Optimal. Leaf size=30 \[ -x+e^{\frac {4}{x}+6 e^{5 \left (5-e^x\right )} x} \log (\log (x)) \]
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Rubi [F] time = 5.01, 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 {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} \left (x+\left (-4 \log (x)+e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)\right ) \log (\log (x))\right )}{x^2 \log (x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+\frac {e^{-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \left (e^{5 e^x} x-4 e^{5 e^x} \log (x) \log (\log (x))+6 e^{25} x^2 \log (x) \log (\log (x))-30 e^{25+x} x^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)}\right ) \, dx\\ &=-x+\int \frac {e^{-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \left (e^{5 e^x} x-4 e^{5 e^x} \log (x) \log (\log (x))+6 e^{25} x^2 \log (x) \log (\log (x))-30 e^{25+x} x^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx\\ &=-x+\int \left (-30 e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x))+\frac {e^{-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \left (e^{5 e^x} x-4 e^{5 e^x} \log (x) \log (\log (x))+6 e^{25} x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)}\right ) \, dx\\ &=-x-30 \int e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x)) \, dx+\int \frac {e^{-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \left (e^{5 e^x} x-4 e^{5 e^x} \log (x) \log (\log (x))+6 e^{25} x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx\\ &=-x-30 \int e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x)) \, dx+\int \left (6 e^{25-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x))+\frac {e^{\frac {4}{x}+6 e^{25-5 e^x} x} (x-4 \log (x) \log (\log (x)))}{x^2 \log (x)}\right ) \, dx\\ &=-x+6 \int e^{25-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x)) \, dx-30 \int e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x)) \, dx+\int \frac {e^{\frac {4}{x}+6 e^{25-5 e^x} x} (x-4 \log (x) \log (\log (x)))}{x^2 \log (x)} \, dx\\ &=-x+6 \int e^{25-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x)) \, dx-30 \int e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x)) \, dx+\int \left (\frac {e^{\frac {4}{x}+6 e^{25-5 e^x} x}}{x \log (x)}-\frac {4 e^{\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x))}{x^2}\right ) \, dx\\ &=-x-4 \int \frac {e^{\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x))}{x^2} \, dx+6 \int e^{25-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x)) \, dx-30 \int e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x)) \, dx+\int \frac {e^{\frac {4}{x}+6 e^{25-5 e^x} x}}{x \log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.29, size = 28, normalized size = 0.93 \begin {gather*} -x+e^{\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x)) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 28, normalized size = 0.93 \begin {gather*} e^{\left (\frac {2 \, {\left (3 \, x^{2} e^{\left (-5 \, e^{x} + 25\right )} + 2\right )}}{x}\right )} \log \left (\log \relax (x)\right ) - x \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 {x^{2} \log \relax (x) + {\left (2 \, {\left (3 \, {\left (5 \, x^{3} e^{x} - x^{2}\right )} e^{\left (-5 \, e^{x} + 25\right )} \log \relax (x) + 2 \, \log \relax (x)\right )} \log \left (\log \relax (x)\right ) - x\right )} e^{\left (\frac {2 \, {\left (3 \, x^{2} e^{\left (-5 \, e^{x} + 25\right )} + 2\right )}}{x}\right )}}{x^{2} \log \relax (x)}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 29, normalized size = 0.97
method | result | size |
risch | \({\mathrm e}^{\frac {6 x^{2} {\mathrm e}^{-5 \,{\mathrm e}^{x}+25}+4}{x}} \ln \left (\ln \relax (x )\right )-x\) | \(29\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -x - \int \frac {{\left (6 \, {\left (5 \, x^{3} e^{\left (x + 25\right )} \log \relax (x) - x^{2} e^{25} \log \relax (x)\right )} e^{\frac {4}{x}} \log \left (\log \relax (x)\right ) + {\left (4 \, \log \relax (x) \log \left (\log \relax (x)\right ) - x\right )} e^{\left (\frac {4}{x} + 5 \, e^{x}\right )}\right )} e^{\left (6 \, x e^{\left (-5 \, e^{x} + 25\right )} - 5 \, e^{x}\right )}}{x^{2} \log \relax (x)}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.63, size = 25, normalized size = 0.83 \begin {gather*} \ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{6\,x\,{\mathrm {e}}^{25}\,{\mathrm {e}}^{-5\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{4/x}-x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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