Integrand size = 18, antiderivative size = 14 \[ \int \frac {1}{2} e^{2 x} (5+2 x+2 \log (2)) \, dx=\frac {1}{2} e^{2 x} (2+x+\log (2)) \]
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Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.86, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 2207, 2225} \[ \int \frac {1}{2} e^{2 x} (5+2 x+2 \log (2)) \, dx=\frac {1}{4} e^{2 x} (2 x+5+\log (4))-\frac {e^{2 x}}{4} \]
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Rule 12
Rule 2207
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int e^{2 x} (5+2 x+2 \log (2)) \, dx \\ & = \frac {1}{4} e^{2 x} (5+2 x+\log (4))-\frac {1}{2} \int e^{2 x} \, dx \\ & = -\frac {e^{2 x}}{4}+\frac {1}{4} e^{2 x} (5+2 x+\log (4)) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {1}{2} e^{2 x} (5+2 x+2 \log (2)) \, dx=\frac {1}{2} e^{2 x} \left (x+\frac {1}{2} (4+\log (4))\right ) \]
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Time = 0.94 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86
method | result | size |
gosper | \(\frac {{\mathrm e}^{2 x} \left (\ln \left (2\right )+2+x \right )}{2}\) | \(12\) |
risch | \(\frac {{\mathrm e}^{2 x} \left (\ln \left (2\right )+2+x \right )}{2}\) | \(12\) |
norman | \(\left (1+\frac {\ln \left (2\right )}{2}\right ) {\mathrm e}^{2 x}+\frac {x \,{\mathrm e}^{2 x}}{2}\) | \(20\) |
derivativedivides | \(\frac {x \,{\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x}+\frac {\ln \left (2\right ) {\mathrm e}^{2 x}}{2}\) | \(21\) |
default | \(\frac {x \,{\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x}+\frac {\ln \left (2\right ) {\mathrm e}^{2 x}}{2}\) | \(21\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x}+\frac {\ln \left (2\right ) {\mathrm e}^{2 x}}{2}\) | \(21\) |
parts | \(\frac {x \,{\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x}+\frac {\ln \left (2\right ) {\mathrm e}^{2 x}}{2}\) | \(21\) |
meijerg | \(-1+\frac {5 \,{\mathrm e}^{2 x}}{4}-\frac {\ln \left (2\right ) \left (1-{\mathrm e}^{2 x}\right )}{2}-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{8}\) | \(32\) |
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Time = 0.24 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {1}{2} e^{2 x} (5+2 x+2 \log (2)) \, dx=\frac {1}{2} \, {\left (x + \log \left (2\right ) + 2\right )} e^{\left (2 \, x\right )} \]
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Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1}{2} e^{2 x} (5+2 x+2 \log (2)) \, dx=\frac {\left (x + \log {\left (2 \right )} + 2\right ) e^{2 x}}{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (11) = 22\).
Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.86 \[ \int \frac {1}{2} e^{2 x} (5+2 x+2 \log (2)) \, dx=\frac {1}{4} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + \frac {1}{2} \, e^{\left (2 \, x\right )} \log \left (2\right ) + \frac {5}{4} \, e^{\left (2 \, x\right )} \]
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Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {1}{2} e^{2 x} (5+2 x+2 \log (2)) \, dx=\frac {1}{2} \, {\left (x + \log \left (2\right ) + 2\right )} e^{\left (2 \, x\right )} \]
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Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {1}{2} e^{2 x} (5+2 x+2 \log (2)) \, dx=\frac {{\mathrm {e}}^{2\,x}\,\left (2\,x+\ln \left (4\right )+4\right )}{4} \]
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