3 x (2 A+B x)-\frac {a^5 (A+2 B x)}{2 x^2}+\frac {5}{6} a b^4 x^2 (3 A+2 B x)+\frac {1}{12} b^5 x^3 (4 A+3 B x)+5 a^3 b (2 A b+a B) \log (x) \]
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Time = 0.39 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.05
method | result | size |
default | \(\frac {b^{5} B \,x^{4}}{4}+\frac {A \,b^{5} x^{3}}{3}+\frac {5 B a \,b^{4} x^{3}}{3}+\frac {5 A a \,b^{4} x^{2}}{2}+5 B \,a^{2} b^{3} x^{2}+10 A \,a^{2} b^{3} x +10 B \,a^{3} b^{2} x +5 a^{3} b \left (2 A b +B a \right ) \ln \left (x \right )-\frac {a^{4} \left (5 A b +B a \right )}{x}-\frac {a^{5} A}{2 x^{2}}\) | \(113\) |
risch | \(\frac {b^{5} B \,x^{4}}{4}+\frac {A \,b^{5} x^{3}}{3}+\frac {5 B a \,b^{4} x^{3}}{3}+\frac {5 A a \,b^{4} x^{2}}{2}+5 B \,a^{2} b^{3} x^{2}+10 A \,a^{2} b^{3} x +10 B \,a^{3} b^{2} x +\frac {\left (-5 a^{4} b A -a^{5} B \right ) x -\frac {a^{5} A}{2}}{x^{2}}+10 A \ln \left (x \right ) a^{3} b^{2}+5 B \ln \left (x \right ) a^{4} b\) | \(119\) |
norman | \(\frac {\left (\frac {1}{3} b^{5} A +\frac {5}{3} a \,b^{4} B \right ) x^{5}+\left (\frac {5}{2} a \,b^{4} A +5 a^{2} b^{3} B \right ) x^{4}+\left (10 a^{2} b^{3} A +10 a^{3} b^{2} B \right ) x^{3}+\left (-5 a^{4} b A -a^{5} B \right ) x -\frac {a^{5} A}{2}+\frac {b^{5} B \,x^{6}}{4}}{x^{2}}+\left (10 a^{3} b^{2} A +5 a^{4} b B \right ) \ln \left (x \right )\) | \(120\) |
parallelrisch | \(\frac {3 b^{5} B \,x^{6}+4 A \,b^{5} x^{5}+20 B a \,b^{4} x^{5}+30 a A \,b^{4} x^{4}+60 B \,a^{2} b^{3} x^{4}+120 A \ln \left (x \right ) x^{2} a^{3} b^{2}+120 a^{2} A \,b^{3} x^{3}+60 B \ln \left (x \right ) x^{2} a^{4} b +120 B \,a^{3} b^{2} x^{3}-60 a^{4} A b x -12 a^{5} B x -6 a^{5} A}{12 x^{2}}\) | \(128\) |
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none
Time = 0.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^5 (A+B x)}{x^3} \, dx=\frac {3 \, B b^{5} x^{6} - 6 \, A a^{5} + 4 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 30 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 120 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 60 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} \log \left (x\right ) - 12 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{12 \, x^{2}} \]
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Time = 0.22 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b x)^5 (A+B x)}{x^3} \, dx=\frac {B b^{5} x^{4}}{4} + 5 a^{3} b \left (2 A b + B a\right ) \log {\left (x \right )} + x^{3} \left (\frac {A b^{5}}{3} + \frac {5 B a b^{4}}{3}\right ) + x^{2} \cdot \left (\frac {5 A a b^{4}}{2} + 5 B a^{2} b^{3}\right ) + x \left (10 A a^{2} b^{3} + 10 B a^{3} b^{2}\right ) + \frac {- A a^{5} + x \left (- 10 A a^{4} b - 2 B a^{5}\right )}{2 x^{2}} \]
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Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b x)^5 (A+B x)}{x^3} \, dx=\frac {1}{4} \, B b^{5} x^{4} + \frac {1}{3} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{3} + \frac {5}{2} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{2} + 10 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x + 5 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} \log \left (x\right ) - \frac {A a^{5} + 2 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{2 \, x^{2}} \]
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Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x)^5 (A+B x)}{x^3} \, dx=\frac {1}{4} \, B b^{5} x^{4} + \frac {5}{3} \, B a b^{4} x^{3} + \frac {1}{3} \, A b^{5} x^{3} + 5 \, B a^{2} b^{3} x^{2} + \frac {5}{2} \, A a b^{4} x^{2} + 10 \, B a^{3} b^{2} x + 10 \, A a^{2} b^{3} x + 5 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac {A a^{5} + 2 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{2 \, x^{2}} \]
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Time = 0.34 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^5 (A+B x)}{x^3} \, dx=\ln \left (x\right )\,\left (5\,B\,a^4\,b+10\,A\,a^3\,b^2\right )-\frac {x\,\left (B\,a^5+5\,A\,b\,a^4\right )+\frac {A\,a^5}{2}}{x^2}+x^3\,\left (\frac {A\,b^5}{3}+\frac {5\,B\,a\,b^4}{3}\right )+\frac {B\,b^5\,x^4}{4}+10\,a^2\,b^2\,x\,\left (A\,b+B\,a\right )+\frac {5\,a\,b^3\,x^2\,\left (A\,b+2\,B\,a\right )}{2} \]
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