Optimal. Leaf size=24 \[ \frac {1}{2} e^{e^{\frac {1}{4}+x}} \left (-5+(3-x+\log (x))^2\right ) \]
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Rubi [F] time = 5.69, 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^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{x} \, dx\\ &=\frac {1}{2} \int \frac {e^{e^{\frac {1}{4}+x}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{x} \, dx\\ &=\frac {1}{2} \int \left (\frac {2 e^{e^{\frac {1}{4}+x}} (-1+x) (-3+x-\log (x))}{x}+e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \left (4-6 x+x^2+6 \log (x)-2 x \log (x)+\log ^2(x)\right )\right ) \, dx\\ &=\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \left (4-6 x+x^2+6 \log (x)-2 x \log (x)+\log ^2(x)\right ) \, dx+\int \frac {e^{e^{\frac {1}{4}+x}} (-1+x) (-3+x-\log (x))}{x} \, dx\\ &=\frac {1}{2} \int \left (4 e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x}-6 e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x+e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2+6 e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log (x)-2 e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \log (x)+e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x)\right ) \, dx+\int \left (\frac {e^{e^{\frac {1}{4}+x}} \left (3-4 x+x^2\right )}{x}-\frac {e^{e^{\frac {1}{4}+x}} (-1+x) \log (x)}{x}\right ) \, dx\\ &=\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2 \, dx+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x) \, dx+2 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \, dx-3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log (x) \, dx+\int \frac {e^{e^{\frac {1}{4}+x}} \left (3-4 x+x^2\right )}{x} \, dx-\int \frac {e^{e^{\frac {1}{4}+x}} (-1+x) \log (x)}{x} \, dx-\int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \log (x) \, dx\\ &=3 e^{e^{\frac {1}{4}+x}} \log (x)-\text {Ei}\left (e^{\frac {1}{4}+x}\right ) \log (x)+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2 \, dx+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x) \, dx+2 \operatorname {Subst}\left (\int e^{\frac {1}{4}+\sqrt [4]{e} x} \, dx,x,e^x\right )-3 \int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx-3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\log (x) \int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx-\log (x) \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\int \left (-4 e^{e^{\frac {1}{4}+x}}+\frac {3 e^{e^{\frac {1}{4}+x}}}{x}+e^{e^{\frac {1}{4}+x}} x\right ) \, dx+\int \frac {\text {Ei}\left (e^{\frac {1}{4}+x}\right )-\int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx}{x} \, dx+\int \frac {\int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx}{x} \, dx\\ &=2 e^{e^{\frac {1}{4}+x}}+3 e^{e^{\frac {1}{4}+x}} \log (x)-\text {Ei}\left (e^{\frac {1}{4}+x}\right ) \log (x)+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2 \, dx+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x) \, dx-3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx-4 \int e^{e^{\frac {1}{4}+x}} \, dx+\log (x) \int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx-\log (x) \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\int e^{e^{\frac {1}{4}+x}} x \, dx+\int \left (\frac {\text {Ei}\left (e^{\frac {1}{4}+x}\right )}{x}-\frac {\int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx}{x}\right ) \, dx+\int \frac {\int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx}{x} \, dx\\ &=2 e^{e^{\frac {1}{4}+x}}+3 e^{e^{\frac {1}{4}+x}} \log (x)-\text {Ei}\left (e^{\frac {1}{4}+x}\right ) \log (x)+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2 \, dx+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x) \, dx-3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx-4 \operatorname {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^{\frac {1}{4}+x}\right )+\log (x) \int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx-\log (x) \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\int e^{e^{\frac {1}{4}+x}} x \, dx+\int \frac {\text {Ei}\left (e^{\frac {1}{4}+x}\right )}{x} \, dx-\int \frac {\int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx}{x} \, dx+\int \frac {\int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx}{x} \, dx\\ &=2 e^{e^{\frac {1}{4}+x}}-4 \text {Ei}\left (e^{\frac {1}{4}+x}\right )+3 e^{e^{\frac {1}{4}+x}} \log (x)-\text {Ei}\left (e^{\frac {1}{4}+x}\right ) \log (x)+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2 \, dx+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x) \, dx-3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\log (x) \int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx-\log (x) \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\int e^{e^{\frac {1}{4}+x}} x \, dx+\int \frac {\text {Ei}\left (e^{\frac {1}{4}+x}\right )}{x} \, dx-\int \frac {\int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx}{x} \, dx+\int \frac {\int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx}{x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 33, normalized size = 1.38 \begin {gather*} \frac {1}{2} e^{e^{\frac {1}{4}+x}} \left (4-6 x+x^2+(6-2 x) \log (x)+\log ^2(x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.77, size = 26, normalized size = 1.08 \begin {gather*} \frac {1}{2} \, {\left (x^{2} - 2 \, {\left (x - 3\right )} \log \relax (x) + \log \relax (x)^{2} - 6 \, x + 4\right )} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.22, size = 56, normalized size = 2.33 \begin {gather*} \frac {1}{2} \, x^{2} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} - x e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \relax (x) + \frac {1}{2} \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \relax (x)^{2} - 3 \, x e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} + 3 \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \relax (x) + 2 \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 29, normalized size = 1.21
method | result | size |
risch | \(\frac {\left (x^{2}-2 x \ln \relax (x )+\ln \relax (x )^{2}-6 x +6 \ln \relax (x )+4\right ) {\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}}{2}\) | \(29\) |
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*} \frac {1}{2} \, {\left (x^{2} - 2 \, {\left (x - 3\right )} \log \relax (x) + \log \relax (x)^{2} - 6 \, x + 4\right )} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} - 4 \, {\rm Ei}\left (e^{\left (x + \frac {1}{4}\right )}\right ) + 4 \, \int e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.51, size = 32, normalized size = 1.33 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{1/4}\,{\mathrm {e}}^x}\,\left (\frac {x^2}{2}-x\,\ln \relax (x)-3\,x+\frac {{\ln \relax (x)}^2}{2}+3\,\ln \relax (x)+2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 64.36, size = 34, normalized size = 1.42 \begin {gather*} \frac {\left (x^{2} - 2 x \log {\relax (x )} - 6 x + \log {\relax (x )}^{2} + 6 \log {\relax (x )} + 4\right ) e^{e^{x + \frac {1}{4}}}}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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