Optimal. Leaf size=34 \[ e^{x \left (-x+\frac {1-e^{2 e^{2 x}}+4 x (5+\log (4))}{4+x}\right )} \]
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Rubi [F]
time = 24.63, 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 {\exp \left (\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}\right ) \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+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 \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}\right ) \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{(4+x)^2} \, dx\\ &=\int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{(4+x)^2} \, dx\\ &=\int \left (\frac {4 \exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x}{-4-x}-\frac {4 \exp \left (2 e^{2 x}+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}+\frac {4 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}+\frac {128 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x}{(4+x)^2}+\frac {4 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x^2}{(4+x)^2}-\frac {2 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x^3}{(4+x)^2}+\frac {4 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x (8+x) \log (4)}{(4+x)^2}\right ) \, dx\\ &=-\left (2 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x^3}{(4+x)^2} \, dx\right )+4 \int \frac {\exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x}{-4-x} \, dx-4 \int \frac {\exp \left (2 e^{2 x}+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+4 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+4 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x^2}{(4+x)^2} \, dx+128 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x}{(4+x)^2} \, dx+(4 \log (4)) \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x (8+x)}{(4+x)^2} \, dx\\ &=-\left (2 \int \left (-8 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )+\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x-\frac {64 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}+\frac {48 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x}\right ) \, dx\right )-4 \int \frac {\exp \left (2 e^{2 x}+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+4 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+4 \int \left (-\exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )+\frac {4 \exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x}\right ) \, dx+4 \int \left (\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )+\frac {16 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}-\frac {8 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x}\right ) \, dx+128 \int \left (-\frac {4 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}+\frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x}\right ) \, dx+(4 \log (4)) \int \left (\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )-\frac {16 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}\right ) \, dx\\ &=-\left (2 \int \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x \, dx\right )-4 \int \exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) \, dx+4 \int \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) \, dx-4 \int \frac {\exp \left (2 e^{2 x}+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+4 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+16 \int \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) \, dx+16 \int \frac {\exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x} \, dx-32 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x} \, dx+64 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx-96 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x} \, dx+128 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+128 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x} \, dx-512 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+(4 \log (4)) \int \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) \, dx-(64 \log (4)) \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 2.79, size = 40, normalized size = 1.18 \begin {gather*} 2^{\frac {8 x^2}{4+x}} e^{-\frac {x \left (-1+e^{2 e^{2 x}}-16 x+x^2\right )}{4+x}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.52, size = 33, normalized size = 0.97
method | result | size |
risch | \({\mathrm e}^{\frac {x \left (8 x \ln \left (2\right )-x^{2}-{\mathrm e}^{2 \,{\mathrm e}^{2 x}}+16 x +1\right )}{4+x}}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.65, size = 57, normalized size = 1.68 \begin {gather*} \frac {1}{4294967296} \, e^{\left (-x^{2} + 8 \, x \log \left (2\right ) + 20 \, x + \frac {4 \, e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x + 4} + \frac {128 \, \log \left (2\right )}{x + 4} + \frac {316}{x + 4} - e^{\left (2 \, e^{\left (2 \, x\right )}\right )} - 79\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 36, normalized size = 1.06 \begin {gather*} e^{\left (-\frac {x^{3} - 8 \, x^{2} \log \left (2\right ) - 16 \, x^{2} + x e^{\left (2 \, e^{\left (2 \, x\right )}\right )} - x}{x + 4}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.48, size = 32, normalized size = 0.94 \begin {gather*} e^{\frac {- x^{3} + 8 x^{2} \log {\left (2 \right )} + 16 x^{2} - x e^{2 e^{2 x}} + x}{x + 4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.50, size = 56, normalized size = 1.65 \begin {gather*} e^{\left (-\frac {x^{3}}{x + 4} + \frac {8 \, x^{2} \log \left (2\right )}{x + 4} + \frac {16 \, x^{2}}{x + 4} - \frac {x e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x + 4} + \frac {x}{x + 4}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.71, size = 59, normalized size = 1.74 \begin {gather*} 2^{\frac {8\,x^2}{x+4}}\,{\mathrm {e}}^{\frac {x}{x+4}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}}{x+4}}\,{\mathrm {e}}^{-\frac {x^3}{x+4}}\,{\mathrm {e}}^{\frac {16\,x^2}{x+4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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