Optimal. Leaf size=26 \[ -e^{e^{\frac {e^{4 e^x+x}}{\left (4-\log \left (x^2\right )\right )^2}}} \]
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Rubi [F] time = 20.18, 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 (4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}\right ) \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right ) \left (-4 \left (1+x+4 e^x x\right )+\left (1+4 e^x\right ) x \log \left (x^2\right )\right )}{x \left (4-\log \left (x^2\right )\right )^3} \, dx\\ &=\int \left (\frac {4 \exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^3}+\frac {4 \exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{x \left (-4+\log \left (x^2\right )\right )^3}-\frac {4 \exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+2 x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2}-\frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right ) \log \left (x^2\right )}{\left (-4+\log \left (x^2\right )\right )^3}\right ) \, dx\\ &=4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^3} \, dx+4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{x \left (-4+\log \left (x^2\right )\right )^3} \, dx-4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+2 x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2} \, dx-\int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right ) \log \left (x^2\right )}{\left (-4+\log \left (x^2\right )\right )^3} \, dx\\ &=4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^3} \, dx+4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{x \left (-4+\log \left (x^2\right )\right )^3} \, dx-4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+2 x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2} \, dx-\int \left (\frac {4 \exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^3}+\frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2}\right ) \, dx\\ &=4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{x \left (-4+\log \left (x^2\right )\right )^3} \, dx-4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+2 x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2} \, dx-\int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.33, size = 24, normalized size = 0.92 \begin {gather*} -e^{e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.72, size = 120, normalized size = 4.62 \begin {gather*} -e^{\left (\frac {4 \, e^{x} \log \left (x^{2}\right )^{2} + {\left (\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16\right )} e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16}\right )} - 32 \, e^{x} \log \left (x^{2}\right ) + e^{\left (x + 4 \, e^{x}\right )} + 64 \, e^{x}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} - \frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} - 4 \, e^{x}\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 (16 \, x e^{\left (2 \, x\right )} + 4 \, {\left (x + 1\right )} e^{x} - {\left (4 \, x e^{\left (2 \, x\right )} + x e^{x}\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} + 4 \, e^{x} + e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16}\right )}\right )}}{x \log \left (x^{2}\right )^{3} - 12 \, x \log \left (x^{2}\right )^{2} + 48 \, x \log \left (x^{2}\right ) - 64 \, x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 1.16, size = 71, normalized size = 2.73
method | result | size |
risch | \(-{\mathrm e}^{{\mathrm e}^{\frac {4 \,{\mathrm e}^{4 \,{\mathrm e}^{x}+x}}{\left (-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+4 \ln \relax (x )-8\right )^{2}}}}\) | \(71\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.07, size = 25, normalized size = 0.96 \begin {gather*} -e^{\left (e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{4 \, {\left (\log \relax (x)^{2} - 4 \, \log \relax (x) + 4\right )}}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.10, size = 28, normalized size = 1.08 \begin {gather*} -{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^{4\,{\mathrm {e}}^x}\,{\mathrm {e}}^x}{{\ln \left (x^2\right )}^2-8\,\ln \left (x^2\right )+16}}} \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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