Integrand size = 90, antiderivative size = 20 \[ \int \frac {160 e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx=\frac {64}{\left (-5+e^{\frac {5 e}{4 x \log (3)}}\right )^2} \]
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Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {12, 6820, 6818} \[ \int \frac {160 e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx=\frac {64}{\left (5-e^{\frac {5 e}{x \log (81)}}\right )^2} \]
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Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = 160 \int \frac {e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx \\ & = 160 \int \frac {e^{1+\frac {5 e}{x \log (81)}}}{\left (-5+e^{\frac {5 e}{x \log (81)}}\right )^3 x^2 \log (3)} \, dx \\ & = \frac {160 \int \frac {e^{1+\frac {5 e}{x \log (81)}}}{\left (-5+e^{\frac {5 e}{x \log (81)}}\right )^3 x^2} \, dx}{\log (3)} \\ & = \frac {64}{\left (5-e^{\frac {5 e}{x \log (81)}}\right )^2} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {160 e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx=\frac {64}{\left (-5+e^{\frac {5 e}{x \log (81)}}\right )^2} \]
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Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {64}{\left ({\mathrm e}^{\frac {5 \,{\mathrm e}}{4 x \ln \left (3\right )}}-5\right )^{2}}\) | \(19\) |
norman | \(\frac {64}{\left ({\mathrm e}^{\frac {{\mathrm e}^{\ln \left (5\right )+1}}{4 x \ln \left (3\right )}}-5\right )^{2}}\) | \(22\) |
derivativedivides | \(\frac {64 \,{\mathrm e}^{2} {\mathrm e}^{-2}}{\left ({\mathrm e}^{\frac {{\mathrm e}^{\ln \left (5\right )+1}}{4 x \ln \left (3\right )}}-5\right )^{2}}\) | \(33\) |
default | \(\frac {64 \,{\mathrm e}^{2} {\mathrm e}^{-2}}{\left ({\mathrm e}^{\frac {{\mathrm e}^{\ln \left (5\right )+1}}{4 x \ln \left (3\right )}}-5\right )^{2}}\) | \(33\) |
parallelrisch | \(\frac {64}{{\mathrm e}^{\frac {5 \,{\mathrm e}}{2 x \ln \left (3\right )}}-10 \,{\mathrm e}^{\frac {{\mathrm e}^{\ln \left (5\right )+1}}{4 x \ln \left (3\right )}}+25}\) | \(41\) |
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Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (21) = 42\).
Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 4.25 \[ \int \frac {160 e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx=-\frac {64 \, e^{\left (2 \, \log \left (5\right ) + 2\right )}}{10 \, e^{\left (\frac {4 \, x \log \left (5\right ) \log \left (3\right ) + 4 \, x \log \left (3\right ) + e^{\left (\log \left (5\right ) + 1\right )}}{4 \, x \log \left (3\right )} + \log \left (5\right ) + 1\right )} - e^{\left (\frac {4 \, x \log \left (5\right ) \log \left (3\right ) + 4 \, x \log \left (3\right ) + e^{\left (\log \left (5\right ) + 1\right )}}{2 \, x \log \left (3\right )}\right )} - 25 \, e^{\left (2 \, \log \left (5\right ) + 2\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {160 e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx=\frac {64}{- 10 e^{\frac {5 e}{4 x \log {\left (3 \right )}}} + e^{\frac {5 e}{2 x \log {\left (3 \right )}}} + 25} \]
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Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (21) = 42\).
Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.95 \[ \int \frac {160 e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx=-\frac {64 \, {\left (e^{\left (\frac {5 \, e}{2 \, x \log \left (3\right )}\right )} - 10 \, e^{\left (\frac {5 \, e}{4 \, x \log \left (3\right )}\right )}\right )}}{25 \, {\left (e^{\left (\frac {5 \, e}{2 \, x \log \left (3\right )}\right )} - 10 \, e^{\left (\frac {5 \, e}{4 \, x \log \left (3\right )}\right )} + 25\right )}} \]
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none
Time = 0.42 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.05 \[ \int \frac {160 e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx=\frac {64 \, e}{25 \, e + e^{\left (\frac {5 \, e}{2 \, x \log \left (3\right )} + 1\right )} - 10 \, e^{\left (\frac {5 \, e}{4 \, x \log \left (3\right )} + 1\right )}} \]
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Time = 9.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {160 e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx=\frac {64}{{\left ({\mathrm {e}}^{\frac {5\,\mathrm {e}}{4\,x\,\ln \left (3\right )}}-5\right )}^2} \]
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