Optimal. Leaf size=24 \[ 4^{\frac {1}{x}} \left (2+e^x+e^{(5-x) x}\right ) \log (\log (\log (5))) \]
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Rubi [A]
time = 0.84, antiderivative size = 37, normalized size of antiderivative = 1.54, number of steps
used = 5, number of rules used = 3, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 6874, 2326}
\begin {gather*} 4^{\frac {1}{x}} e^{5 x-x^2} \log (\log (\log (5)))+4^{\frac {1}{x}} \left (e^x+2\right ) \log (\log (\log (5))) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2326
Rule 6874
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\log (\log (\log (5))) \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )-2 \log (4)\right )}{x^2} \, dx\\ &=\log (\log (\log (5))) \int \left (\frac {4^{\frac {1}{x}} e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )}{x^2}+\frac {4^{\frac {1}{x}} \left (e^x x^2-2 \log (4)-e^x \log (4)\right )}{x^2}\right ) \, dx\\ &=\log (\log (\log (5))) \int \frac {4^{\frac {1}{x}} e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )}{x^2} \, dx+\log (\log (\log (5))) \int \frac {4^{\frac {1}{x}} \left (e^x x^2-2 \log (4)-e^x \log (4)\right )}{x^2} \, dx\\ &=4^{\frac {1}{x}} e^{5 x-x^2} \log (\log (\log (5)))+4^{\frac {1}{x}} \left (2+e^x\right ) \log (\log (\log (5)))\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.95, size = 37, normalized size = 1.54 \begin {gather*} 4^{\frac {1}{x}} e^{-x^2} \left (e^{5 x}+2 e^{x^2}+e^{x+x^2}\right ) \log (\log (\log (5))) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.56, size = 22, normalized size = 0.92
method | result | size |
risch | \(\ln \left (\ln \left (\ln \left (5\right )\right )\right ) \left ({\mathrm e}^{x}+2+{\mathrm e}^{-\left (x -5\right ) x}\right ) 4^{\frac {1}{x}}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 41, normalized size = 1.71 \begin {gather*} {\left ({\left (e^{\left (x^{2} + x\right )} + e^{\left (5 \, x\right )}\right )} e^{\left (-x^{2} + \frac {2 \, \log \left (2\right )}{x}\right )} + 2^{\frac {2}{x} + 1}\right )} \log \left (\log \left (\log \left (5\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 26, normalized size = 1.08 \begin {gather*} 2^{\frac {2}{x}} {\left (e^{\left (-x^{2} + 5 \, x\right )} + e^{x} + 2\right )} \log \left (\log \left (\log \left (5\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 4.66, size = 49, normalized size = 2.04 \begin {gather*} \left (e^{x} \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )} + 2 \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )}\right ) e^{\frac {2 \log {\left (2 \right )}}{x}} + e^{\frac {2 \log {\left (2 \right )}}{x}} e^{- x^{2} + 5 x} \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 8.92, size = 50, normalized size = 2.08 \begin {gather*} 2\,2^{2/x}\,\ln \left (\ln \left (\ln \left (5\right )\right )\right )+2^{2/x}\,\ln \left (\ln \left (\ln \left (5\right )\right )\right )\,{\mathrm {e}}^{5\,x-x^2}+2^{2/x}\,\ln \left (\ln \left (\ln \left (5\right )\right )\right )\,{\mathrm {e}}^x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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