Optimal. Leaf size=28 \[ e^{\frac {e^{-x} (5+x)}{x+\frac {4 \left (2+\log \left (x^2\right )\right )}{x^3}}}+x \]
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Rubi [F] time = 57.52, 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 (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\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^{-x} \left (e^x \left (8+x^4\right )^2+8 e^x \left (8+x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )-e^{\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-80+16 x+8 x^2+5 x^4+5 x^5+x^6+4 \left (-15+x+x^2\right ) \log \left (x^2\right )\right )\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx\\ &=\int \left (1-\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-80+16 x+8 x^2+5 x^4+5 x^5+x^6-60 \log \left (x^2\right )+4 x \log \left (x^2\right )+4 x^2 \log \left (x^2\right )\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=x-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-80+16 x+8 x^2+5 x^4+5 x^5+x^6-60 \log \left (x^2\right )+4 x \log \left (x^2\right )+4 x^2 \log \left (x^2\right )\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx\\ &=x-\int \left (\frac {4 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (10+2 x+5 x^4+x^5\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-15+x+x^2\right )}{8+x^4+4 \log \left (x^2\right )}\right ) \, dx\\ &=x-4 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (10+2 x+5 x^4+x^5\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-15+x+x^2\right )}{8+x^4+4 \log \left (x^2\right )} \, dx\\ &=x-4 \int \left (\frac {10 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {2 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {5 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^6}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^7}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}\right ) \, dx-\int \left (-\frac {15 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{8+x^4+4 \log \left (x^2\right )}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{8+x^4+4 \log \left (x^2\right )}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^4}{8+x^4+4 \log \left (x^2\right )}\right ) \, dx\\ &=x-4 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^7}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-8 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx+15 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{8+x^4+4 \log \left (x^2\right )} \, dx-20 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^6}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-40 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{8+x^4+4 \log \left (x^2\right )} \, dx-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^4}{8+x^4+4 \log \left (x^2\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.30, size = 29, normalized size = 1.04 \begin {gather*} e^{\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}}+x \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.99, size = 32, normalized size = 1.14 \begin {gather*} x + e^{\left (\frac {x^{4} + 5 \, x^{3}}{{\left (x^{4} + 8\right )} e^{x} + 4 \, e^{x} \log \left (x^{2}\right )}\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 {8 \, {\left (x^{4} + 8\right )} e^{x} \log \left (x^{2}\right ) + 16 \, e^{x} \log \left (x^{2}\right )^{2} + {\left (x^{8} + 16 \, x^{4} + 64\right )} e^{x} - {\left (x^{8} + 5 \, x^{7} + 5 \, x^{6} + 8 \, x^{4} + 16 \, x^{3} - 80 \, x^{2} + 4 \, {\left (x^{4} + x^{3} - 15 \, x^{2}\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {x^{4} + 5 \, x^{3}}{{\left (x^{4} + 8\right )} e^{x} + 4 \, e^{x} \log \left (x^{2}\right )}\right )}}{8 \, {\left (x^{4} + 8\right )} e^{x} \log \left (x^{2}\right ) + 16 \, e^{x} \log \left (x^{2}\right )^{2} + {\left (x^{8} + 16 \, x^{4} + 64\right )} e^{x}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.67, size = 75, normalized size = 2.68
method | result | size |
risch | \(x +{\mathrm e}^{\frac {x^{3} \left (5+x \right ) {\mathrm e}^{-x}}{-2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-2 i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+x^{4}+8 \ln \relax (x )+8}}\) | \(75\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.43, size = 116, normalized size = 4.14 \begin {gather*} {\left (x e^{\left (\frac {8 \, \log \relax (x)}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \relax (x)} + \frac {8}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \relax (x)}\right )} + e^{\left (\frac {5 \, x^{3}}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \relax (x)} + e^{\left (-x\right )}\right )}\right )} e^{\left (-\frac {8 \, \log \relax (x)}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \relax (x)} - \frac {8}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \relax (x)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.76, size = 56, normalized size = 2.00 \begin {gather*} x+{\mathrm {e}}^{\frac {x^4}{8\,{\mathrm {e}}^x+x^4\,{\mathrm {e}}^x+4\,\ln \left (x^2\right )\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{\frac {5\,x^3}{8\,{\mathrm {e}}^x+x^4\,{\mathrm {e}}^x+4\,\ln \left (x^2\right )\,{\mathrm {e}}^x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.20, size = 29, normalized size = 1.04 \begin {gather*} x + e^{\frac {x^{4} + 5 x^{3}}{\left (x^{4} + 8\right ) e^{x} + 4 e^{x} \log {\left (x^{2} \right )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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