Optimal. Leaf size=33 \[ \frac {\log (x)}{2+\frac {x^2}{\log \left (-\left (1+e^{e^x}\right )^2+\frac {3+x}{4}\right )}} \]
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Rubi [F] time = 46.34, 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 \log (x)+8 e^{e^x+x} x^3 \log (x)+8 e^{2 e^x+x} x^3 \log (x)+\left (x^2-x^3+\left (-2 x^2+2 x^3\right ) \log (x)+e^{e^x} \left (8 x^2-16 x^2 \log (x)\right )+e^{2 e^x} \left (4 x^2-8 x^2 \log (x)\right )\right ) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )+\left (2+16 e^{e^x}+8 e^{2 e^x}-2 x\right ) \log ^2\left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{x^5+8 e^{e^x} x^5+4 e^{2 e^x} x^5-x^6+\left (4 x^3+32 e^{e^x} x^3+16 e^{2 e^x} x^3-4 x^4\right ) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )+\left (4 x+32 e^{e^x} x+16 e^{2 e^x} x-4 x^2\right ) \log ^2\left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, 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 {\log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{x^3+2 x \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}+\frac {x \log (x) \left (-x+8 e^{e^x+x} x+8 e^{2 e^x+x} x+\log (16)-2 \left (8 e^{e^x}+4 e^{2 e^x}-x\right ) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )-2 \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}\right ) \, dx\\ &=\int \frac {\log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{x^3+2 x \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx+\int \frac {x \log (x) \left (-x+8 e^{e^x+x} x+8 e^{2 e^x+x} x+\log (16)-2 \left (8 e^{e^x}+4 e^{2 e^x}-x\right ) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )-2 \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx\\ &=\int \left (\frac {1}{2 x}-\frac {x}{2 \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )}\right ) \, dx+\int \left (-\frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}+\frac {8 e^{e^x+x} \left (1+e^{e^x}\right ) x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}+\frac {x \log (16) \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}-\frac {2 \left (8 e^{e^x}+4 e^{2 e^x}-x\right ) x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}-\frac {2 x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}\right ) \, dx\\ &=\frac {\log (x)}{2}-\frac {1}{2} \int \frac {x}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx-2 \int \frac {\left (8 e^{e^x}+4 e^{2 e^x}-x\right ) x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-2 \int \frac {x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{e^x+x} \left (1+e^{e^x}\right ) x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+\log (16) \int \frac {x \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx\\ &=\frac {\log (x)}{2}-\frac {1}{2} \int \frac {x}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx-2 \int \left (\frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}-\frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}\right ) \, dx-2 \int \frac {x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \left (\frac {e^{e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}+\frac {e^{2 e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}\right ) \, dx+\log (16) \int \frac {x \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx\\ &=\frac {\log (x)}{2}-\frac {1}{2} \int \frac {x}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx-2 \int \frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+2 \int \frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-2 \int \frac {x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{2 e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+\log (16) \int \frac {x \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx\\ &=\frac {\log (x)}{2}-\frac {1}{2} \int \frac {x}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx+2 \int \frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-2 \int \left (-\frac {x^3 \log (x)}{2 \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}+\frac {x \log (x)}{2 \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )}\right ) \, dx-2 \int \frac {x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{2 e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+\log (16) \int \frac {x \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx\\ &=\frac {\log (x)}{2}-\frac {1}{2} \int \frac {x}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx+2 \int \frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-2 \int \frac {x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{2 e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+\log (16) \int \frac {x \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+\int \frac {x^3 \log (x)}{\left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x \log (x)}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.65, size = 187, normalized size = 5.67 \begin {gather*} \frac {\log (x) \left (2 \left (8 e^{e^x+x}+8 e^{2 e^x+x}+8 e^{e^x} x+4 e^{2 e^x} x-x^2\right ) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )-(-1+x) \left (\log (16)-2 \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )}{2 \left (-1+8 e^{e^x+x}+8 e^{2 e^x+x}+x+8 e^{e^x} x+4 e^{2 e^x} x-x^2\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.76, size = 83, normalized size = 2.52 \begin {gather*} \frac {\log \left (\frac {1}{4} \, {\left ({\left (x - 1\right )} e^{\left (2 \, x\right )} - 4 \, e^{\left (2 \, x + 2 \, e^{x}\right )} - 8 \, e^{\left (2 \, x + e^{x}\right )}\right )} e^{\left (-2 \, x\right )}\right ) \log \relax (x)}{x^{2} + 2 \, \log \left (\frac {1}{4} \, {\left ({\left (x - 1\right )} e^{\left (2 \, x\right )} - 4 \, e^{\left (2 \, x + 2 \, e^{x}\right )} - 8 \, e^{\left (2 \, x + e^{x}\right )}\right )} e^{\left (-2 \, x\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 3.95, size = 57, normalized size = 1.73 \begin {gather*} -\frac {2 \, \log \relax (2) \log \relax (x) - \log \left (x - 4 \, e^{\left (2 \, e^{x}\right )} - 8 \, e^{\left (e^{x}\right )} - 1\right ) \log \relax (x)}{x^{2} - 4 \, \log \relax (2) + 2 \, \log \left (x - 4 \, e^{\left (2 \, e^{x}\right )} - 8 \, e^{\left (e^{x}\right )} - 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.17, size = 39, normalized size = 1.18
method | result | size |
risch | \(\frac {\ln \relax (x )}{2}-\frac {x^{2} \ln \relax (x )}{2 \left (x^{2}+2 \ln \left (-{\mathrm e}^{2 \,{\mathrm e}^{x}}-2 \,{\mathrm e}^{{\mathrm e}^{x}}+\frac {x}{4}-\frac {1}{4}\right )\right )}\) | \(39\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.44, size = 40, normalized size = 1.21 \begin {gather*} -\frac {x^{2} \log \relax (x)}{2 \, {\left (x^{2} - 4 \, \log \relax (2) + 2 \, \log \left (x - 4 \, e^{\left (2 \, e^{x}\right )} - 8 \, e^{\left (e^{x}\right )} - 1\right )\right )}} + \frac {1}{2} \, \log \relax (x) \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 = -0.03 \begin {gather*} \int \frac {{\ln \left (\frac {x}{4}-2\,{\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^{2\,{\mathrm {e}}^x}-\frac {1}{4}\right )}^2\,\left (16\,{\mathrm {e}}^{{\mathrm {e}}^x}-2\,x+8\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+2\right )-\ln \left (\frac {x}{4}-2\,{\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^{2\,{\mathrm {e}}^x}-\frac {1}{4}\right )\,\left (\ln \relax (x)\,\left (2\,x^2-2\,x^3\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (16\,x^2\,\ln \relax (x)-8\,x^2\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (8\,x^2\,\ln \relax (x)-4\,x^2\right )-x^2+x^3\right )-x^3\,\ln \relax (x)+8\,x^3\,{\mathrm {e}}^{x+{\mathrm {e}}^x}\,\ln \relax (x)+8\,x^3\,{\mathrm {e}}^{x+2\,{\mathrm {e}}^x}\,\ln \relax (x)}{{\ln \left (\frac {x}{4}-2\,{\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^{2\,{\mathrm {e}}^x}-\frac {1}{4}\right )}^2\,\left (4\,x+32\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+16\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}-4\,x^2\right )+8\,x^5\,{\mathrm {e}}^{{\mathrm {e}}^x}+\ln \left (\frac {x}{4}-2\,{\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^{2\,{\mathrm {e}}^x}-\frac {1}{4}\right )\,\left (32\,x^3\,{\mathrm {e}}^{{\mathrm {e}}^x}+4\,x^3-4\,x^4+16\,x^3\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}\right )+x^5-x^6+4\,x^5\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.48, size = 39, normalized size = 1.18 \begin {gather*} - \frac {x^{2} \log {\relax (x )}}{2 x^{2} + 4 \log {\left (\frac {x}{4} - e^{2 e^{x}} - 2 e^{e^{x}} - \frac {1}{4} \right )}} + \frac {\log {\relax (x )}}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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