Optimal. Leaf size=31 \[ e^{\frac {x^2+\log \left (\log \left (e^{e^x} x\right )\right )}{8 e^{-3/x}+x^2}} \]
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Rubi [F] time = 34.45, 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 {\exp \left (\frac {e^{3/x} x^2+e^{3/x} \log \left (\log \left (e^{e^x} x\right )\right )}{8+e^{3/x} x^2}\right ) \left (8 e^{3/x} x+e^{6/x} x^3+e^x \left (8 e^{3/x} x^2+e^{6/x} x^4\right )+e^{3/x} \left (-24 x^2+16 x^3\right ) \log \left (e^{e^x} x\right )+\left (-24 e^{3/x}-2 e^{6/x} x^3\right ) \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )}{\left (64 x^2+16 e^{3/x} x^4+e^{6/x} x^6\right ) \log \left (e^{e^x} x\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x \left (1+e^x x\right ) \left (8+e^{3/x} x^2\right )-2 \log \left (e^{e^x} x\right ) \left (4 (3-2 x) x^2+\left (12+e^{3/x} x^3\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx\\ &=\int \left (\frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2}-\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (-8 x-e^{3/x} x^3+24 x^2 \log \left (e^{e^x} x\right )-16 x^3 \log \left (e^{e^x} x\right )+24 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )+2 e^{3/x} x^3 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2}\right ) \, dx\\ &=\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx-\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (-8 x-e^{3/x} x^3+24 x^2 \log \left (e^{e^x} x\right )-16 x^3 \log \left (e^{e^x} x\right )+24 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )+2 e^{3/x} x^3 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx\\ &=\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx-\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (-x \left (8+e^{3/x} x^2\right )+2 \log \left (e^{e^x} x\right ) \left (4 (3-2 x) x^2+\left (12+e^{3/x} x^3\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx\\ &=\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx-\int \left (-\frac {8 e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} (-3+2 x) \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2}+\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (-1+2 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )}{x \left (8+e^{3/x} x^2\right )}\right ) \, dx\\ &=8 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} (-3+2 x) \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx+\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx-\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (-1+2 \log \left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\\ &=8 \int \left (-\frac {3 e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2}+\frac {2 e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x \left (8+e^{3/x} x^2\right )^2}\right ) \, dx+\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx-\int \left (-\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{x \left (8+e^{3/x} x^2\right )}+\frac {2 e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )}\right ) \, dx\\ &=-\left (2 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\right )+16 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x \left (8+e^{3/x} x^2\right )^2} \, dx-24 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \left (x^2+\log \left (\log \left (e^{e^x} x\right )\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx+\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx+\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\\ &=-\left (2 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\right )+16 \int \left (\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} x \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{\left (8+e^{3/x} x^2\right )^2}+\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )^2}\right ) \, dx-24 \int \left (\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{\left (8+e^{3/x} x^2\right )^2}+\frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2}\right ) \, dx+\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx+\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\\ &=-\left (2 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\right )+16 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} x \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{\left (8+e^{3/x} x^2\right )^2} \, dx+16 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x \left (8+e^{3/x} x^2\right )^2} \, dx-24 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{\left (8+e^{3/x} x^2\right )^2} \, dx-24 \int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \log \left (\log \left (e^{e^x} x\right )\right )}{x^2 \left (8+e^{3/x} x^2\right )^2} \, dx+\int \frac {e^{\frac {3}{x}+x+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{8+e^{3/x} x^2} \, dx+\int \frac {e^{\frac {3}{x}+\frac {e^{3/x} x^2}{8+e^{3/x} x^2}} \log ^{-1+\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right )}{x \left (8+e^{3/x} x^2\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.42, size = 54, normalized size = 1.74 \begin {gather*} e^{1-\frac {8}{8+e^{3/x} x^2}} \log ^{\frac {e^{3/x}}{8+e^{3/x} x^2}}\left (e^{e^x} x\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.91, size = 41, normalized size = 1.32 \begin {gather*} e^{\left (\frac {x^{2} e^{\frac {3}{x}} + e^{\frac {3}{x}} \log \left (\log \left (x e^{\left (e^{x}\right )}\right )\right )}{x^{2} e^{\frac {3}{x}} + 8}\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 {{\left (x^{3} e^{\frac {6}{x}} + 8 \, {\left (2 \, x^{3} - 3 \, x^{2}\right )} e^{\frac {3}{x}} \log \left (x e^{\left (e^{x}\right )}\right ) - 2 \, {\left (x^{3} e^{\frac {6}{x}} + 12 \, e^{\frac {3}{x}}\right )} \log \left (x e^{\left (e^{x}\right )}\right ) \log \left (\log \left (x e^{\left (e^{x}\right )}\right )\right ) + {\left (x^{4} e^{\frac {6}{x}} + 8 \, x^{2} e^{\frac {3}{x}}\right )} e^{x} + 8 \, x e^{\frac {3}{x}}\right )} e^{\left (\frac {x^{2} e^{\frac {3}{x}} + e^{\frac {3}{x}} \log \left (\log \left (x e^{\left (e^{x}\right )}\right )\right )}{x^{2} e^{\frac {3}{x}} + 8}\right )}}{{\left (x^{6} e^{\frac {6}{x}} + 16 \, x^{4} e^{\frac {3}{x}} + 64 \, x^{2}\right )} \log \left (x e^{\left (e^{x}\right )}\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.33, size = 81, normalized size = 2.61
method | result | size |
risch | \({\mathrm e}^{\frac {{\mathrm e}^{\frac {3}{x}} \left (x^{2}+\ln \left (\ln \relax (x )+\ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{x}}\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{x}}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{x}}\right )+\mathrm {csgn}\left (i {\mathrm e}^{{\mathrm e}^{x}}\right )\right )}{2}\right )\right )}{x^{2} {\mathrm e}^{\frac {3}{x}}+8}}\) | \(81\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.74, size = 46, normalized size = 1.48 \begin {gather*} e^{\left (\frac {e^{\frac {3}{x}} \log \left (e^{x} + \log \relax (x)\right )}{x^{2} e^{\frac {3}{x}} + 8} - \frac {8}{x^{2} e^{\frac {3}{x}} + 8} + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.67, size = 53, normalized size = 1.71 \begin {gather*} {\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{3/x}}{x^2\,{\mathrm {e}}^{3/x}+8}}\,{\left ({\mathrm {e}}^x+\ln \relax (x)\right )}^{\frac {{\mathrm {e}}^{3/x}}{x^2\,{\mathrm {e}}^{3/x}+8}} \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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