Integrand size = 35, antiderivative size = 21 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} e^{x-\left (\frac {5}{2}-x^2\right )^2} \]
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Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {12, 6838} \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} e^{\frac {1}{4} \left (-4 x^4+20 x^2+4 x-25\right )} \]
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Rule 12
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {1}{7} \int e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx \\ & = \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} e^{-\frac {25}{4}+x+5 x^2-x^4} \]
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Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81
method | result | size |
gosper | \(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\) | \(17\) |
derivativedivides | \(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\) | \(17\) |
default | \(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\) | \(17\) |
norman | \(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\) | \(17\) |
risch | \(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\) | \(17\) |
parallelrisch | \(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\) | \(17\) |
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Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} \, e^{\left (-x^{4} + 5 \, x^{2} + x - \frac {25}{4}\right )} \]
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Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {e^{- x^{4} + 5 x^{2} + x - \frac {25}{4}}}{7} \]
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Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} \, e^{\left (-x^{4} + 5 \, x^{2} + x - \frac {25}{4}\right )} \]
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Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} \, e^{\left (-x^{4} + 5 \, x^{2} + x - \frac {25}{4}\right )} \]
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Time = 8.88 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {{\mathrm {e}}^{-\frac {25}{4}}\,{\mathrm {e}}^{-x^4}\,{\mathrm {e}}^{5\,x^2}\,{\mathrm {e}}^x}{7} \]
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