Optimal. Leaf size=28 \[ e^{2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}} \]
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Rubi [F]
time = 20.24, 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 {2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}} e^{2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}} \left (-\log (2)+e^{3-x} \log (2)\right )}{16+e^{6-2 x}+8 x+x^2+e^{3-x} (8+2 x)+\left (8+2 e^{3-x}+2 x\right ) (i \pi +\log (5-\log (5)))+(i \pi +\log (5-\log (5)))^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 {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x} \left (-e^3+e^x\right ) \log (2)}{\left (i e^3-e^x (\pi -i (4+x+\log (5-\log (5))))\right )^2} \, dx\\ &=\log (2) \int \frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x} \left (-e^3+e^x\right )}{\left (i e^3-e^x (\pi -i (4+x+\log (5-\log (5))))\right )^2} \, dx\\ &=\log (2) \int \left (\frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{(4 i-\pi +i x+i \log (5-\log (5))) \left (i e^3+i e^x x+4 i e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )}+\frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{3+2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x} (5 i-\pi +i x+i \log (5-\log (5)))}{(4 i-\pi +i x+i \log (5-\log (5))) \left (e^3+e^x x+4 e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )^2}\right ) \, dx\\ &=\log (2) \int \frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{(4 i-\pi +i x+i \log (5-\log (5))) \left (i e^3+i e^x x+4 i e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )} \, dx+\log (2) \int \frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{3+2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x} (5 i-\pi +i x+i \log (5-\log (5)))}{(4 i-\pi +i x+i \log (5-\log (5))) \left (e^3+e^x x+4 e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )^2} \, dx\\ &=\log (2) \int \frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{(4 i-\pi +i x+i \log (5-\log (5))) \left (i e^3+i e^x x+4 i e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )} \, dx+\log (2) \int \left (-\frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{3+2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{\left (i e^3+i e^x x+4 i e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )^2}+\frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{3+2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{(-4-i \pi -x-\log (5-\log (5))) \left (i e^3+i e^x x+4 i e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )^2}\right ) \, dx\\ &=-\left (\log (2) \int \frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{3+2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{\left (i e^3+i e^x x+4 i e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )^2} \, dx\right )+\log (2) \int \frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{3+2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{(-4-i \pi -x-\log (5-\log (5))) \left (i e^3+i e^x x+4 i e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )^2} \, dx+\log (2) \int \frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{(4 i-\pi +i x+i \log (5-\log (5))) \left (i e^3+i e^x x+4 i e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )} \, dx\\ &=\log (2) \int \frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{3+2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{(-4-i \pi -x-\log (5-\log (5))) \left (i e^3+i e^x x+4 i e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )^2} \, dx+\log (2) \int \frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{(4 i-\pi +i x+i \log (5-\log (5))) \left (i e^3+i e^x x+4 i e^x \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )\right )} \, dx-\log (2) \int \frac {2^{\frac {1}{e^{3-x}+x+4 \left (1+\frac {1}{4} (i \pi +\log (5-\log (5)))\right )}} e^{3+2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}+x}}{\left (i e^3-e^x (\pi -i (4+x+\log (5-\log (5))))\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.09, size = 28, normalized size = 1.00 \begin {gather*} e^{2^{\frac {1}{4+e^{3-x}+i \pi +x+\log (5-\log (5))}}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.20, size = 20, normalized size = 0.71
method | result | size |
risch | \({\mathrm e}^{2^{\frac {1}{\ln \left (\ln \left (5\right )-5\right )+{\mathrm e}^{3-x}+4+x}}}\) | \(20\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.69, size = 22, normalized size = 0.79 \begin {gather*} e^{\left (2^{\frac {e^{x}}{{\left (x + \log \left (\log \left (5\right ) - 5\right ) + 4\right )} e^{x} + e^{3}}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 19, normalized size = 0.68 \begin {gather*} e^{\left (2^{\left (\frac {1}{x + e^{\left (-x + 3\right )} + \log \left (\log \left (5\right ) - 5\right ) + 4}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 19, normalized size = 0.68 \begin {gather*} e^{\left (2^{\left (\frac {1}{x + e^{\left (-x + 3\right )} + \log \left (\log \left (5\right ) - 5\right ) + 4}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.94, size = 20, normalized size = 0.71 \begin {gather*} {\mathrm {e}}^{2^{\frac {1}{x+\ln \left (\ln \left (5\right )-5\right )+{\mathrm {e}}^{-x}\,{\mathrm {e}}^3+4}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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