Integrand size = 18, antiderivative size = 34 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^5} \, dx=\frac {7}{108 (2+3 x)^4}-\frac {37}{81 (2+3 x)^3}+\frac {5}{27 (2+3 x)^2} \]
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Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^5} \, dx=\frac {5}{27 (3 x+2)^2}-\frac {37}{81 (3 x+2)^3}+\frac {7}{108 (3 x+2)^4} \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {7}{9 (2+3 x)^5}+\frac {37}{9 (2+3 x)^4}-\frac {10}{9 (2+3 x)^3}\right ) \, dx \\ & = \frac {7}{108 (2+3 x)^4}-\frac {37}{81 (2+3 x)^3}+\frac {5}{27 (2+3 x)^2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^5} \, dx=\frac {-35+276 x+540 x^2}{324 (2+3 x)^4} \]
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Time = 0.71 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.59
method | result | size |
gosper | \(\frac {540 x^{2}+276 x -35}{324 \left (2+3 x \right )^{4}}\) | \(20\) |
risch | \(\frac {\frac {5}{3} x^{2}+\frac {23}{27} x -\frac {35}{324}}{\left (2+3 x \right )^{4}}\) | \(20\) |
norman | \(\frac {\frac {25}{8} x^{2}+\frac {3}{2} x +\frac {35}{24} x^{3}+\frac {35}{64} x^{4}}{\left (2+3 x \right )^{4}}\) | \(28\) |
default | \(\frac {7}{108 \left (2+3 x \right )^{4}}-\frac {37}{81 \left (2+3 x \right )^{3}}+\frac {5}{27 \left (2+3 x \right )^{2}}\) | \(29\) |
parallelrisch | \(\frac {105 x^{4}+280 x^{3}+600 x^{2}+288 x}{192 \left (2+3 x \right )^{4}}\) | \(29\) |
meijerg | \(\frac {3 x \left (\frac {27}{8} x^{3}+9 x^{2}+9 x +4\right )}{128 \left (1+\frac {3 x}{2}\right )^{4}}-\frac {x^{2} \left (\frac {9}{4} x^{2}+6 x +6\right )}{384 \left (1+\frac {3 x}{2}\right )^{4}}-\frac {5 x^{3} \left (\frac {3 x}{2}+4\right )}{192 \left (1+\frac {3 x}{2}\right )^{4}}\) | \(66\) |
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Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^5} \, dx=\frac {540 \, x^{2} + 276 \, x - 35}{324 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^5} \, dx=- \frac {- 540 x^{2} - 276 x + 35}{26244 x^{4} + 69984 x^{3} + 69984 x^{2} + 31104 x + 5184} \]
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Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^5} \, dx=\frac {540 \, x^{2} + 276 \, x - 35}{324 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^5} \, dx=\frac {5}{27 \, {\left (3 \, x + 2\right )}^{2}} - \frac {37}{81 \, {\left (3 \, x + 2\right )}^{3}} + \frac {7}{108 \, {\left (3 \, x + 2\right )}^{4}} \]
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Time = 1.42 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^5} \, dx=\frac {5}{27\,{\left (3\,x+2\right )}^2}-\frac {37}{81\,{\left (3\,x+2\right )}^3}+\frac {7}{108\,{\left (3\,x+2\right )}^4} \]
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