Optimal. Leaf size=28 \[ \frac {x}{\frac {3}{-2+x}+\frac {e^{8 e^x}}{x^2}}-\log (x) \]
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Rubi [F] time = 6.39, 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 {-9 x^4-6 x^5+6 x^6+e^{16 e^x} \left (-4+4 x-x^2\right )+e^{8 e^x} \left (12 x^2+6 x^3-12 x^4+3 x^5+e^x \left (-32 x^4+32 x^5-8 x^6\right )\right )}{9 x^5+e^{16 e^x} \left (4 x-4 x^2+x^3\right )+e^{8 e^x} \left (-12 x^3+6 x^4\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 {-e^{16 e^x} (-2+x)^2-8 e^{8 e^x+x} (-2+x)^2 x^4+3 x^4 \left (-3-2 x+2 x^2\right )+3 e^{8 e^x} x^2 \left (4+2 x-4 x^2+x^3\right )}{x \left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx\\ &=\int \left (-\frac {8 e^{8 e^x+x} (-2+x)^2 x^3}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2}+\frac {-4 e^{16 e^x}+4 e^{16 e^x} x+12 e^{8 e^x} x^2-e^{16 e^x} x^2+6 e^{8 e^x} x^3-9 x^4-12 e^{8 e^x} x^4-6 x^5+3 e^{8 e^x} x^5+6 x^6}{x \left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2}\right ) \, dx\\ &=-\left (8 \int \frac {e^{8 e^x+x} (-2+x)^2 x^3}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2} \, dx\right )+\int \frac {-4 e^{16 e^x}+4 e^{16 e^x} x+12 e^{8 e^x} x^2-e^{16 e^x} x^2+6 e^{8 e^x} x^3-9 x^4-12 e^{8 e^x} x^4-6 x^5+3 e^{8 e^x} x^5+6 x^6}{x \left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2} \, dx\\ &=-\left (8 \int \left (\frac {4 e^{8 e^x+x} x^3}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2}-\frac {4 e^{8 e^x+x} x^4}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2}+\frac {e^{8 e^x+x} x^5}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2}\right ) \, dx\right )+\int \frac {-e^{16 e^x} (-2+x)^2+3 x^4 \left (-3-2 x+2 x^2\right )+3 e^{8 e^x} x^2 \left (4+2 x-4 x^2+x^3\right )}{x \left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx\\ &=-\left (8 \int \frac {e^{8 e^x+x} x^5}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2} \, dx\right )-32 \int \frac {e^{8 e^x+x} x^3}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2} \, dx+32 \int \frac {e^{8 e^x+x} x^4}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2} \, dx+\int \left (-\frac {1}{x}-\frac {3 (-4+x) x^4}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2}+\frac {3 (-2+x) x^2}{-2 e^{8 e^x}+e^{8 e^x} x+3 x^2}\right ) \, dx\\ &=-\log (x)-3 \int \frac {(-4+x) x^4}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2} \, dx+3 \int \frac {(-2+x) x^2}{-2 e^{8 e^x}+e^{8 e^x} x+3 x^2} \, dx-8 \int \frac {e^{8 e^x+x} x^5}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx-32 \int \frac {e^{8 e^x+x} x^3}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx+32 \int \frac {e^{8 e^x+x} x^4}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx\\ &=-\log (x)-3 \int \frac {(-4+x) x^4}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx+3 \int \frac {(-2+x) x^2}{e^{8 e^x} (-2+x)+3 x^2} \, dx-8 \int \frac {e^{8 e^x+x} x^5}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx-32 \int \frac {e^{8 e^x+x} x^3}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx+32 \int \frac {e^{8 e^x+x} x^4}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx\\ &=-\log (x)-3 \int \left (-\frac {4 x^4}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2}+\frac {x^5}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2}\right ) \, dx+3 \int \left (-\frac {2 x^2}{-2 e^{8 e^x}+e^{8 e^x} x+3 x^2}+\frac {x^3}{-2 e^{8 e^x}+e^{8 e^x} x+3 x^2}\right ) \, dx-8 \int \frac {e^{8 e^x+x} x^5}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx-32 \int \frac {e^{8 e^x+x} x^3}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx+32 \int \frac {e^{8 e^x+x} x^4}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx\\ &=-\log (x)-3 \int \frac {x^5}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2} \, dx+3 \int \frac {x^3}{-2 e^{8 e^x}+e^{8 e^x} x+3 x^2} \, dx-6 \int \frac {x^2}{-2 e^{8 e^x}+e^{8 e^x} x+3 x^2} \, dx-8 \int \frac {e^{8 e^x+x} x^5}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx+12 \int \frac {x^4}{\left (-2 e^{8 e^x}+e^{8 e^x} x+3 x^2\right )^2} \, dx-32 \int \frac {e^{8 e^x+x} x^3}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx+32 \int \frac {e^{8 e^x+x} x^4}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx\\ &=-\log (x)-3 \int \frac {x^5}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx+3 \int \frac {x^3}{e^{8 e^x} (-2+x)+3 x^2} \, dx-6 \int \frac {x^2}{e^{8 e^x} (-2+x)+3 x^2} \, dx-8 \int \frac {e^{8 e^x+x} x^5}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx+12 \int \frac {x^4}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx-32 \int \frac {e^{8 e^x+x} x^3}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx+32 \int \frac {e^{8 e^x+x} x^4}{\left (e^{8 e^x} (-2+x)+3 x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 31, normalized size = 1.11 \begin {gather*} \frac {(-2+x) x^3}{e^{8 e^x} (-2+x)+3 x^2}-\log (x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 46, normalized size = 1.64 \begin {gather*} \frac {x^{4} - 2 \, x^{3} - 3 \, x^{2} \log \relax (x) - {\left (x - 2\right )} e^{\left (8 \, e^{x}\right )} \log \relax (x)}{3 \, x^{2} + {\left (x - 2\right )} e^{\left (8 \, e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.28, size = 58, normalized size = 2.07 \begin {gather*} \frac {x^{4} - 2 \, x^{3} - 3 \, x^{2} \log \relax (x) - x e^{\left (8 \, e^{x}\right )} \log \relax (x) + 2 \, e^{\left (8 \, e^{x}\right )} \log \relax (x)}{3 \, x^{2} + x e^{\left (8 \, e^{x}\right )} - 2 \, e^{\left (8 \, e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 35, normalized size = 1.25
method | result | size |
risch | \(-\ln \relax (x )+\frac {\left (x -2\right ) x^{3}}{{\mathrm e}^{8 \,{\mathrm e}^{x}} x +3 x^{2}-2 \,{\mathrm e}^{8 \,{\mathrm e}^{x}}}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 32, normalized size = 1.14 \begin {gather*} \frac {x^{4} - 2 \, x^{3}}{3 \, x^{2} + {\left (x - 2\right )} e^{\left (8 \, e^{x}\right )}} - \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.10, size = 60, normalized size = 2.14 \begin {gather*} -\frac {3\,x^2\,\ln \relax (x)+2\,x^3-x^4-2\,{\mathrm {e}}^{8\,{\mathrm {e}}^x}\,\ln \relax (x)+x\,{\mathrm {e}}^{8\,{\mathrm {e}}^x}\,\ln \relax (x)}{x\,{\mathrm {e}}^{8\,{\mathrm {e}}^x}-2\,{\mathrm {e}}^{8\,{\mathrm {e}}^x}+3\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.24, size = 26, normalized size = 0.93 \begin {gather*} - \log {\relax (x )} + \frac {x^{4} - 2 x^{3}}{3 x^{2} + \left (x - 2\right ) e^{8 e^{x}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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