Optimal. Leaf size=22 \[ 2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \]
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Rubi [F] time = 4.98, 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 {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \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} &=\int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log ^2(x)} \, dx\\ &=\int \left (\frac {15\ 2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2 \log (2)}{e^5 \left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)}+\frac {5\ 2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \log (2) \left (-e^5+e^5 \log (x)-2 x^2 \log (x)\right )}{e^5 \left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)}\right ) \, dx\\ &=\frac {(5 \log (2)) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \left (-e^5+e^5 \log (x)-2 x^2 \log (x)\right )}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)} \, dx}{e^5}+\frac {(15 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)} \, dx}{e^5}\\ &=\frac {(5 \log (2)) \int \left (-\frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} e^5}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)}+\frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} e^5}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}-\frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}\right ) \, dx}{e^5}+\frac {(15 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)} \, dx}{e^5}\\ &=-\left ((5 \log (2)) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}}}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)} \, dx\right )+(5 \log (2)) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}}}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)} \, dx-\frac {(5 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)} \, dx}{e^5}+\frac {(15 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)} \, dx}{e^5}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 30, normalized size = 1.36 \begin {gather*} \frac {5\ 2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \log (2)}{\log (32)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 20, normalized size = 0.91 \begin {gather*} 2^{\frac {5 \, x}{{\left (e^{\left (x^{2} e^{\left (-5\right )}\right )} + 3\right )} \log \relax (x)}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 22, normalized size = 1.00 \begin {gather*} 2^{\frac {5 \, x}{e^{\left (x^{2} e^{\left (-5\right )}\right )} \log \relax (x) + 3 \, \log \relax (x)}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 23, normalized size = 1.05
method | result | size |
risch | \(2^{\frac {5 x}{\ln \relax (x ) \left ({\mathrm e}^{x^{2} {\mathrm e}^{-5}}+3\right )}}\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 20, normalized size = 0.91 \begin {gather*} 2^{\frac {5 \, x}{{\left (e^{\left (x^{2} e^{\left (-5\right )}\right )} + 3\right )} \log \relax (x)}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.64, size = 23, normalized size = 1.05 \begin {gather*} {\mathrm {e}}^{\frac {5\,x\,\ln \relax (2)}{3\,\ln \relax (x)+{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-5}}\,\ln \relax (x)}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.13, size = 26, normalized size = 1.18 \begin {gather*} e^{\frac {10 x \log {\relax (2 )}}{2 e^{\frac {x^{2}}{e^{5}}} \log {\relax (x )} + 6 \log {\relax (x )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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