Optimal. Leaf size=30 \[ \frac {\log \left (e^{e^{e^x}-x}\right )}{x \left (2+x \log \left (3+\frac {3 x}{2}\right )\right )} \]
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Rubi [F]
time = 4.58, 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 {x^3+\left (2 x^2+x^3\right ) \log \left (\frac {1}{2} (6+3 x)\right )+e^{e^x} \left (-4-2 x-x^2+e^x \left (4 x+2 x^2\right )+\left (-4 x-2 x^2+e^x \left (2 x^2+x^3\right )\right ) \log \left (\frac {1}{2} (6+3 x)\right )\right )}{8 x^2+4 x^3+\left (8 x^3+4 x^4\right ) \log \left (\frac {1}{2} (6+3 x)\right )+\left (2 x^4+x^5\right ) \log ^2\left (\frac {1}{2} (6+3 x)\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 {x^3+2 e^{e^x+x} x (2+x)-e^{e^x} \left (4+2 x+x^2\right )+x (2+x) \left (-2 e^{e^x}+x+e^{e^x+x} x\right ) \log \left (\frac {3 (2+x)}{2}\right )}{x^2 (2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\\ &=\int \left (\frac {x}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}-\frac {e^{e^x} \left (4+2 x+x^2\right )}{x^2 (2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}+\frac {e^{e^x+x}}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )}+\frac {\log \left (3+\frac {3 x}{2}\right )}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}-\frac {2 e^{e^x} \log \left (3+\frac {3 x}{2}\right )}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{e^x} \log \left (3+\frac {3 x}{2}\right )}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\right )+\int \frac {x}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx-\int \frac {e^{e^x} \left (4+2 x+x^2\right )}{x^2 (2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx+\int \frac {e^{e^x+x}}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )} \, dx+\int \frac {\log \left (3+\frac {3 x}{2}\right )}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\\ &=-\left (2 \int \frac {e^{e^x} \log \left (3+\frac {3 x}{2}\right )}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\right )+\int \frac {e^{e^x+x}}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )} \, dx+\int \left (\frac {1}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}-\frac {2}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}\right ) \, dx-\int \left (\frac {2 e^{e^x}}{x^2 \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}+\frac {e^{e^x}}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}\right ) \, dx+\int \frac {\log \left (3+\frac {3 x}{2}\right )}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\\ &=-\left (2 \int \frac {e^{e^x}}{x^2 \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\right )-2 \int \frac {1}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx-2 \int \frac {e^{e^x} \log \left (3+\frac {3 x}{2}\right )}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx+\int \frac {1}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx-\int \frac {e^{e^x}}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx+\int \frac {e^{e^x+x}}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )} \, dx+\int \frac {\log \left (3+\frac {3 x}{2}\right )}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.07, size = 27, normalized size = 0.90 \begin {gather*} \frac {e^{e^x}-x}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 124.29, size = 35, normalized size = 1.17
method | result | size |
risch | \(-\frac {1}{2+x \ln \left (3+\frac {3 x}{2}\right )}+\frac {{\mathrm e}^{{\mathrm e}^{x}}}{\left (2+x \ln \left (3+\frac {3 x}{2}\right )\right ) x}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.57, size = 34, normalized size = 1.13 \begin {gather*} -\frac {x - e^{\left (e^{x}\right )}}{x^{2} {\left (\log \left (3\right ) - \log \left (2\right )\right )} + x^{2} \log \left (x + 2\right ) + 2 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 25, normalized size = 0.83 \begin {gather*} -\frac {x - e^{\left (e^{x}\right )}}{x^{2} \log \left (\frac {3}{2} \, x + 3\right ) + 2 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.15, size = 32, normalized size = 1.07 \begin {gather*} \frac {e^{e^{x}}}{x^{2} \log {\left (\frac {3 x}{2} + 3 \right )} + 2 x} - \frac {1}{x \log {\left (\frac {3 x}{2} + 3 \right )} + 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.46, size = 25, normalized size = 0.83 \begin {gather*} -\frac {x - e^{\left (e^{x}\right )}}{x^{2} \log \left (\frac {3}{2} \, x + 3\right ) + 2 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.49, size = 24, normalized size = 0.80 \begin {gather*} -\frac {x-{\mathrm {e}}^{{\mathrm {e}}^x}}{x\,\left (x\,\ln \left (\frac {3\,x}{2}+3\right )+2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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