Integrand size = 36, antiderivative size = 17 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\frac {16}{9 x^2 \left (-\frac {5}{2}+\log (3)\right )}} \]
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Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 12, 6838} \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{-\frac {32}{9 \left (5 x^2-2 x^2 \log (3)\right )}} \]
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Rule 6
Rule 12
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{x^3 (-45+18 \log (3))} \, dx \\ & = \frac {64 \int \frac {e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{x^3} \, dx}{9 (5-\log (9))} \\ & = e^{-\frac {32}{9 \left (5 x^2-2 x^2 \log (3)\right )}} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\frac {32}{9 x^2 (-5+\log (9))}} \]
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Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88
method | result | size |
gosper | \({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}\) | \(15\) |
derivativedivides | \({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}\) | \(15\) |
default | \({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}\) | \(15\) |
risch | \({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}\) | \(15\) |
norman | \({\mathrm e}^{\frac {32}{18 x^{2} \ln \left (3\right )-45 x^{2}}}\) | \(19\) |
parallelrisch | \(\frac {18 \ln \left (3\right ) {\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}-45 \,{\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}}{18 \ln \left (3\right )-45}\) | \(46\) |
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Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\left (\frac {32}{9 \, {\left (2 \, x^{2} \log \left (3\right ) - 5 \, x^{2}\right )}}\right )} \]
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Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\frac {32}{- 45 x^{2} + 18 x^{2} \log {\left (3 \right )}}} \]
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Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\left (\frac {32}{9 \, x^{2} {\left (2 \, \log \left (3\right ) - 5\right )}}\right )} \]
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\left (\frac {32}{9 \, {\left (2 \, x^{2} \log \left (3\right ) - 5 \, x^{2}\right )}}\right )} \]
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Time = 9.34 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx={\mathrm {e}}^{\frac {32}{9\,x^2\,\left (2\,\ln \left (3\right )-5\right )}} \]
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