Optimal. Leaf size=25 \[ -x-\frac {e^5 \log (\log (5 x))}{3 \left (4+\frac {2}{x}\right )} \]
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Rubi [F] time = 0.17, 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 {e^5 (-1-2 x)+\left (-6-24 x-24 x^2\right ) \log (5 x)-e^5 \log (5 x) \log (\log (5 x))}{\left (6+24 x+24 x^2\right ) \log (5 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 {e^5 (-1-2 x)+\left (-6-24 x-24 x^2\right ) \log (5 x)-e^5 \log (5 x) \log (\log (5 x))}{6 (1+2 x)^2 \log (5 x)} \, dx\\ &=\frac {1}{6} \int \frac {e^5 (-1-2 x)+\left (-6-24 x-24 x^2\right ) \log (5 x)-e^5 \log (5 x) \log (\log (5 x))}{(1+2 x)^2 \log (5 x)} \, dx\\ &=\frac {1}{6} \int \left (-6-\frac {e^5}{(1+2 x) \log (5 x)}-\frac {e^5 \log (\log (5 x))}{(1+2 x)^2}\right ) \, dx\\ &=-x-\frac {1}{6} e^5 \int \frac {1}{(1+2 x) \log (5 x)} \, dx-\frac {1}{6} e^5 \int \frac {\log (\log (5 x))}{(1+2 x)^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.15, size = 24, normalized size = 0.96 \begin {gather*} -\frac {x \left (6+12 x+e^5 \log (\log (5 x))\right )}{6+12 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.19, size = 27, normalized size = 1.08 \begin {gather*} -\frac {x e^{5} \log \left (\log \left (5 \, x\right )\right ) + 12 \, x^{2} + 6 \, x}{6 \, {\left (2 \, x + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 27, normalized size = 1.08 \begin {gather*} -\frac {x e^{5} \log \left (\log \left (5 \, x\right )\right ) + 12 \, x^{2} + 6 \, x}{6 \, {\left (2 \, x + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 28, normalized size = 1.12
method | result | size |
norman | \(\frac {-x -\frac {{\mathrm e}^{5} x \ln \left (\ln \left (5 x \right )\right )}{6}-2 x^{2}}{2 x +1}\) | \(28\) |
risch | \(\frac {{\mathrm e}^{5} \ln \left (\ln \left (5 x \right )\right )}{24 x +12}-x -\frac {{\mathrm e}^{5} \ln \left (\ln \left (5 x \right )\right )}{12}\) | \(30\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.50, size = 28, normalized size = 1.12 \begin {gather*} -\frac {x e^{5} \log \left (\log \relax (5) + \log \relax (x)\right ) + 12 \, x^{2} + 6 \, x}{6 \, {\left (2 \, x + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.55, size = 23, normalized size = 0.92 \begin {gather*} -\frac {x\,\left (12\,x+{\mathrm {e}}^5\,\ln \left (\ln \left (5\,x\right )\right )+6\right )}{6\,\left (2\,x+1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.39, size = 27, normalized size = 1.08 \begin {gather*} - x - \frac {e^{5} \log {\left (\log {\left (5 x \right )} \right )}}{12} + \frac {e^{5} \log {\left (\log {\left (5 x \right )} \right )}}{24 x + 12} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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