Integrand size = 24, antiderivative size = 18 \[ \int \frac {e^{e^x} \left (-1+e^x \left (-17+e^2-x\right )\right )}{\log (3)} \, dx=\frac {e^{e^x} \left (-17+e^2-x\right )}{\log (3)} \]
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Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 2326} \[ \int \frac {e^{e^x} \left (-1+e^x \left (-17+e^2-x\right )\right )}{\log (3)} \, dx=-\frac {e^{e^x} \left (x-e^2+17\right )}{\log (3)} \]
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Rule 12
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {\int e^{e^x} \left (-1+e^x \left (-17+e^2-x\right )\right ) \, dx}{\log (3)} \\ & = -\frac {e^{e^x} \left (17-e^2+x\right )}{\log (3)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^x} \left (-1+e^x \left (-17+e^2-x\right )\right )}{\log (3)} \, dx=\frac {e^{e^x} \left (-17+e^2-x\right )}{\log (3)} \]
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Time = 0.33 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89
method | result | size |
risch | \(\frac {\left ({\mathrm e}^{2}-x -17\right ) {\mathrm e}^{{\mathrm e}^{x}}}{\ln \left (3\right )}\) | \(16\) |
norman | \(\frac {\left ({\mathrm e}^{2}-17\right ) {\mathrm e}^{{\mathrm e}^{x}}}{\ln \left (3\right )}-\frac {x \,{\mathrm e}^{{\mathrm e}^{x}}}{\ln \left (3\right )}\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{x}}-x \,{\mathrm e}^{{\mathrm e}^{x}}-17 \,{\mathrm e}^{{\mathrm e}^{x}}}{\ln \left (3\right )}\) | \(24\) |
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none
Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e^{e^x} \left (-1+e^x \left (-17+e^2-x\right )\right )}{\log (3)} \, dx=-\frac {{\left (x - e^{2} + 17\right )} e^{\left (e^{x}\right )}}{\log \left (3\right )} \]
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Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {e^{e^x} \left (-1+e^x \left (-17+e^2-x\right )\right )}{\log (3)} \, dx=\frac {\left (- x - 17 + e^{2}\right ) e^{e^{x}}}{\log {\left (3 \right )}} \]
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\[ \int \frac {e^{e^x} \left (-1+e^x \left (-17+e^2-x\right )\right )}{\log (3)} \, dx=\int { -\frac {{\left ({\left (x - e^{2} + 17\right )} e^{x} + 1\right )} e^{\left (e^{x}\right )}}{\log \left (3\right )} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (16) = 32\).
Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.83 \[ \int \frac {e^{e^x} \left (-1+e^x \left (-17+e^2-x\right )\right )}{\log (3)} \, dx=-\frac {{\left (x e^{\left (x + e^{x}\right )} - e^{\left (x + e^{x} + 2\right )} + 17 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}}{\log \left (3\right )} \]
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Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e^{e^x} \left (-1+e^x \left (-17+e^2-x\right )\right )}{\log (3)} \, dx=-\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x-{\mathrm {e}}^2+17\right )}{\ln \left (3\right )} \]
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