Optimal. Leaf size=23 \[ \frac {4}{4+e^{e^x}+2 x+\frac {16 \log (2+x)}{x}} \]
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Rubi [F] time = 4.41, 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 {-64 x-16 x^2-8 x^3+e^{e^x+x} \left (-8 x^2-4 x^3\right )+(128+64 x) \log (2+x)}{32 x^2+48 x^3+24 x^4+4 x^5+e^{2 e^x} \left (2 x^2+x^3\right )+\left (256 x+256 x^2+64 x^3\right ) \log (2+x)+(512+256 x) \log ^2(2+x)+e^{e^x} \left (16 x^2+16 x^3+4 x^4+\left (64 x+32 x^2\right ) \log (2+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 {-4 x \left (16+2 \left (2+e^{e^x+x}\right ) x+\left (2+e^{e^x+x}\right ) x^2\right )+64 (2+x) \log (2+x)}{(2+x) \left (x \left (4+e^{e^x}+2 x\right )+16 \log (2+x)\right )^2} \, dx\\ &=\int \left (-\frac {4 e^{e^x+x} x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {64 x}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {16 x^2}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {8 x^3}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}+\frac {64 \log (2+x)}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {e^{e^x+x} x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx\right )-8 \int \frac {x^3}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx-16 \int \frac {x^2}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx-64 \int \frac {x}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx+64 \int \frac {\log (2+x)}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx\\ &=-\left (4 \int \frac {e^{e^x+x} x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx\right )-8 \int \left (\frac {4}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {2 x}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}+\frac {x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {8}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}\right ) \, dx-16 \int \left (-\frac {2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}+\frac {x}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}+\frac {4}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}\right ) \, dx+64 \int \frac {\log (2+x)}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx-64 \int \left (\frac {1}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {2}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {e^{e^x+x} x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx\right )-8 \int \frac {x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx-64 \int \frac {1}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx+64 \int \frac {\log (2+x)}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx+128 \int \frac {1}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 2.02, size = 24, normalized size = 1.04 \begin {gather*} \frac {4 x}{x \left (4+e^{e^x}+2 x\right )+16 \log (2+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.83, size = 34, normalized size = 1.48 \begin {gather*} \frac {4 \, x e^{x}}{x e^{\left (x + e^{x}\right )} + 2 \, {\left (x^{2} + 2 \, x\right )} e^{x} + 16 \, e^{x} \log \left (x + 2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.42, size = 25, normalized size = 1.09 \begin {gather*} \frac {4 \, x}{2 \, x^{2} + x e^{\left (e^{x}\right )} + 4 \, x + 16 \, \log \left (x + 2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 26, normalized size = 1.13
method | result | size |
risch | \(\frac {4 x}{x \,{\mathrm e}^{{\mathrm e}^{x}}+2 x^{2}+16 \ln \left (2+x \right )+4 x}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.76, size = 25, normalized size = 1.09 \begin {gather*} \frac {4 \, x}{2 \, x^{2} + x e^{\left (e^{x}\right )} + 4 \, x + 16 \, \log \left (x + 2\right )} \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 = -0.04 \begin {gather*} \int -\frac {64\,x+16\,x^2+8\,x^3-\ln \left (x+2\right )\,\left (64\,x+128\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^x\,\left (4\,x^3+8\,x^2\right )}{{\ln \left (x+2\right )}^2\,\left (256\,x+512\right )+\ln \left (x+2\right )\,\left (64\,x^3+256\,x^2+256\,x\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (x^3+2\,x^2\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (\ln \left (x+2\right )\,\left (32\,x^2+64\,x\right )+16\,x^2+16\,x^3+4\,x^4\right )+32\,x^2+48\,x^3+24\,x^4+4\,x^5} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.46, size = 24, normalized size = 1.04 \begin {gather*} \frac {4 x}{2 x^{2} + x e^{e^{x}} + 4 x + 16 \log {\left (x + 2 \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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