Integrand size = 68, antiderivative size = 24 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x-\frac {15 \left (2 x-\frac {1}{e^3 x+\log (\log (2))}\right )}{x} \]
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Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {1608, 27, 1634} \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x-\frac {15 e^3}{\log (\log (2)) \left (e^3 x+\log (\log (2))\right )}+\frac {15}{x \log (\log (2))} \]
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Rule 27
Rule 1608
Rule 1634
Rubi steps \begin{align*} \text {integral}& = \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{x^2 \left (e^6 x^2+2 e^3 x \log (\log (2))+\log ^2(\log (2))\right )} \, dx \\ & = \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{x^2 \left (e^3 x+\log (\log (2))\right )^2} \, dx \\ & = \int \left (1-\frac {15}{x^2 \log (\log (2))}+\frac {15 e^6}{\log (\log (2)) \left (e^3 x+\log (\log (2))\right )^2}\right ) \, dx \\ & = x+\frac {15}{x \log (\log (2))}-\frac {15 e^3}{\log (\log (2)) \left (e^3 x+\log (\log (2))\right )} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x+\frac {15}{x \log (\log (2))}-\frac {15 e^3}{\log (\log (2)) \left (e^3 x+\log (\log (2))\right )} \]
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Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75
method | result | size |
risch | \(x +\frac {15}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \left (2\right )\right )\right )}\) | \(18\) |
norman | \(\frac {15+x^{3} {\mathrm e}^{3}+x^{2} \ln \left (\ln \left (2\right )\right )}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \left (2\right )\right )\right )}\) | \(30\) |
gosper | \(\frac {\left (x^{3} {\mathrm e}^{6}-x \ln \left (\ln \left (2\right )\right )^{2}+15 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-3}}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \left (2\right )\right )\right )}\) | \(40\) |
parallelrisch | \(\frac {\left (x^{3} {\mathrm e}^{6}-x \ln \left (\ln \left (2\right )\right )^{2}+15 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-3}}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \left (2\right )\right )\right )}\) | \(40\) |
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\frac {x^{3} e^{3} + x^{2} \log \left (\log \left (2\right )\right ) + 15}{x^{2} e^{3} + x \log \left (\log \left (2\right )\right )} \]
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Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x + \frac {15}{x^{2} e^{3} + x \log {\left (\log {\left (2 \right )} \right )}} \]
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Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x + \frac {15}{x^{2} e^{3} + x \log \left (\log \left (2\right )\right )} \]
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Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x + \frac {15}{x^{2} e^{3} + x \log \left (\log \left (2\right )\right )} \]
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Time = 9.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x+\frac {15}{x\,\left (\ln \left (\ln \left (2\right )\right )+x\,{\mathrm {e}}^3\right )} \]
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