Integrand size = 16, antiderivative size = 107 \[ \int \frac {(a+b x)^5 (A+B x)}{x^5} \, dx=-\frac {a^5 A}{4 x^4}-\frac {a^4 (5 A b+a B)}{3 x^3}-\frac {5 a^3 b (2 A b+a B)}{2 x^2}-\frac {10 a^2 b^2 (A b+a B)}{x}+b^4 (A b+5 a B) x+\frac {1}{2} b^5 B x^2+5 a b^3 (A b+2 a B) \log (x) \]
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Time = 0.04 (sec) , antiderivative size = 107, 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} \[ \int \frac {(a+b x)^5 (A+B x)}{x^5} \, dx=-\frac {a^5 A}{4 x^4}-\frac {a^4 (a B+5 A b)}{3 x^3}-\frac {5 a^3 b (a B+2 A b)}{2 x^2}-\frac {10 a^2 b^2 (a B+A b)}{x}+b^4 x (5 a B+A b)+5 a b^3 \log (x) (2 a B+A b)+\frac {1}{2} b^5 B x^2 \]
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Rule 77
Rubi steps \begin{align*} \text {integral}& = \int \left (b^4 (A b+5 a B)+\frac {a^5 A}{x^5}+\frac {a^4 (5 A b+a B)}{x^4}+\frac {5 a^3 b (2 A b+a B)}{x^3}+\frac {10 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\right ) \, dx \\ & = -\frac {a^5 A}{4 x^4}-\frac {a^4 (5 A b+a B)}{3 x^3}-\frac {5 a^3 b (2 A b+a B)}{2 x^2}-\frac {10 a^2 b^2 (A b+a B)}{x}+b^4 (A b+5 a B) x+\frac {1}{2} b^5 B x^2+5 a b^3 (A b+2 a B) \log (x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b x)^5 (A+B x)}{x^5} \, dx=-\frac {10 a^2 A b^3}{x}+5 a b^4 B x+\frac {1}{2} b^5 x (2 A+B x)-\frac {5 a^3 b^2 (A+2 B x)}{x^2}-\frac {5 a^4 b (2 A+3 B x)}{6 x^3}-\frac {a^5 (3 A+4 B x)}{12 x^4}+5 a b^3 (A b+2 a B) \log (x) \]
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Time = 0.39 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94
method | result | size |
default | \(\frac {b^{5} B \,x^{2}}{2}+A \,b^{5} x +5 B a \,b^{4} x +5 a \,b^{3} \left (A b +2 B a \right ) \ln \left (x \right )-\frac {a^{4} \left (5 A b +B a \right )}{3 x^{3}}-\frac {10 a^{2} b^{2} \left (A b +B a \right )}{x}-\frac {5 a^{3} b \left (2 A b +B a \right )}{2 x^{2}}-\frac {a^{5} A}{4 x^{4}}\) | \(101\) |
risch | \(\frac {b^{5} B \,x^{2}}{2}+A \,b^{5} x +5 B a \,b^{4} x +\frac {\left (-10 a^{2} b^{3} A -10 a^{3} b^{2} B \right ) x^{3}+\left (-5 a^{3} b^{2} A -\frac {5}{2} a^{4} b B \right ) x^{2}+\left (-\frac {5}{3} a^{4} b A -\frac {1}{3} a^{5} B \right ) x -\frac {a^{5} A}{4}}{x^{4}}+5 A \ln \left (x \right ) a \,b^{4}+10 B \ln \left (x \right ) a^{2} b^{3}\) | \(116\) |
norman | \(\frac {\left (-5 a^{3} b^{2} A -\frac {5}{2} a^{4} b B \right ) x^{2}+\left (-\frac {5}{3} a^{4} b A -\frac {1}{3} a^{5} B \right ) x +\left (b^{5} A +5 a \,b^{4} B \right ) x^{5}+\left (-10 a^{2} b^{3} A -10 a^{3} b^{2} B \right ) x^{3}-\frac {a^{5} A}{4}+\frac {b^{5} B \,x^{6}}{2}}{x^{4}}+\left (5 a \,b^{4} A +10 a^{2} b^{3} B \right ) \ln \left (x \right )\) | \(119\) |
parallelrisch | \(\frac {6 b^{5} B \,x^{6}+60 A \ln \left (x \right ) x^{4} a \,b^{4}+12 A \,b^{5} x^{5}+120 B \ln \left (x \right ) x^{4} a^{2} b^{3}+60 B a \,b^{4} x^{5}-120 a^{2} A \,b^{3} x^{3}-120 B \,a^{3} b^{2} x^{3}-60 a^{3} A \,b^{2} x^{2}-30 B \,a^{4} b \,x^{2}-20 a^{4} A b x -4 a^{5} B x -3 a^{5} A}{12 x^{4}}\) | \(128\) |
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Time = 0.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b x)^5 (A+B x)}{x^5} \, dx=\frac {6 \, B b^{5} x^{6} - 3 \, A a^{5} + 12 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 60 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} \log \left (x\right ) - 120 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} - 30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 4 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{12 \, x^{4}} \]
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Time = 0.68 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x)^5 (A+B x)}{x^5} \, dx=\frac {B b^{5} x^{2}}{2} + 5 a b^{3} \left (A b + 2 B a\right ) \log {\left (x \right )} + x \left (A b^{5} + 5 B a b^{4}\right ) + \frac {- 3 A a^{5} + x^{3} \left (- 120 A a^{2} b^{3} - 120 B a^{3} b^{2}\right ) + x^{2} \left (- 60 A a^{3} b^{2} - 30 B a^{4} b\right ) + x \left (- 20 A a^{4} b - 4 B a^{5}\right )}{12 x^{4}} \]
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Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^5 (A+B x)}{x^5} \, dx=\frac {1}{2} \, B b^{5} x^{2} + {\left (5 \, B a b^{4} + A b^{5}\right )} x + 5 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} \log \left (x\right ) - \frac {3 \, A a^{5} + 120 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 4 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{12 \, x^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^5 (A+B x)}{x^5} \, dx=\frac {1}{2} \, B b^{5} x^{2} + 5 \, B a b^{4} x + A b^{5} x + 5 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} \log \left ({\left | x \right |}\right ) - \frac {3 \, A a^{5} + 120 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 4 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{12 \, x^{4}} \]
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Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^5 (A+B x)}{x^5} \, dx=x\,\left (A\,b^5+5\,B\,a\,b^4\right )+\ln \left (x\right )\,\left (10\,B\,a^2\,b^3+5\,A\,a\,b^4\right )-\frac {x\,\left (\frac {B\,a^5}{3}+\frac {5\,A\,b\,a^4}{3}\right )+\frac {A\,a^5}{4}+x^2\,\left (\frac {5\,B\,a^4\,b}{2}+5\,A\,a^3\,b^2\right )+x^3\,\left (10\,B\,a^3\,b^2+10\,A\,a^2\,b^3\right )}{x^4}+\frac {B\,b^5\,x^2}{2} \]
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