Integrand size = 16, antiderivative size = 108 \[ \int \frac {(a+b x)^5 (A+B x)}{x^4} \, dx=-\frac {a^5 A}{3 x^3}-\frac {a^4 (5 A b+a B)}{2 x^2}-\frac {5 a^3 b (2 A b+a B)}{x}+5 a b^3 (A b+2 a B) x+\frac {1}{2} b^4 (A b+5 a B) x^2+\frac {1}{3} b^5 B x^3+10 a^2 b^2 (A b+a B) \log (x) \]
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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} \[ \int \frac {(a+b x)^5 (A+B x)}{x^4} \, dx=-\frac {a^5 A}{3 x^3}-\frac {a^4 (a B+5 A b)}{2 x^2}-\frac {5 a^3 b (a B+2 A b)}{x}+10 a^2 b^2 \log (x) (a B+A b)+\frac {1}{2} b^4 x^2 (5 a B+A b)+5 a b^3 x (2 a B+A b)+\frac {1}{3} b^5 B x^3 \]
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Rule 77
Rubi steps \begin{align*} \text {integral}& = \int \left (5 a b^3 (A b+2 a B)+\frac {a^5 A}{x^4}+\frac {a^4 (5 A b+a B)}{x^3}+\frac {5 a^3 b (2 A b+a B)}{x^2}+\frac {10 a^2 b^2 (A b+a B)}{x}+b^4 (A b+5 a B) x+b^5 B x^2\right ) \, dx \\ & = -\frac {a^5 A}{3 x^3}-\frac {a^4 (5 A b+a B)}{2 x^2}-\frac {5 a^3 b (2 A b+a B)}{x}+5 a b^3 (A b+2 a B) x+\frac {1}{2} b^4 (A b+5 a B) x^2+\frac {1}{3} b^5 B x^3+10 a^2 b^2 (A b+a B) \log (x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b x)^5 (A+B x)}{x^4} \, dx=\frac {-60 a^3 A b^2 x^2+60 a^2 b^3 B x^4+15 a b^4 x^4 (2 A+B x)-15 a^4 b x (A+2 B x)+b^5 x^5 (3 A+2 B x)-a^5 (2 A+3 B x)+60 a^2 b^2 (A b+a B) x^3 \log (x)}{6 x^3} \]
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Time = 0.39 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.99
method | result | size |
default | \(\frac {b^{5} B \,x^{3}}{3}+\frac {A \,b^{5} x^{2}}{2}+\frac {5 B a \,b^{4} x^{2}}{2}+5 A a \,b^{4} x +10 B \,a^{2} b^{3} x +10 a^{2} b^{2} \left (A b +B a \right ) \ln \left (x \right )-\frac {a^{5} A}{3 x^{3}}-\frac {5 a^{3} b \left (2 A b +B a \right )}{x}-\frac {a^{4} \left (5 A b +B a \right )}{2 x^{2}}\) | \(107\) |
risch | \(\frac {b^{5} B \,x^{3}}{3}+\frac {A \,b^{5} x^{2}}{2}+\frac {5 B a \,b^{4} x^{2}}{2}+5 A a \,b^{4} x +10 B \,a^{2} b^{3} x +\frac {\left (-10 a^{3} b^{2} A -5 a^{4} b B \right ) x^{2}+\left (-\frac {5}{2} a^{4} b A -\frac {1}{2} a^{5} B \right ) x -\frac {a^{5} A}{3}}{x^{3}}+10 A \ln \left (x \right ) a^{2} b^{3}+10 B \ln \left (x \right ) a^{3} b^{2}\) | \(118\) |
norman | \(\frac {\left (\frac {1}{2} b^{5} A +\frac {5}{2} a \,b^{4} B \right ) x^{5}+\left (-\frac {5}{2} a^{4} b A -\frac {1}{2} a^{5} B \right ) x +\left (5 a \,b^{4} A +10 a^{2} b^{3} B \right ) x^{4}+\left (-10 a^{3} b^{2} A -5 a^{4} b B \right ) x^{2}-\frac {a^{5} A}{3}+\frac {b^{5} B \,x^{6}}{3}}{x^{3}}+\left (10 a^{2} b^{3} A +10 a^{3} b^{2} B \right ) \ln \left (x \right )\) | \(120\) |
parallelrisch | \(\frac {2 b^{5} B \,x^{6}+3 A \,b^{5} x^{5}+15 B a \,b^{4} x^{5}+60 A \ln \left (x \right ) x^{3} a^{2} b^{3}+30 a A \,b^{4} x^{4}+60 B \ln \left (x \right ) x^{3} a^{3} b^{2}+60 B \,a^{2} b^{3} x^{4}-60 a^{3} A \,b^{2} x^{2}-30 B \,a^{4} b \,x^{2}-15 a^{4} A b x -3 a^{5} B x -2 a^{5} A}{6 x^{3}}\) | \(128\) |
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Time = 0.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^5 (A+B x)}{x^4} \, dx=\frac {2 \, B b^{5} x^{6} - 2 \, A a^{5} + 3 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 30 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 60 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} \log \left (x\right ) - 30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 3 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{6 \, x^{3}} \]
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Time = 0.39 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b x)^5 (A+B x)}{x^4} \, dx=\frac {B b^{5} x^{3}}{3} + 10 a^{2} b^{2} \left (A b + B a\right ) \log {\left (x \right )} + x^{2} \left (\frac {A b^{5}}{2} + \frac {5 B a b^{4}}{2}\right ) + x \left (5 A a b^{4} + 10 B a^{2} b^{3}\right ) + \frac {- 2 A a^{5} + x^{2} \left (- 60 A a^{3} b^{2} - 30 B a^{4} b\right ) + x \left (- 15 A a^{4} b - 3 B a^{5}\right )}{6 x^{3}} \]
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Time = 0.19 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^5 (A+B x)}{x^4} \, dx=\frac {1}{3} \, B b^{5} x^{3} + \frac {1}{2} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{2} + 5 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x + 10 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} \log \left (x\right ) - \frac {2 \, A a^{5} + 30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 3 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{6 \, x^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^5 (A+B x)}{x^4} \, dx=\frac {1}{3} \, B b^{5} x^{3} + \frac {5}{2} \, B a b^{4} x^{2} + \frac {1}{2} \, A b^{5} x^{2} + 10 \, B a^{2} b^{3} x + 5 \, A a b^{4} x + 10 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, A a^{5} + 30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 3 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{6 \, x^{3}} \]
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Time = 0.05 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^5 (A+B x)}{x^4} \, dx=x^2\,\left (\frac {A\,b^5}{2}+\frac {5\,B\,a\,b^4}{2}\right )-\frac {x\,\left (\frac {B\,a^5}{2}+\frac {5\,A\,b\,a^4}{2}\right )+\frac {A\,a^5}{3}+x^2\,\left (5\,B\,a^4\,b+10\,A\,a^3\,b^2\right )}{x^3}+\ln \left (x\right )\,\left (10\,B\,a^3\,b^2+10\,A\,a^2\,b^3\right )+\frac {B\,b^5\,x^3}{3}+5\,a\,b^3\,x\,\left (A\,b+2\,B\,a\right ) \]
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