Optimal. Leaf size=32 \[ e^{5 \left (-x+\frac {x}{e^{2+x+25 \left (x+x^2\right )^2}+\log (\log (3 x))}\right )} \]
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Rubi [F] time = 152.41, 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 {\exp \left (\frac {5 x-5 e^{2+x+25 x^2+50 x^3+25 x^4} x-5 x \log (\log (3 x))}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-5+\left (-5 e^{4+2 x+50 x^2+100 x^3+50 x^4}+e^{2+x+25 x^2+50 x^3+25 x^4} \left (5-5 x-250 x^2-750 x^3-500 x^4\right )\right ) \log (3 x)+\left (5-10 e^{2+x+25 x^2+50 x^3+25 x^4}\right ) \log (3 x) \log (\log (3 x))-5 \log (3 x) \log ^2(\log (3 x))\right )}{e^{4+2 x+50 x^2+100 x^3+50 x^4} \log (3 x)+2 e^{2+x+25 x^2+50 x^3+25 x^4} \log (3 x) \log (\log (3 x))+\log (3 x) \log ^2(\log (3 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 {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-5+\left (-5 e^{4+2 x+50 x^2+100 x^3+50 x^4}+e^{2+x+25 x^2+50 x^3+25 x^4} \left (5-5 x-250 x^2-750 x^3-500 x^4\right )\right ) \log (3 x)+\left (5-10 e^{2+x+25 x^2+50 x^3+25 x^4}\right ) \log (3 x) \log (\log (3 x))-5 \log (3 x) \log ^2(\log (3 x))\right )}{\log (3 x) \left (e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )^2} \, dx\\ &=\int \left (-5 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right )-\frac {5 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-1+x+50 x^2+150 x^3+100 x^4\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}+\frac {5 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-1+x \log (3 x) \log (\log (3 x))+50 x^2 \log (3 x) \log (\log (3 x))+150 x^3 \log (3 x) \log (\log (3 x))+100 x^4 \log (3 x) \log (\log (3 x))\right )}{\log (3 x) \left (e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )^2}\right ) \, dx\\ &=-\left (5 \int \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \, dx\right )-5 \int \frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-1+x+50 x^2+150 x^3+100 x^4\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))} \, dx+5 \int \frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-1+x \log (3 x) \log (\log (3 x))+50 x^2 \log (3 x) \log (\log (3 x))+150 x^3 \log (3 x) \log (\log (3 x))+100 x^4 \log (3 x) \log (\log (3 x))\right )}{\log (3 x) \left (e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )^2} \, dx\\ &=-\left (5 \int \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \, dx\right )+5 \int \frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-1+x \left (1+50 x+150 x^2+100 x^3\right ) \log (3 x) \log (\log (3 x))\right )}{\log (3 x) \left (e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )^2} \, dx-5 \int \left (-\frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}+\frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) x}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}+\frac {50 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) x^2}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}+\frac {150 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) x^3}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}+\frac {100 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) x^4}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [B] time = 1.70, size = 107, normalized size = 3.34 \begin {gather*} e^{-5 x-\frac {5 \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}\right ) x}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}} \log ^{\frac {5 x}{\log (\log (3 x))}-\frac {5 x}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}}(3 x) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.84, size = 62, normalized size = 1.94 \begin {gather*} e^{\left (-\frac {5 \, {\left (x e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )} + x \log \left (\log \left (3 \, x\right )\right ) - x\right )}}{e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )} + \log \left (\log \left (3 \, x\right )\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 58, normalized size = 1.81
method | result | size |
risch | \({\mathrm e}^{-\frac {5 x \left (\ln \left (\ln \left (3 x \right )\right )+{\mathrm e}^{25 x^{4}+50 x^{3}+25 x^{2}+x +2}-1\right )}{\ln \left (\ln \left (3 x \right )\right )+{\mathrm e}^{25 x^{4}+50 x^{3}+25 x^{2}+x +2}}}\) | \(58\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.97, size = 120, normalized size = 3.75 \begin {gather*} e^{\left (-\frac {5 \, x e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )}}{e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )} + \log \left (\log \relax (3) + \log \relax (x)\right )} - \frac {5 \, x \log \left (\log \relax (3) + \log \relax (x)\right )}{e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )} + \log \left (\log \relax (3) + \log \relax (x)\right )} + \frac {5 \, x}{e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )} + \log \left (\log \relax (3) + \log \relax (x)\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.07, size = 138, normalized size = 4.31 \begin {gather*} \frac {{\mathrm {e}}^{\frac {5\,x}{\ln \left (\ln \relax (3)+\ln \relax (x)\right )+{\mathrm {e}}^2\,{\mathrm {e}}^{25\,x^2}\,{\mathrm {e}}^{25\,x^4}\,{\mathrm {e}}^{50\,x^3}\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {5\,x\,{\mathrm {e}}^2\,{\mathrm {e}}^{25\,x^2}\,{\mathrm {e}}^{25\,x^4}\,{\mathrm {e}}^{50\,x^3}\,{\mathrm {e}}^x}{\ln \left (\ln \relax (3)+\ln \relax (x)\right )+{\mathrm {e}}^2\,{\mathrm {e}}^{25\,x^2}\,{\mathrm {e}}^{25\,x^4}\,{\mathrm {e}}^{50\,x^3}\,{\mathrm {e}}^x}}}{{\left (\ln \relax (3)+\ln \relax (x)\right )}^{\frac {5\,x}{\ln \left (\ln \relax (3)+\ln \relax (x)\right )+{\mathrm {e}}^2\,{\mathrm {e}}^{25\,x^2}\,{\mathrm {e}}^{25\,x^4}\,{\mathrm {e}}^{50\,x^3}\,{\mathrm {e}}^x}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 35.11, size = 65, normalized size = 2.03 \begin {gather*} e^{\frac {- 5 x e^{25 x^{4} + 50 x^{3} + 25 x^{2} + x + 2} - 5 x \log {\left (\log {\left (3 x \right )} \right )} + 5 x}{e^{25 x^{4} + 50 x^{3} + 25 x^{2} + x + 2} + \log {\left (\log {\left (3 x \right )} \right )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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