Optimal. Leaf size=24 \[ x-\frac {15 \left (2 x-\frac {1}{e^3 x+\log (\log (2))}\right )}{x} \]
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Rubi [A] time = 0.10, antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 4, number of rules used = 3, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {1594, 27, 1620} \begin {gather*} x-\frac {15 e^3}{\log (\log (2)) \left (e^3 x+\log (\log (2))\right )}+\frac {15}{x \log (\log (2))} \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 1594
Rule 1620
Rubi steps
\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 33, normalized size = 1.38 \begin {gather*} x+\frac {15}{x \log (\log (2))}-\frac {15 e^3}{\log (\log (2)) \left (e^3 x+\log (\log (2))\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.65, size = 30, normalized size = 1.25 \begin {gather*} \frac {x^{3} e^{3} + x^{2} \log \left (\log \relax (2)\right ) + 15}{x^{2} e^{3} + x \log \left (\log \relax (2)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 18, normalized size = 0.75
method | result | size |
risch | \(x +\frac {15}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \relax (2)\right )\right )}\) | \(18\) |
norman | \(\frac {15+x^{3} {\mathrm e}^{3}+x^{2} \ln \left (\ln \relax (2)\right )}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \relax (2)\right )\right )}\) | \(30\) |
gosper | \(\frac {\left (x^{3} {\mathrm e}^{6}-x \ln \left (\ln \relax (2)\right )^{2}+15 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-3}}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \relax (2)\right )\right )}\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 18, normalized size = 0.75 \begin {gather*} x + \frac {15}{x^{2} e^{3} + x \log \left (\log \relax (2)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.57, size = 17, normalized size = 0.71 \begin {gather*} x+\frac {15}{x\,\left (\ln \left (\ln \relax (2)\right )+x\,{\mathrm {e}}^3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.22, size = 15, normalized size = 0.62 \begin {gather*} x + \frac {15}{x^{2} e^{3} + x \log {\left (\log {\relax (2 )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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