Integrand size = 29, antiderivative size = 17 \[ \int \frac {e^{e^{e^3+x}} \left (2 x+e^{e^3+x} x^2\right )}{\log (5)} \, dx=\frac {e^{e^{e^3+x}} x^2}{\log (5)} \]
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Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {12, 2326} \[ \int \frac {e^{e^{e^3+x}} \left (2 x+e^{e^3+x} x^2\right )}{\log (5)} \, dx=\frac {e^{e^{x+e^3}} x^2}{\log (5)} \]
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Rule 12
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {\int e^{e^{e^3+x}} \left (2 x+e^{e^3+x} x^2\right ) \, dx}{\log (5)} \\ & = \frac {e^{e^{e^3+x}} x^2}{\log (5)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{e^3+x}} \left (2 x+e^{e^3+x} x^2\right )}{\log (5)} \, dx=\frac {e^{e^{e^3+x}} x^2}{\log (5)} \]
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Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}+x}}}{\ln \left (5\right )}\) | \(15\) |
risch | \(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}+x}}}{\ln \left (5\right )}\) | \(15\) |
parallelrisch | \(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}+x}}}{\ln \left (5\right )}\) | \(15\) |
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none
Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{e^{e^3+x}} \left (2 x+e^{e^3+x} x^2\right )}{\log (5)} \, dx=\frac {x^{2} e^{\left (e^{\left (x + e^{3}\right )}\right )}}{\log \left (5\right )} \]
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Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{e^{e^3+x}} \left (2 x+e^{e^3+x} x^2\right )}{\log (5)} \, dx=\frac {x^{2} e^{e^{x + e^{3}}}}{\log {\left (5 \right )}} \]
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none
Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{e^{e^3+x}} \left (2 x+e^{e^3+x} x^2\right )}{\log (5)} \, dx=\frac {x^{2} e^{\left (e^{\left (x + e^{3}\right )}\right )}}{\log \left (5\right )} \]
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none
Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{e^{e^3+x}} \left (2 x+e^{e^3+x} x^2\right )}{\log (5)} \, dx=\frac {x^{2} e^{\left (e^{\left (x + e^{3}\right )}\right )}}{\log \left (5\right )} \]
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Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^{e^3+x}} \left (2 x+e^{e^3+x} x^2\right )}{\log (5)} \, dx=\frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^3}\,{\mathrm {e}}^x}}{\ln \left (5\right )} \]
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