Optimal. Leaf size=22 \[ -2 x-\frac {\left (5+x^2\right )^2}{64 \log \left (\frac {1}{\log (x)}\right )} \]
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Rubi [F]
time = 0.65, 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 {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{64} \int \frac {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx\\ &=\frac {1}{64} \int \frac {-\frac {\left (5+x^2\right )^2}{x \log (x)}-4 \log \left (\frac {1}{\log (x)}\right ) \left (5 x+x^3+32 \log \left (\frac {1}{\log (x)}\right )\right )}{\log ^2\left (\frac {1}{\log (x)}\right )} \, dx\\ &=\frac {1}{64} \int \left (-128-\frac {\left (5+x^2\right )^2}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}-\frac {4 x \left (5+x^2\right )}{\log \left (\frac {1}{\log (x)}\right )}\right ) \, dx\\ &=-2 x-\frac {1}{64} \int \frac {\left (5+x^2\right )^2}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {1}{16} \int \frac {x \left (5+x^2\right )}{\log \left (\frac {1}{\log (x)}\right )} \, dx\\ &=-2 x-\frac {1}{64} \int \left (\frac {25}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}+\frac {10 x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )}+\frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )}\right ) \, dx-\frac {1}{16} \int \left (\frac {5 x}{\log \left (\frac {1}{\log (x)}\right )}+\frac {x^3}{\log \left (\frac {1}{\log (x)}\right )}\right ) \, dx\\ &=-2 x-\frac {1}{64} \int \frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {1}{16} \int \frac {x^3}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{32} \int \frac {x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{16} \int \frac {x}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {25}{64} \int \frac {1}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx\\ &=-2 x-\frac {1}{64} \int \frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {1}{16} \int \frac {x^3}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{32} \int \frac {x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{16} \int \frac {x}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {25}{64} \text {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {1}{x}\right )} \, dx,x,\log (x)\right )\\ &=-2 x-\frac {1}{64} \int \frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {1}{16} \int \frac {x^3}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{32} \int \frac {x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{16} \int \frac {x}{\log \left (\frac {1}{\log (x)}\right )} \, dx+\frac {25}{64} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {1}{\log (x)}\right )\right )\\ &=-2 x-\frac {25}{64 \log \left (\frac {1}{\log (x)}\right )}-\frac {1}{64} \int \frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {1}{16} \int \frac {x^3}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{32} \int \frac {x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{16} \int \frac {x}{\log \left (\frac {1}{\log (x)}\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.11, size = 22, normalized size = 1.00 \begin {gather*} -2 x-\frac {\left (5+x^2\right )^2}{64 \log \left (\frac {1}{\log (x)}\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.66, size = 22, normalized size = 1.00
method | result | size |
risch | \(-2 x +\frac {x^{4}+10 x^{2}+25}{64 \ln \left (\ln \left (x \right )\right )}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.31, size = 27, normalized size = 1.23 \begin {gather*} -2 \, x + \frac {x^{4} + 10 \, x^{2}}{64 \, \log \left (\log \left (x\right )\right )} + \frac {25}{64 \, \log \left (\log \left (x\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 27, normalized size = 1.23 \begin {gather*} -\frac {x^{4} + 10 \, x^{2} + 128 \, x \log \left (\frac {1}{\log \left (x\right )}\right ) + 25}{64 \, \log \left (\frac {1}{\log \left (x\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.06, size = 22, normalized size = 1.00 \begin {gather*} - 2 x + \frac {- x^{4} - 10 x^{2} - 25}{64 \log {\left (\frac {1}{\log {\left (x \right )}} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.42, size = 21, normalized size = 0.95 \begin {gather*} -2 \, x + \frac {x^{4} + 10 \, x^{2} + 25}{64 \, \log \left (\log \left (x\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.60, size = 27, normalized size = 1.23 \begin {gather*} -\frac {128\,x\,\ln \left (\frac {1}{\ln \left (x\right )}\right )+10\,x^2+x^4+25}{64\,\ln \left (\frac {1}{\ln \left (x\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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