Optimal. Leaf size=24 \[ -1+e^{e^{15} x \left (x+e^{-x} x\right )}+\frac {\log (x)}{x} \]
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Rubi [F] time = 1.69, 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^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1+e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} \left (2+2 e^x-x\right ) x^3-\log (x)}{x^2} \, dx\\ &=\int \left (e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} \left (2+2 e^x-x\right ) x+\frac {1-\log (x)}{x^2}\right ) \, dx\\ &=\int e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} \left (2+2 e^x-x\right ) x \, dx+\int \frac {1-\log (x)}{x^2} \, dx\\ &=\frac {\log (x)}{x}+\int \left (2 e^{15+e^{15-x} \left (1+e^x\right ) x^2} x-e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} (-2+x) x\right ) \, dx\\ &=\frac {\log (x)}{x}+2 \int e^{15+e^{15-x} \left (1+e^x\right ) x^2} x \, dx-\int e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} (-2+x) x \, dx\\ &=\frac {\log (x)}{x}+2 \int e^{15+e^{15-x} \left (1+e^x\right ) x^2} x \, dx-\int \left (-2 e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} x+e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} x^2\right ) \, dx\\ &=\frac {\log (x)}{x}+2 \int e^{15+e^{15-x} \left (1+e^x\right ) x^2} x \, dx+2 \int e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} x \, dx-\int e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} x^2 \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.43, size = 25, normalized size = 1.04 \begin {gather*} e^{e^{15-x} \left (1+e^x\right ) x^2}+\frac {\log (x)}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 32, normalized size = 1.33 \begin {gather*} \frac {x e^{\left ({\left (x^{2} e^{30} + x^{2} e^{\left (x + 30\right )}\right )} e^{\left (-x - 15\right )}\right )} + \log \relax (x)}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 25, normalized size = 1.04 \begin {gather*} \frac {\log \relax (x)}{x} + e^{\left (x^{2} e^{15} + x^{2} e^{\left (-x + 15\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 24, normalized size = 1.00
method | result | size |
risch | \(\frac {\ln \relax (x )}{x}+{\mathrm e}^{x^{2} \left ({\mathrm e}^{x +15}+{\mathrm e}^{15}\right ) {\mathrm e}^{-x}}\) | \(24\) |
default | \({\mathrm e}^{x^{2} \left ({\mathrm e}^{15} {\mathrm e}^{x}+{\mathrm e}^{15}\right ) {\mathrm e}^{-x}}+\frac {\ln \relax (x )}{x}\) | \(29\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 34, normalized size = 1.42 \begin {gather*} \frac {x e^{\left (x^{2} e^{15} + x^{2} e^{\left (-x + 15\right )}\right )} + \log \relax (x) + 1}{x} - \frac {1}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.46, size = 26, normalized size = 1.08 \begin {gather*} \frac {\ln \relax (x)}{x}+{\mathrm {e}}^{x^2\,{\mathrm {e}}^{15}}\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{15}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.53, size = 26, normalized size = 1.08 \begin {gather*} e^{\left (x^{2} e^{15} e^{x} + x^{2} e^{15}\right ) e^{- x}} + \frac {\log {\relax (x )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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