Optimal. Leaf size=24 \[ e^{-4+e^{\frac {5}{\left (x^2+\frac {x}{\log (5)}\right ) \log (x)}}} \]
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Rubi [F]
time = 8.62, 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 {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{x^2 \left (1+2 x \log (5)+x^2 \log ^2(5)\right ) \log ^2(x)} \, dx\\ &=\int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{x^2 (1+x \log (5))^2 \log ^2(x)} \, dx\\ &=\int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \log (5) (-1-x \log (5)-\log (x)-x \log (25) \log (x))}{x^2 (1+x \log (5))^2 \log ^2(x)} \, dx\\ &=\log (5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} (-1-x \log (5)-\log (x)-x \log (25) \log (x))}{x^2 (1+x \log (5))^2 \log ^2(x)} \, dx\\ &=\log (5) \int \left (-\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 (1+x \log (5)) \log ^2(x)}+\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} (-1-x \log (25))}{x^2 (1+x \log (5))^2 \log (x)}\right ) \, dx\\ &=-\left (\log (5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 (1+x \log (5)) \log ^2(x)} \, dx\right )+\log (5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} (-1-x \log (25))}{x^2 (1+x \log (5))^2 \log (x)} \, dx\\ &=-\left (\log (5) \int \left (\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 \log ^2(x)}-\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \log (5)}{x \log ^2(x)}+\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \log ^2(5)}{(1+x \log (5)) \log ^2(x)}\right ) \, dx\right )+\log (5) \int \left (-\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 \log (x)}+\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \log ^2(5)}{(1+x \log (5))^2 \log (x)}\right ) \, dx\\ &=-\left (\log (5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 \log ^2(x)} \, dx\right )-\log (5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 \log (x)} \, dx+\log ^2(5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x \log ^2(x)} \, dx-\log ^3(5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{(1+x \log (5)) \log ^2(x)} \, dx+\log ^3(5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{(1+x \log (5))^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F]
time = 1.78, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [A]
time = 0.23, size = 22, normalized size = 0.92
method | result | size |
risch | \({\mathrm e}^{3125^{\frac {1}{x \left (x \ln \left (5\right )+1\right ) \ln \left (x \right )}}-4}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.68, size = 34, normalized size = 1.42 \begin {gather*} e^{\left (e^{\left (-\frac {5 \, \log \left (5\right )^{2}}{{\left (x \log \left (5\right ) + 1\right )} \log \left (x\right )} + \frac {5 \, \log \left (5\right )}{x \log \left (x\right )}\right )} - 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 21, normalized size = 0.88 \begin {gather*} e^{\left (5^{\frac {5}{{\left (x^{2} \log \left (5\right ) + x\right )} \log \left (x\right )}} - 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 22, normalized size = 0.92 \begin {gather*} e^{\left (5^{\frac {5}{x^{2} \log \left (5\right ) \log \left (x\right ) + x \log \left (x\right )}} - 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.13, size = 23, normalized size = 0.96 \begin {gather*} {\mathrm {e}}^{5^{\frac {5}{x\,\ln \left (x\right )+x^2\,\ln \left (5\right )\,\ln \left (x\right )}}}\,{\mathrm {e}}^{-4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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