Optimal. Leaf size=28 \[ \log \left (e^{4/x}-e^{-x}+4 x-\frac {x}{5+\log (x)}\right ) \]
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Rubi [F] time = 26.04, 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 {25 x^2+e^x \left (-100 e^{4/x}+96 x^2\right )+\left (10 x^2+e^x \left (-40 e^{4/x}+39 x^2\right )\right ) \log (x)+\left (x^2+e^x \left (-4 e^{4/x}+4 x^2\right )\right ) \log ^2(x)}{-25 x^2+e^x \left (25 e^{4/x} x^2+95 x^3\right )+\left (-10 x^2+e^x \left (10 e^{4/x} x^2+39 x^3\right )\right ) \log (x)+\left (-x^2+e^x \left (e^{4/x} x^2+4 x^3\right )\right ) \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-25 x^2-e^x \left (-100 e^{4/x}+96 x^2\right )-\left (10 x^2+e^x \left (-40 e^{4/x}+39 x^2\right )\right ) \log (x)-\left (x^2+e^x \left (-4 e^{4/x}+4 x^2\right )\right ) \log ^2(x)}{x^2 (5+\log (x)) \left (5-5 e^{\frac {4}{x}+x}-19 e^x x+\log (x)-e^{\frac {4}{x}+x} \log (x)-4 e^x x \log (x)\right )} \, dx\\ &=\int \left (\frac {-100 e^{4/x}+96 x^2-40 e^{4/x} \log (x)+39 x^2 \log (x)-4 e^{4/x} \log ^2(x)+4 x^2 \log ^2(x)}{x^2 (5+\log (x)) \left (5 e^{4/x}+19 x+e^{4/x} \log (x)+4 x \log (x)\right )}+\frac {-100 e^{4/x}+96 x^2+25 e^{4/x} x^2+95 x^3-40 e^{4/x} \log (x)+39 x^2 \log (x)+10 e^{4/x} x^2 \log (x)+39 x^3 \log (x)-4 e^{4/x} \log ^2(x)+4 x^2 \log ^2(x)+e^{4/x} x^2 \log ^2(x)+4 x^3 \log ^2(x)}{x^2 \left (5 e^{4/x}+19 x+e^{4/x} \log (x)+4 x \log (x)\right ) \left (-5+5 e^{\frac {4}{x}+x}+19 e^x x-\log (x)+e^{\frac {4}{x}+x} \log (x)+4 e^x x \log (x)\right )}\right ) \, dx\\ &=\int \frac {-100 e^{4/x}+96 x^2-40 e^{4/x} \log (x)+39 x^2 \log (x)-4 e^{4/x} \log ^2(x)+4 x^2 \log ^2(x)}{x^2 (5+\log (x)) \left (5 e^{4/x}+19 x+e^{4/x} \log (x)+4 x \log (x)\right )} \, dx+\int \frac {-100 e^{4/x}+96 x^2+25 e^{4/x} x^2+95 x^3-40 e^{4/x} \log (x)+39 x^2 \log (x)+10 e^{4/x} x^2 \log (x)+39 x^3 \log (x)-4 e^{4/x} \log ^2(x)+4 x^2 \log ^2(x)+e^{4/x} x^2 \log ^2(x)+4 x^3 \log ^2(x)}{x^2 \left (5 e^{4/x}+19 x+e^{4/x} \log (x)+4 x \log (x)\right ) \left (-5+5 e^{\frac {4}{x}+x}+19 e^x x-\log (x)+e^{\frac {4}{x}+x} \log (x)+4 e^x x \log (x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 54, normalized size = 1.93 \begin {gather*} -x-\log (5+\log (x))+\log \left (5-5 e^{\frac {4}{x}+x}-19 e^x x+\log (x)-e^{\frac {4}{x}+x} \log (x)-4 e^x x \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.94, size = 105, normalized size = 3.75 \begin {gather*} -x + \log \left (4 \, x + e^{\frac {4}{x}}\right ) + \log \left (\frac {{\left (19 \, x + 5 \, e^{\frac {4}{x}}\right )} e^{x} + {\left ({\left (4 \, x + e^{\frac {4}{x}}\right )} e^{x} - 1\right )} \log \relax (x) - 5}{{\left (4 \, x + e^{\frac {4}{x}}\right )} e^{x} - 1}\right ) + \log \left (\frac {{\left (4 \, x + e^{\frac {4}{x}}\right )} e^{x} - 1}{4 \, x + e^{\frac {4}{x}}}\right ) - \log \left (\log \relax (x) + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.34, size = 55, normalized size = 1.96 \begin {gather*} -x + \log \left (4 \, x e^{x} \log \relax (x) + 19 \, x e^{x} + e^{\left (\frac {x^{2} + 4}{x}\right )} \log \relax (x) + 5 \, e^{\left (\frac {x^{2} + 4}{x}\right )} - \log \relax (x) - 5\right ) - \log \left (\log \relax (x) + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.06, size = 72, normalized size = 2.57
method | result | size |
risch | \(\ln \left ({\mathrm e}^{\frac {4}{x}}+\left (4 \,{\mathrm e}^{x} x -1\right ) {\mathrm e}^{-x}\right )+\ln \left (\ln \relax (x )+\frac {19 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{\frac {x^{2}+4}{x}}-5}{4 \,{\mathrm e}^{x} x +{\mathrm e}^{\frac {x^{2}+4}{x}}-1}\right )-\ln \left (5+\ln \relax (x )\right )\) | \(72\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 43, normalized size = 1.54 \begin {gather*} \log \left (\frac {{\left ({\left (\log \relax (x) + 5\right )} e^{\left (x + \frac {4}{x}\right )} + {\left (4 \, x \log \relax (x) + 19 \, x\right )} e^{x} - \log \relax (x) - 5\right )} e^{\left (-x\right )}}{\log \relax (x) + 5}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {{\ln \relax (x)}^2\,\left ({\mathrm {e}}^x\,\left (4\,{\mathrm {e}}^{4/x}-4\,x^2\right )-x^2\right )+\ln \relax (x)\,\left ({\mathrm {e}}^x\,\left (40\,{\mathrm {e}}^{4/x}-39\,x^2\right )-10\,x^2\right )+{\mathrm {e}}^x\,\left (100\,{\mathrm {e}}^{4/x}-96\,x^2\right )-25\,x^2}{\ln \relax (x)\,\left ({\mathrm {e}}^x\,\left (10\,x^2\,{\mathrm {e}}^{4/x}+39\,x^3\right )-10\,x^2\right )+{\mathrm {e}}^x\,\left (25\,x^2\,{\mathrm {e}}^{4/x}+95\,x^3\right )-25\,x^2+{\ln \relax (x)}^2\,\left ({\mathrm {e}}^x\,\left (x^2\,{\mathrm {e}}^{4/x}+4\,x^3\right )-x^2\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 6.37, size = 60, normalized size = 2.14 \begin {gather*} - x + \log {\left (\frac {4 x \log {\relax (x )} + 19 x}{\log {\relax (x )} + 5} + e^{\frac {4}{x}} \right )} + \log {\left (\frac {- \log {\relax (x )} - 5}{4 x \log {\relax (x )} + 19 x + e^{\frac {4}{x}} \log {\relax (x )} + 5 e^{\frac {4}{x}}} + e^{x} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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