Integrand size = 66, antiderivative size = 23 \[ \int \frac {e^{\frac {x}{\log (x)}} \log \left (\frac {1}{2} \left (4-e^5+2 e^{\frac {x}{\log (x)}}\right )\right ) (-4+4 \log (x))}{2 e^{\frac {x}{\log (x)}} \log ^2(x)+\left (4-e^5\right ) \log ^2(x)} \, dx=\log ^2\left (e^{\frac {x}{\log (x)}}+\frac {1}{2} \left (4-e^5\right )\right ) \]
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Time = 0.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 1, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6873, 12, 6816, 6818} \[ \int \frac {e^{\frac {x}{\log (x)}} \log \left (\frac {1}{2} \left (4-e^5+2 e^{\frac {x}{\log (x)}}\right )\right ) (-4+4 \log (x))}{2 e^{\frac {x}{\log (x)}} \log ^2(x)+\left (4-e^5\right ) \log ^2(x)} \, dx=\log ^2\left (\frac {1}{2} \left (2 e^{\frac {x}{\log (x)}}+4-e^5\right )\right ) \]
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Rule 12
Rule 6816
Rule 6818
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \log ^2\left (\frac {1}{2} \left (4-e^5+2 e^{\frac {x}{\log (x)}}\right )\right ) \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {x}{\log (x)}} \log \left (\frac {1}{2} \left (4-e^5+2 e^{\frac {x}{\log (x)}}\right )\right ) (-4+4 \log (x))}{2 e^{\frac {x}{\log (x)}} \log ^2(x)+\left (4-e^5\right ) \log ^2(x)} \, dx=\log ^2\left (2-\frac {e^5}{2}+e^{\frac {x}{\log (x)}}\right ) \]
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Time = 8.39 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\ln \left ({\mathrm e}^{\frac {x}{\ln \left (x \right )}}+2-\frac {{\mathrm e}^{5}}{2}\right )^{2}\) | \(17\) |
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Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {x}{\log (x)}} \log \left (\frac {1}{2} \left (4-e^5+2 e^{\frac {x}{\log (x)}}\right )\right ) (-4+4 \log (x))}{2 e^{\frac {x}{\log (x)}} \log ^2(x)+\left (4-e^5\right ) \log ^2(x)} \, dx=\log \left (-\frac {1}{2} \, e^{5} + e^{\frac {x}{\log \left (x\right )}} + 2\right )^{2} \]
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Time = 0.72 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\frac {x}{\log (x)}} \log \left (\frac {1}{2} \left (4-e^5+2 e^{\frac {x}{\log (x)}}\right )\right ) (-4+4 \log (x))}{2 e^{\frac {x}{\log (x)}} \log ^2(x)+\left (4-e^5\right ) \log ^2(x)} \, dx=\log {\left (e^{\frac {x}{\log {\left (x \right )}}} - \frac {e^{5}}{2} + 2 \right )}^{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (16) = 32\).
Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30 \[ \int \frac {e^{\frac {x}{\log (x)}} \log \left (\frac {1}{2} \left (4-e^5+2 e^{\frac {x}{\log (x)}}\right )\right ) (-4+4 \log (x))}{2 e^{\frac {x}{\log (x)}} \log ^2(x)+\left (4-e^5\right ) \log ^2(x)} \, dx=2 \, \log \left (-\frac {1}{2} \, e^{5} + e^{\frac {x}{\log \left (x\right )}} + 2\right ) \log \left (-e^{5} + 2 \, e^{\frac {x}{\log \left (x\right )}} + 4\right ) - \log \left (-e^{5} + 2 \, e^{\frac {x}{\log \left (x\right )}} + 4\right )^{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (16) = 32\).
Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {x}{\log (x)}} \log \left (\frac {1}{2} \left (4-e^5+2 e^{\frac {x}{\log (x)}}\right )\right ) (-4+4 \log (x))}{2 e^{\frac {x}{\log (x)}} \log ^2(x)+\left (4-e^5\right ) \log ^2(x)} \, dx=-2 \, \log \left (2\right ) \log \left (-e^{5} + 2 \, e^{\frac {x}{\log \left (x\right )}} + 4\right ) + \log \left (-e^{5} + 2 \, e^{\frac {x}{\log \left (x\right )}} + 4\right )^{2} \]
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Time = 10.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {x}{\log (x)}} \log \left (\frac {1}{2} \left (4-e^5+2 e^{\frac {x}{\log (x)}}\right )\right ) (-4+4 \log (x))}{2 e^{\frac {x}{\log (x)}} \log ^2(x)+\left (4-e^5\right ) \log ^2(x)} \, dx={\ln \left ({\mathrm {e}}^{\frac {x}{\ln \left (x\right )}}-\frac {{\mathrm {e}}^5}{2}+2\right )}^2 \]
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