Integrand size = 16, antiderivative size = 104 \[ \int \frac {(a+b x)^5 (A+B x)}{x^6} \, dx=-\frac {a^5 A}{5 x^5}-\frac {a^4 (5 A b+a B)}{4 x^4}-\frac {5 a^3 b (2 A b+a B)}{3 x^3}-\frac {5 a^2 b^2 (A b+a B)}{x^2}-\frac {5 a b^3 (A b+2 a B)}{x}+b^5 B x+b^4 (A b+5 a B) \log (x) \]
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Time = 0.04 (sec) , antiderivative size = 104, 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} \[ \int \frac {(a+b x)^5 (A+B x)}{x^6} \, dx=-\frac {a^5 A}{5 x^5}-\frac {a^4 (a B+5 A b)}{4 x^4}-\frac {5 a^3 b (a B+2 A b)}{3 x^3}-\frac {5 a^2 b^2 (a B+A b)}{x^2}+b^4 \log (x) (5 a B+A b)-\frac {5 a b^3 (2 a B+A b)}{x}+b^5 B x \]
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Rule 77
Rubi steps \begin{align*} \text {integral}& = \int \left (b^5 B+\frac {a^5 A}{x^6}+\frac {a^4 (5 A b+a B)}{x^5}+\frac {5 a^3 b (2 A b+a B)}{x^4}+\frac {10 a^2 b^2 (A b+a B)}{x^3}+\frac {5 a b^3 (A b+2 a B)}{x^2}+\frac {b^4 (A b+5 a B)}{x}\right ) \, dx \\ & = -\frac {a^5 A}{5 x^5}-\frac {a^4 (5 A b+a B)}{4 x^4}-\frac {5 a^3 b (2 A b+a B)}{3 x^3}-\frac {5 a^2 b^2 (A b+a B)}{x^2}-\frac {5 a b^3 (A b+2 a B)}{x}+b^5 B x+b^4 (A b+5 a B) \log (x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.02 \[ \int \frac {(a+b x)^5 (A+B x)}{x^6} \, dx=-\frac {5 a A b^4}{x}+b^5 B x-\frac {5 a^2 b^3 (A+2 B x)}{x^2}-\frac {5 a^3 b^2 (2 A+3 B x)}{3 x^3}-\frac {5 a^4 b (3 A+4 B x)}{12 x^4}-\frac {a^5 (4 A+5 B x)}{20 x^5}+b^4 (A b+5 a B) \log (x) \]
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Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {a^{5} A}{5 x^{5}}-\frac {a^{4} \left (5 A b +B a \right )}{4 x^{4}}-\frac {5 a^{3} b \left (2 A b +B a \right )}{3 x^{3}}-\frac {5 a^{2} b^{2} \left (A b +B a \right )}{x^{2}}-\frac {5 a \,b^{3} \left (A b +2 B a \right )}{x}+b^{5} B x +b^{4} \left (A b +5 B a \right ) \ln \left (x \right )\) | \(99\) |
risch | \(b^{5} B x +\frac {\left (-5 a \,b^{4} A -10 a^{2} b^{3} B \right ) x^{4}+\left (-5 a^{2} b^{3} A -5 a^{3} b^{2} B \right ) x^{3}+\left (-\frac {10}{3} a^{3} b^{2} A -\frac {5}{3} a^{4} b B \right ) x^{2}+\left (-\frac {5}{4} a^{4} b A -\frac {1}{4} a^{5} B \right ) x -\frac {a^{5} A}{5}}{x^{5}}+A \ln \left (x \right ) b^{5}+5 B \ln \left (x \right ) a \,b^{4}\) | \(116\) |
norman | \(\frac {\left (-\frac {10}{3} a^{3} b^{2} A -\frac {5}{3} a^{4} b B \right ) x^{2}+\left (-\frac {5}{4} a^{4} b A -\frac {1}{4} a^{5} B \right ) x +\left (-5 a \,b^{4} A -10 a^{2} b^{3} B \right ) x^{4}+\left (-5 a^{2} b^{3} A -5 a^{3} b^{2} B \right ) x^{3}+b^{5} B \,x^{6}-\frac {a^{5} A}{5}}{x^{5}}+\left (b^{5} A +5 a \,b^{4} B \right ) \ln \left (x \right )\) | \(118\) |
parallelrisch | \(\frac {60 A \ln \left (x \right ) x^{5} b^{5}+300 B \ln \left (x \right ) x^{5} a \,b^{4}+60 b^{5} B \,x^{6}-300 a A \,b^{4} x^{4}-600 B \,a^{2} b^{3} x^{4}-300 a^{2} A \,b^{3} x^{3}-300 B \,a^{3} b^{2} x^{3}-200 a^{3} A \,b^{2} x^{2}-100 B \,a^{4} b \,x^{2}-75 a^{4} A b x -15 a^{5} B x -12 a^{5} A}{60 x^{5}}\) | \(128\) |
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Time = 0.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b x)^5 (A+B x)}{x^6} \, dx=\frac {60 \, B b^{5} x^{6} + 60 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} \log \left (x\right ) - 12 \, A a^{5} - 300 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} - 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} - 100 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 15 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{60 \, x^{5}} \]
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Time = 1.08 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b x)^5 (A+B x)}{x^6} \, dx=B b^{5} x + b^{4} \left (A b + 5 B a\right ) \log {\left (x \right )} + \frac {- 12 A a^{5} + x^{4} \left (- 300 A a b^{4} - 600 B a^{2} b^{3}\right ) + x^{3} \left (- 300 A a^{2} b^{3} - 300 B a^{3} b^{2}\right ) + x^{2} \left (- 200 A a^{3} b^{2} - 100 B a^{4} b\right ) + x \left (- 75 A a^{4} b - 15 B a^{5}\right )}{60 x^{5}} \]
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Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^5 (A+B x)}{x^6} \, dx=B b^{5} x + {\left (5 \, B a b^{4} + A b^{5}\right )} \log \left (x\right ) - \frac {12 \, A a^{5} + 300 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 100 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 15 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{60 \, x^{5}} \]
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Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^5 (A+B x)}{x^6} \, dx=B b^{5} x + {\left (5 \, B a b^{4} + A b^{5}\right )} \log \left ({\left | x \right |}\right ) - \frac {12 \, A a^{5} + 300 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 100 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 15 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{60 \, x^{5}} \]
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Time = 0.36 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^5 (A+B x)}{x^6} \, dx=\ln \left (x\right )\,\left (A\,b^5+5\,B\,a\,b^4\right )-\frac {x\,\left (\frac {B\,a^5}{4}+\frac {5\,A\,b\,a^4}{4}\right )+\frac {A\,a^5}{5}+x^4\,\left (10\,B\,a^2\,b^3+5\,A\,a\,b^4\right )+x^2\,\left (\frac {5\,B\,a^4\,b}{3}+\frac {10\,A\,a^3\,b^2}{3}\right )+x^3\,\left (5\,B\,a^3\,b^2+5\,A\,a^2\,b^3\right )}{x^5}+B\,b^5\,x \]
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