Optimal. Leaf size=30 \[ e^{e^{5 \left (-5 \left (2+e^{3 x}\right )+x\right )} x}-\frac {1}{5} \log \left (\frac {100}{x^2}\right ) \]
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Rubi [F] time = 0.74, 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+\exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) \left (5 x+25 x^2-375 e^{3 x} x^2\right )}{5 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 {2+\exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) \left (5 x+25 x^2-375 e^{3 x} x^2\right )}{x} \, dx\\ &=\frac {1}{5} \int \left (\frac {2}{x}-5 \exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) \left (-1-5 x+75 e^{3 x} x\right )\right ) \, dx\\ &=\frac {2 \log (x)}{5}-\int \exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) \left (-1-5 x+75 e^{3 x} x\right ) \, dx\\ &=\frac {2 \log (x)}{5}-\int \left (-\exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right )-5 \exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) x+75 \exp \left (-50-25 e^{3 x}+8 x+e^{-50-25 e^{3 x}+5 x} x\right ) x\right ) \, dx\\ &=\frac {2 \log (x)}{5}+5 \int \exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) x \, dx-75 \int \exp \left (-50-25 e^{3 x}+8 x+e^{-50-25 e^{3 x}+5 x} x\right ) x \, dx+\int \exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.44, size = 25, normalized size = 0.83 \begin {gather*} e^{e^{-50-25 e^{3 x}+5 x} x}+\frac {2 \log (x)}{5} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.56, size = 59, normalized size = 1.97 \begin {gather*} \frac {1}{5} \, {\left (2 \, e^{\left (5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )} \log \relax (x) + 5 \, e^{\left (x e^{\left (5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )} + 5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )}\right )} e^{\left (-5 \, x + 25 \, e^{\left (3 \, x\right )} + 50\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*} \int -\frac {5 \, {\left (75 \, x^{2} e^{\left (3 \, x\right )} - 5 \, x^{2} - x\right )} e^{\left (x e^{\left (5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )} + 5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )} - 2}{5 \, x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 21, normalized size = 0.70
method | result | size |
norman | \({\mathrm e}^{x \,{\mathrm e}^{-25 \,{\mathrm e}^{3 x}+5 x -50}}+\frac {2 \ln \relax (x )}{5}\) | \(21\) |
risch | \({\mathrm e}^{x \,{\mathrm e}^{-25 \,{\mathrm e}^{3 x}+5 x -50}}+\frac {2 \ln \relax (x )}{5}\) | \(21\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.66, size = 20, normalized size = 0.67 \begin {gather*} e^{\left (x e^{\left (5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )}\right )} + \frac {2}{5} \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.59, size = 21, normalized size = 0.70 \begin {gather*} {\mathrm {e}}^{x\,{\mathrm {e}}^{-25\,{\mathrm {e}}^{3\,x}}\,{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{-50}}+\frac {2\,\ln \relax (x)}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.41, size = 22, normalized size = 0.73 \begin {gather*} e^{x e^{5 x - 25 e^{3 x} - 50}} + \frac {2 \log {\relax (x )}}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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