Integrand size = 53, antiderivative size = 23 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {1}{9} e^{3^{e^{-x} x}+x} x^2 (1+x) \]
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Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {12, 2326} \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {e^{3^{e^{-x} x}} \left (x^2-x^4\right )}{9 \left (e^{-x}-e^{-x} x\right )} \]
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Rule 12
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} \int e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx \\ & = \frac {e^{3^{e^{-x} x}} \left (x^2-x^4\right )}{9 \left (e^{-x}-e^{-x} x\right )} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {1}{9} e^{3^{e^{-x} x}+x} x^2 (1+x) \]
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Time = 4.79 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
risch | \(\frac {x^{2} \left (1+x \right ) {\mathrm e}^{x +3^{x \,{\mathrm e}^{-x}}}}{9}\) | \(20\) |
parallelrisch | \(\frac {x^{3} {\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x \ln \left (3\right ) {\mathrm e}^{-x}}}}{9}+\frac {{\mathrm e}^{{\mathrm e}^{x \ln \left (3\right ) {\mathrm e}^{-x}}} {\mathrm e}^{x} x^{2}}{9}\) | \(36\) |
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {1}{9} \, {\left (x^{3} + x^{2}\right )} e^{\left (3^{x e^{\left (-x\right )}} + x\right )} \]
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Time = 0.94 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {\left (x^{3} e^{x} + x^{2} e^{x}\right ) e^{e^{x e^{- x} \log {\left (3 \right )}}}}{9} \]
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Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {1}{9} \, {\left (x^{3} + x^{2}\right )} e^{\left (3^{x e^{\left (-x\right )}} + x\right )} \]
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\[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\int { -\frac {1}{9} \, {\left ({\left (x^{4} - x^{2}\right )} 3^{x e^{\left (-x\right )}} \log \left (3\right ) - {\left (x^{3} + 4 \, x^{2} + 2 \, x\right )} e^{x}\right )} e^{\left (3^{x e^{\left (-x\right )}}\right )} \,d x } \]
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Time = 12.60 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {x^2\,{\mathrm {e}}^{3^{x\,{\mathrm {e}}^{-x}}}\,{\mathrm {e}}^x\,\left (x+1\right )}{9} \]
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