Optimal. Leaf size=31 \[ \frac {1}{5} e^{\frac {2 x}{\log (x)}} \left (-e^{-15+x-x \log (2 x)}+x\right )^2 \]
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Rubi [F] time = 6.61, 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 {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{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} &=\frac {1}{5} \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{\log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {2 e^{\frac {2 x}{\log (x)}} x \left (-x+x \log (x)+\log ^2(x)\right )}{\log ^2(x)}-\frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \left (1-\log (x)+\log ^2(x) \log (2 x)\right )}{\log ^2(x)}+\frac {2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x} \left (2 x-2 x \log (x)-\log ^2(x)+x \log ^2(x) \log (2 x)\right )}{\log ^2(x)}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \left (1-\log (x)+\log ^2(x) \log (2 x)\right )}{\log ^2(x)} \, dx\right )+\frac {1}{5} \int \frac {2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x} \left (2 x-2 x \log (x)-\log ^2(x)+x \log ^2(x) \log (2 x)\right )}{\log ^2(x)} \, dx+\frac {2}{5} \int \frac {e^{\frac {2 x}{\log (x)}} x \left (-x+x \log (x)+\log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=\frac {e^{\frac {2 x}{\log (x)}} x (x-x \log (x))}{5 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x)}+\frac {1}{5} \int \left (\frac {2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x} \left (2 x-2 x \log (x)-\log ^2(x)\right )}{\log ^2(x)}+2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x} \log (2 x)\right ) \, dx-\frac {1}{5} \int \left (\frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} (1-\log (x))}{\log ^2(x)}+2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \log (2 x)\right ) \, dx\\ &=\frac {e^{\frac {2 x}{\log (x)}} x (x-x \log (x))}{5 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x)}-\frac {1}{5} \int \frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} (1-\log (x))}{\log ^2(x)} \, dx+\frac {1}{5} \int \frac {2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x} \left (2 x-2 x \log (x)-\log ^2(x)\right )}{\log ^2(x)} \, dx+\frac {1}{5} \int 2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x} \log (2 x) \, dx-\frac {1}{5} \int 2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \log (2 x) \, dx\\ &=\frac {e^{\frac {2 x}{\log (x)}} x (x-x \log (x))}{5 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x)}+\frac {1}{5} \int \left (-2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x}+\frac {2^{2-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x}}{\log ^2(x)}-\frac {2^{2-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x}}{\log (x)}\right ) \, dx-\frac {1}{5} \int \left (\frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x}}{\log ^2(x)}-\frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x}}{\log (x)}\right ) \, dx+\frac {1}{5} \int 2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x} \log (2 x) \, dx-\frac {1}{5} \int 2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \log (2 x) \, dx\\ &=\frac {e^{\frac {2 x}{\log (x)}} x (x-x \log (x))}{5 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x)}-\frac {1}{5} \int 2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x} \, dx+\frac {1}{5} \int \frac {2^{2-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x}}{\log ^2(x)} \, dx-\frac {1}{5} \int \frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x}}{\log ^2(x)} \, dx-\frac {1}{5} \int \frac {2^{2-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x}}{\log (x)} \, dx+\frac {1}{5} \int \frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x}}{\log (x)} \, dx+\frac {1}{5} \int 2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x} \log (2 x) \, dx-\frac {1}{5} \int 2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \log (2 x) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 0.73, size = 110, normalized size = 3.55 \begin {gather*} \frac {1}{5} \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{\log ^2(x)} \, dx \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 47, normalized size = 1.52 \begin {gather*} \frac {1}{5} \, {\left (x^{2} - 2 \, x e^{\left (-x \log \relax (2) - x \log \relax (x) + x - 15\right )} + e^{\left (-2 \, x \log \relax (2) - 2 \, x \log \relax (x) + 2 \, x - 30\right )}\right )} e^{\left (\frac {2 \, x}{\log \relax (x)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.26, size = 83, normalized size = 2.68 \begin {gather*} \frac {1}{5} \, x^{2} e^{\left (\frac {2 \, x}{\log \relax (x)}\right )} - \frac {2}{5} \, x e^{\left (-\frac {x \log \relax (2) \log \relax (x) + x \log \relax (x)^{2} - x \log \relax (x) - 2 \, x + 15 \, \log \relax (x)}{\log \relax (x)}\right )} + \frac {1}{5} \, e^{\left (-\frac {2 \, {\left (x \log \relax (2) \log \relax (x) + x \log \relax (x)^{2} - x \log \relax (x) - x + 15 \, \log \relax (x)\right )}}{\log \relax (x)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.11, size = 75, normalized size = 2.42
method | result | size |
risch | \(\frac {x^{2} {\mathrm e}^{\frac {2 x}{\ln \relax (x )}}}{5}-\frac {2 \left (\frac {1}{2}\right )^{x} x^{-x} x \,{\mathrm e}^{\frac {x \ln \relax (x )-15 \ln \relax (x )+2 x}{\ln \relax (x )}}}{5}+\frac {2^{-2 x} x^{-2 x} {\mathrm e}^{\frac {2 x \ln \relax (x )-30 \ln \relax (x )+2 x}{\ln \relax (x )}}}{5}\) | \(75\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.49, size = 54, normalized size = 1.74 \begin {gather*} {\mathrm {e}}^{\frac {2\,x}{\ln \relax (x)}}\,\left (\frac {x^2}{5}+\frac {{\mathrm {e}}^{2\,x-30}}{5\,2^{2\,x}\,x^{2\,x}}-\frac {2\,x\,{\mathrm {e}}^{x-15}}{5\,2^x\,x^x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 33.25, size = 66, normalized size = 2.13 \begin {gather*} \frac {x^{2} e^{\frac {2 x}{\log {\relax (x )}}}}{5} - \frac {2 x e^{\frac {2 x}{\log {\relax (x )}}} e^{- x \left (\log {\relax (x )} + \log {\relax (2 )}\right ) + x - 15}}{5} + \frac {e^{\frac {2 x}{\log {\relax (x )}}} e^{- 2 x \left (\log {\relax (x )} + \log {\relax (2 )}\right ) + 2 x - 30}}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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