Optimal. Leaf size=21 \[ e^{x+\frac {1}{2} e^4 \left (x+\frac {5 e x}{\log (x)}\right )} \]
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Rubi [F] time = 0.62, 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 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right ) \left (-5 e^5+5 e^5 \log (x)+\left (2+e^4\right ) \log ^2(x)\right )}{2 \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {\exp \left (\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right ) \left (-5 e^5+5 e^5 \log (x)+\left (2+e^4\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=\frac {1}{2} \int \left (2 \exp \left (\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right ) \left (1+\frac {e^4}{2}\right )-\frac {5 \exp \left (5+\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right )}{\log ^2(x)}+\frac {5 \exp \left (5+\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right )}{\log (x)}\right ) \, dx\\ &=-\left (\frac {5}{2} \int \frac {\exp \left (5+\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right )}{\log ^2(x)} \, dx\right )+\frac {5}{2} \int \frac {\exp \left (5+\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right )}{\log (x)} \, dx+\frac {1}{2} \left (2+e^4\right ) \int \exp \left (\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right ) \, dx\\ &=-\left (\frac {5}{2} \int \frac {\exp \left (\frac {5 e^5 x+10 \log (x)+2 \left (1+\frac {e^4}{2}\right ) x \log (x)}{2 \log (x)}\right )}{\log ^2(x)} \, dx\right )+\frac {5}{2} \int \frac {\exp \left (\frac {5 e^5 x+10 \log (x)+2 \left (1+\frac {e^4}{2}\right ) x \log (x)}{2 \log (x)}\right )}{\log (x)} \, dx+\frac {1}{2} \left (2+e^4\right ) \int \exp \left (\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.29, size = 21, normalized size = 1.00 \begin {gather*} e^{\frac {1}{2} x \left (2+e^4+\frac {5 e^5}{\log (x)}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.86, size = 24, normalized size = 1.14 \begin {gather*} e^{\left (\frac {5 \, x e^{5} + {\left (x e^{4} + 2 \, x\right )} \log \relax (x)}{2 \, \log \relax (x)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 17, normalized size = 0.81 \begin {gather*} e^{\left (\frac {1}{2} \, x e^{4} + x + \frac {5 \, x e^{5}}{2 \, \log \relax (x)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 23, normalized size = 1.10
method | result | size |
risch | \({\mathrm e}^{\frac {x \left ({\mathrm e}^{4} \ln \relax (x )+5 \,{\mathrm e}^{5}+2 \ln \relax (x )\right )}{2 \ln \relax (x )}}\) | \(23\) |
norman | \({\mathrm e}^{\frac {\left (x \,{\mathrm e} \,{\mathrm e}^{3}+2 x \right ) \ln \relax (x )+5 x \,{\mathrm e}^{2} {\mathrm e}^{3}}{2 \ln \relax (x )}}\) | \(31\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 17, normalized size = 0.81 \begin {gather*} e^{\left (\frac {1}{2} \, x e^{4} + x + \frac {5 \, x e^{5}}{2 \, \log \relax (x)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.58, size = 19, normalized size = 0.90 \begin {gather*} {\mathrm {e}}^{\frac {x\,{\mathrm {e}}^4}{2}}\,{\mathrm {e}}^x\,{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^5}{2\,\ln \relax (x)}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.38, size = 26, normalized size = 1.24 \begin {gather*} e^{\frac {\frac {5 x e^{5}}{2} + \frac {\left (2 x + x e^{4}\right ) \log {\relax (x )}}{2}}{\log {\relax (x )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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