Optimal. Leaf size=17 \[ \frac {e^{\frac {4}{x^4}}}{\log (\log (6+2 x))} \]
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Rubi [F] time = 12.18, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-e^{\frac {4}{x^4}} x^5+e^{\frac {4}{x^4}} (-48-16 x) \log (6+2 x) \log (\log (6+2 x))}{\left (3 x^5+x^6\right ) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^{\frac {4}{x^4}} x^5+e^{\frac {4}{x^4}} (-48-16 x) \log (6+2 x) \log (\log (6+2 x))}{x^5 (3+x) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx\\ &=\int \left (\frac {e^{\frac {4}{x^4}} \left (-x^5-48 \log (2 (3+x)) \log (\log (2 (3+x)))-16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{3 x^5 \log (6+2 x) \log ^2(\log (6+2 x))}+\frac {e^{\frac {4}{x^4}} \left (-x^5-48 \log (2 (3+x)) \log (\log (2 (3+x)))-16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{27 x^3 \log (6+2 x) \log ^2(\log (6+2 x))}+\frac {e^{\frac {4}{x^4}} \left (-x^5-48 \log (2 (3+x)) \log (\log (2 (3+x)))-16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{243 x \log (6+2 x) \log ^2(\log (6+2 x))}+\frac {e^{\frac {4}{x^4}} \left (x^5+48 \log (2 (3+x)) \log (\log (2 (3+x)))+16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{9 x^4 \log (6+2 x) \log ^2(\log (6+2 x))}+\frac {e^{\frac {4}{x^4}} \left (x^5+48 \log (2 (3+x)) \log (\log (2 (3+x)))+16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{81 x^2 \log (6+2 x) \log ^2(\log (6+2 x))}+\frac {e^{\frac {4}{x^4}} \left (x^5+48 \log (2 (3+x)) \log (\log (2 (3+x)))+16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{243 (3+x) \log (6+2 x) \log ^2(\log (6+2 x))}\right ) \, dx\\ &=\frac {1}{243} \int \frac {e^{\frac {4}{x^4}} \left (-x^5-48 \log (2 (3+x)) \log (\log (2 (3+x)))-16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{x \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{243} \int \frac {e^{\frac {4}{x^4}} \left (x^5+48 \log (2 (3+x)) \log (\log (2 (3+x)))+16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{(3+x) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{81} \int \frac {e^{\frac {4}{x^4}} \left (x^5+48 \log (2 (3+x)) \log (\log (2 (3+x)))+16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{x^2 \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{27} \int \frac {e^{\frac {4}{x^4}} \left (-x^5-48 \log (2 (3+x)) \log (\log (2 (3+x)))-16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{x^3 \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{9} \int \frac {e^{\frac {4}{x^4}} \left (x^5+48 \log (2 (3+x)) \log (\log (2 (3+x)))+16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{x^4 \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{3} \int \frac {e^{\frac {4}{x^4}} \left (-x^5-48 \log (2 (3+x)) \log (\log (2 (3+x)))-16 x \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{x^5 \log (6+2 x) \log ^2(\log (6+2 x))} \, dx\\ &=\frac {1}{243} \int \frac {e^{\frac {4}{x^4}} \left (-x^5-16 (3+x) \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{x \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{243} \int \frac {e^{\frac {4}{x^4}} \left (x^5+16 (3+x) \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{(3+x) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{81} \int \frac {e^{\frac {4}{x^4}} \left (x^5+16 (3+x) \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{x^2 \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{27} \int \frac {e^{\frac {4}{x^4}} \left (-x^5-16 (3+x) \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{x^3 \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{9} \int \frac {e^{\frac {4}{x^4}} \left (x^5+16 (3+x) \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{x^4 \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{3} \int \frac {e^{\frac {4}{x^4}} \left (-x^5-16 (3+x) \log (2 (3+x)) \log (\log (2 (3+x)))\right )}{x^5 \log (6+2 x) \log ^2(\log (6+2 x))} \, dx\\ &=\frac {1}{243} \int \left (-\frac {e^{\frac {4}{x^4}} x^4}{\log (6+2 x) \log ^2(\log (6+2 x))}-\frac {16 e^{\frac {4}{x^4}}}{\log (\log (6+2 x))}-\frac {48 e^{\frac {4}{x^4}}}{x \log (\log (6+2 x))}\right ) \, dx+\frac {1}{243} \int \left (\frac {e^{\frac {4}{x^4}} x^5}{(3+x) \log (6+2 x) \log ^2(\log (6+2 x))}+\frac {48 e^{\frac {4}{x^4}}}{(3+x) \log (\log (6+2 x))}+\frac {16 e^{\frac {4}{x^4}} x}{(3+x) \log (\log (6+2 x))}\right ) \, dx+\frac {1}{81} \int \left (\frac {e^{\frac {4}{x^4}} x^3}{\log (6+2 x) \log ^2(\log (6+2 x))}+\frac {48 e^{\frac {4}{x^4}}}{x^2 \log (\log (6+2 x))}+\frac {16 e^{\frac {4}{x^4}}}{x \log (\log (6+2 x))}\right ) \, dx+\frac {1}{27} \int \left (-\frac {e^{\frac {4}{x^4}} x^2}{\log (6+2 x) \log ^2(\log (6+2 x))}-\frac {48 e^{\frac {4}{x^4}}}{x^3 \log (\log (6+2 x))}-\frac {16 e^{\frac {4}{x^4}}}{x^2 \log (\log (6+2 x))}\right ) \, dx+\frac {1}{9} \int \left (\frac {e^{\frac {4}{x^4}} x}{\log (6+2 x) \log ^2(\log (6+2 x))}+\frac {48 e^{\frac {4}{x^4}}}{x^4 \log (\log (6+2 x))}+\frac {16 e^{\frac {4}{x^4}}}{x^3 \log (\log (6+2 x))}\right ) \, dx+\frac {1}{3} \int \left (-\frac {e^{\frac {4}{x^4}}}{\log (6+2 x) \log ^2(\log (6+2 x))}-\frac {48 e^{\frac {4}{x^4}}}{x^5 \log (\log (6+2 x))}-\frac {16 e^{\frac {4}{x^4}}}{x^4 \log (\log (6+2 x))}\right ) \, dx\\ &=-\left (\frac {1}{243} \int \frac {e^{\frac {4}{x^4}} x^4}{\log (6+2 x) \log ^2(\log (6+2 x))} \, dx\right )+\frac {1}{243} \int \frac {e^{\frac {4}{x^4}} x^5}{(3+x) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{81} \int \frac {e^{\frac {4}{x^4}} x^3}{\log (6+2 x) \log ^2(\log (6+2 x))} \, dx-\frac {1}{27} \int \frac {e^{\frac {4}{x^4}} x^2}{\log (6+2 x) \log ^2(\log (6+2 x))} \, dx-\frac {16}{243} \int \frac {e^{\frac {4}{x^4}}}{\log (\log (6+2 x))} \, dx+\frac {16}{243} \int \frac {e^{\frac {4}{x^4}} x}{(3+x) \log (\log (6+2 x))} \, dx+\frac {1}{9} \int \frac {e^{\frac {4}{x^4}} x}{\log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {16}{81} \int \frac {e^{\frac {4}{x^4}}}{(3+x) \log (\log (6+2 x))} \, dx-\frac {1}{3} \int \frac {e^{\frac {4}{x^4}}}{\log (6+2 x) \log ^2(\log (6+2 x))} \, dx-16 \int \frac {e^{\frac {4}{x^4}}}{x^5 \log (\log (6+2 x))} \, dx\\ &=\frac {1}{243} \int \left (\frac {81 e^{\frac {4}{x^4}}}{\log (6+2 x) \log ^2(\log (6+2 x))}+\frac {243 e^{\frac {4}{x^4}}}{(-3-x) \log (6+2 x) \log ^2(\log (6+2 x))}-\frac {27 e^{\frac {4}{x^4}} x}{\log (6+2 x) \log ^2(\log (6+2 x))}+\frac {9 e^{\frac {4}{x^4}} x^2}{\log (6+2 x) \log ^2(\log (6+2 x))}-\frac {3 e^{\frac {4}{x^4}} x^3}{\log (6+2 x) \log ^2(\log (6+2 x))}+\frac {e^{\frac {4}{x^4}} x^4}{\log (6+2 x) \log ^2(\log (6+2 x))}\right ) \, dx-\frac {1}{243} \int \frac {e^{\frac {4}{x^4}} x^4}{\log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {1}{81} \int \frac {e^{\frac {4}{x^4}} x^3}{\log (6+2 x) \log ^2(\log (6+2 x))} \, dx-\frac {1}{27} \int \frac {e^{\frac {4}{x^4}} x^2}{\log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {16}{243} \int \left (\frac {e^{\frac {4}{x^4}}}{\log (\log (6+2 x))}+\frac {3 e^{\frac {4}{x^4}}}{(-3-x) \log (\log (6+2 x))}\right ) \, dx-\frac {16}{243} \int \frac {e^{\frac {4}{x^4}}}{\log (\log (6+2 x))} \, dx+\frac {1}{9} \int \frac {e^{\frac {4}{x^4}} x}{\log (6+2 x) \log ^2(\log (6+2 x))} \, dx+\frac {16}{81} \int \frac {e^{\frac {4}{x^4}}}{(3+x) \log (\log (6+2 x))} \, dx-\frac {1}{3} \int \frac {e^{\frac {4}{x^4}}}{\log (6+2 x) \log ^2(\log (6+2 x))} \, dx-16 \int \frac {e^{\frac {4}{x^4}}}{x^5 \log (\log (6+2 x))} \, dx\\ &=\frac {16}{81} \int \frac {e^{\frac {4}{x^4}}}{(-3-x) \log (\log (6+2 x))} \, dx+\frac {16}{81} \int \frac {e^{\frac {4}{x^4}}}{(3+x) \log (\log (6+2 x))} \, dx-16 \int \frac {e^{\frac {4}{x^4}}}{x^5 \log (\log (6+2 x))} \, dx+\int \frac {e^{\frac {4}{x^4}}}{(-3-x) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.03, size = 17, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {4}{x^4}}}{\log (\log (2 (3+x)))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 16, normalized size = 0.94 \begin {gather*} \frac {e^{\left (\frac {4}{x^{4}}\right )}}{\log \left (\log \left (2 \, x + 6\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 16, normalized size = 0.94 \begin {gather*} \frac {e^{\left (\frac {4}{x^{4}}\right )}}{\log \left (\log \left (2 \, x + 6\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.45, size = 17, normalized size = 1.00
method | result | size |
risch | \(\frac {{\mathrm e}^{\frac {4}{x^{4}}}}{\ln \left (\ln \left (2 x +6\right )\right )}\) | \(17\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.57, size = 17, normalized size = 1.00 \begin {gather*} \frac {e^{\left (\frac {4}{x^{4}}\right )}}{\log \left (\log \relax (2) + \log \left (x + 3\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int -\frac {x^5\,{\mathrm {e}}^{\frac {4}{x^4}}+\ln \left (\ln \left (2\,x+6\right )\right )\,{\mathrm {e}}^{\frac {4}{x^4}}\,\ln \left (2\,x+6\right )\,\left (16\,x+48\right )}{{\ln \left (\ln \left (2\,x+6\right )\right )}^2\,\ln \left (2\,x+6\right )\,\left (x^6+3\,x^5\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.53, size = 14, normalized size = 0.82 \begin {gather*} \frac {e^{\frac {4}{x^{4}}}}{\log {\left (\log {\left (2 x + 6 \right )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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