Optimal. Leaf size=27 \[ x+4 (4+x)+\frac {\left (e^{e^x+2 x}+\log ^4(x)\right )^2}{x} \]
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Rubi [F] time = 1.65, 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 {5 x^2+e^{2 e^x+4 x} \left (-1+4 x+2 e^x x\right )+8 \log ^7(x)-\log ^8(x)+e^{e^x+2 x} \left (8 \log ^3(x)+\left (-2+4 x+2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 e^{2 e^x+5 x}}{x}+\frac {e^{2 \left (e^x+2 x\right )} (-1+4 x)}{x^2}+\frac {2 e^{e^x+3 x} \log ^4(x)}{x}+\frac {2 e^{e^x+2 x} \log ^3(x) (4-\log (x)+2 x \log (x))}{x^2}+\frac {5 x^2+8 \log ^7(x)-\log ^8(x)}{x^2}\right ) \, dx\\ &=2 \int \frac {e^{2 e^x+5 x}}{x} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+2 \int \frac {e^{e^x+2 x} \log ^3(x) (4-\log (x)+2 x \log (x))}{x^2} \, dx+\int \frac {e^{2 \left (e^x+2 x\right )} (-1+4 x)}{x^2} \, dx+\int \frac {5 x^2+8 \log ^7(x)-\log ^8(x)}{x^2} \, dx\\ &=2 \int \frac {e^{2 e^x+5 x}}{x} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+2 \int \left (\frac {4 e^{e^x+2 x} \log ^3(x)}{x^2}+\frac {e^{e^x+2 x} (-1+2 x) \log ^4(x)}{x^2}\right ) \, dx+\int \left (-\frac {e^{2 \left (e^x+2 x\right )}}{x^2}+\frac {4 e^{2 \left (e^x+2 x\right )}}{x}\right ) \, dx+\int \left (5+\frac {8 \log ^7(x)}{x^2}-\frac {\log ^8(x)}{x^2}\right ) \, dx\\ &=5 x+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+2 \int \frac {e^{e^x+2 x} (-1+2 x) \log ^4(x)}{x^2} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx+8 \int \frac {\log ^7(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx-\int \frac {\log ^8(x)}{x^2} \, dx\\ &=5 x-\frac {8 \log ^7(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+2 \int \left (-\frac {e^{e^x+2 x} \log ^4(x)}{x^2}+\frac {2 e^{e^x+2 x} \log ^4(x)}{x}\right ) \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-8 \int \frac {\log ^7(x)}{x^2} \, dx+56 \int \frac {\log ^6(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x-\frac {56 \log ^6(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-56 \int \frac {\log ^6(x)}{x^2} \, dx+336 \int \frac {\log ^5(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x-\frac {336 \log ^5(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-336 \int \frac {\log ^5(x)}{x^2} \, dx+1680 \int \frac {\log ^4(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x-\frac {1680 \log ^4(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-1680 \int \frac {\log ^4(x)}{x^2} \, dx+6720 \int \frac {\log ^3(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x-\frac {6720 \log ^3(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-6720 \int \frac {\log ^3(x)}{x^2} \, dx+20160 \int \frac {\log ^2(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x-\frac {20160 \log ^2(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-20160 \int \frac {\log ^2(x)}{x^2} \, dx+40320 \int \frac {\log (x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=-\frac {40320}{x}+5 x-\frac {40320 \log (x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-40320 \int \frac {\log (x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.42, size = 40, normalized size = 1.48 \begin {gather*} \frac {e^{2 e^x+4 x}+5 x^2+2 e^{e^x+2 x} \log ^4(x)+\log ^8(x)}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.97, size = 36, normalized size = 1.33 \begin {gather*} \frac {\log \relax (x)^{8} + 2 \, e^{\left (2 \, x + e^{x}\right )} \log \relax (x)^{4} + 5 \, x^{2} + e^{\left (4 \, x + 2 \, e^{x}\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 36, normalized size = 1.33 \begin {gather*} \frac {\log \relax (x)^{8} + 2 \, e^{\left (2 \, x + e^{x}\right )} \log \relax (x)^{4} + 5 \, x^{2} + e^{\left (4 \, x + 2 \, e^{x}\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 42, normalized size = 1.56
method | result | size |
risch | \(\frac {\ln \relax (x )^{8}}{x}+5 x +\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}+4 x}}{x}+\frac {2 \ln \relax (x )^{4} {\mathrm e}^{{\mathrm e}^{x}+2 x}}{x}\) | \(42\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 35, normalized size = 1.30 \begin {gather*} 5 \, x + \frac {\log \relax (x)^{8} + 2 \, e^{\left (2 \, x + e^{x}\right )} \log \relax (x)^{4} + e^{\left (4 \, x + 2 \, e^{x}\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.78, size = 41, normalized size = 1.52 \begin {gather*} 5\,x+\frac {{\mathrm {e}}^{4\,x+2\,{\mathrm {e}}^x}}{x}+\frac {{\ln \relax (x)}^8}{x}+\frac {2\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^x}\,{\ln \relax (x)}^4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.44, size = 41, normalized size = 1.52 \begin {gather*} 5 x + \frac {\log {\relax (x )}^{8}}{x} + \frac {2 x e^{2 x + e^{x}} \log {\relax (x )}^{4} + x e^{4 x + 2 e^{x}}}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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