Integrand size = 16, antiderivative size = 44 \[ \int \frac {(a+b x)^3 (A+B x)}{x^6} \, dx=-\frac {A (a+b x)^4}{5 a x^5}+\frac {(A b-5 a B) (a+b x)^4}{20 a^2 x^4} \]
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Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {79, 37} \[ \int \frac {(a+b x)^3 (A+B x)}{x^6} \, dx=\frac {(a+b x)^4 (A b-5 a B)}{20 a^2 x^4}-\frac {A (a+b x)^4}{5 a x^5} \]
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Rule 37
Rule 79
Rubi steps \begin{align*} \text {integral}& = -\frac {A (a+b x)^4}{5 a x^5}+\frac {(-A b+5 a B) \int \frac {(a+b x)^3}{x^5} \, dx}{5 a} \\ & = -\frac {A (a+b x)^4}{5 a x^5}+\frac {(A b-5 a B) (a+b x)^4}{20 a^2 x^4} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b x)^3 (A+B x)}{x^6} \, dx=-\frac {10 b^3 x^3 (A+2 B x)+10 a b^2 x^2 (2 A+3 B x)+5 a^2 b x (3 A+4 B x)+a^3 (4 A+5 B x)}{20 x^5} \]
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Time = 0.40 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.50
method | result | size |
default | \(-\frac {a b \left (A b +B a \right )}{x^{3}}-\frac {b^{3} B}{x}-\frac {b^{2} \left (A b +3 B a \right )}{2 x^{2}}-\frac {a^{2} \left (3 A b +B a \right )}{4 x^{4}}-\frac {a^{3} A}{5 x^{5}}\) | \(66\) |
norman | \(\frac {-b^{3} B \,x^{4}+\left (-\frac {1}{2} b^{3} A -\frac {3}{2} a \,b^{2} B \right ) x^{3}+\left (-a \,b^{2} A -a^{2} b B \right ) x^{2}+\left (-\frac {3}{4} a^{2} b A -\frac {1}{4} a^{3} B \right ) x -\frac {a^{3} A}{5}}{x^{5}}\) | \(74\) |
risch | \(\frac {-b^{3} B \,x^{4}+\left (-\frac {1}{2} b^{3} A -\frac {3}{2} a \,b^{2} B \right ) x^{3}+\left (-a \,b^{2} A -a^{2} b B \right ) x^{2}+\left (-\frac {3}{4} a^{2} b A -\frac {1}{4} a^{3} B \right ) x -\frac {a^{3} A}{5}}{x^{5}}\) | \(74\) |
gosper | \(-\frac {20 b^{3} B \,x^{4}+10 A \,b^{3} x^{3}+30 B a \,b^{2} x^{3}+20 a A \,b^{2} x^{2}+20 B \,a^{2} b \,x^{2}+15 a^{2} A b x +5 a^{3} B x +4 a^{3} A}{20 x^{5}}\) | \(76\) |
parallelrisch | \(-\frac {20 b^{3} B \,x^{4}+10 A \,b^{3} x^{3}+30 B a \,b^{2} x^{3}+20 a A \,b^{2} x^{2}+20 B \,a^{2} b \,x^{2}+15 a^{2} A b x +5 a^{3} B x +4 a^{3} A}{20 x^{5}}\) | \(76\) |
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Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.66 \[ \int \frac {(a+b x)^3 (A+B x)}{x^6} \, dx=-\frac {20 \, B b^{3} x^{4} + 4 \, A a^{3} + 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 20 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{20 \, x^{5}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (37) = 74\).
Time = 0.70 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.86 \[ \int \frac {(a+b x)^3 (A+B x)}{x^6} \, dx=\frac {- 4 A a^{3} - 20 B b^{3} x^{4} + x^{3} \left (- 10 A b^{3} - 30 B a b^{2}\right ) + x^{2} \left (- 20 A a b^{2} - 20 B a^{2} b\right ) + x \left (- 15 A a^{2} b - 5 B a^{3}\right )}{20 x^{5}} \]
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Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.66 \[ \int \frac {(a+b x)^3 (A+B x)}{x^6} \, dx=-\frac {20 \, B b^{3} x^{4} + 4 \, A a^{3} + 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 20 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{20 \, x^{5}} \]
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Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.70 \[ \int \frac {(a+b x)^3 (A+B x)}{x^6} \, dx=-\frac {20 \, B b^{3} x^{4} + 30 \, B a b^{2} x^{3} + 10 \, A b^{3} x^{3} + 20 \, B a^{2} b x^{2} + 20 \, A a b^{2} x^{2} + 5 \, B a^{3} x + 15 \, A a^{2} b x + 4 \, A a^{3}}{20 \, x^{5}} \]
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Time = 0.33 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b x)^3 (A+B x)}{x^6} \, dx=-\frac {x^2\,\left (B\,a^2\,b+A\,a\,b^2\right )+x\,\left (\frac {B\,a^3}{4}+\frac {3\,A\,b\,a^2}{4}\right )+\frac {A\,a^3}{5}+x^3\,\left (\frac {A\,b^3}{2}+\frac {3\,B\,a\,b^2}{2}\right )+B\,b^3\,x^4}{x^5} \]
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