Optimal. Leaf size=22 \[ \log \left (\frac {5 x \left (1+x^2\right )}{e^6 \left (-2+\frac {5}{\log (x)}\right )}\right ) \]
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Rubi [A]
time = 0.94, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps
used = 10, number of rules used = 6, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6873, 6857,
457, 78, 2339, 29} \begin {gather*} \log \left (x^2+1\right )+\log (x)-\log (5-2 \log (x))+\log (\log (x)) \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 78
Rule 457
Rule 2339
Rule 6857
Rule 6873
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5+5 x^2-\left (-5-15 x^2\right ) \log (x)-\left (2+6 x^2\right ) \log ^2(x)}{x \left (1+x^2\right ) (5-2 \log (x)) \log (x)} \, dx\\ &=\int \left (\frac {1+3 x^2}{x \left (1+x^2\right )}+\frac {1}{x \log (x)}-\frac {2}{x (-5+2 \log (x))}\right ) \, dx\\ &=-\left (2 \int \frac {1}{x (-5+2 \log (x))} \, dx\right )+\int \frac {1+3 x^2}{x \left (1+x^2\right )} \, dx+\int \frac {1}{x \log (x)} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {1+3 x}{x (1+x)} \, dx,x,x^2\right )+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )-\text {Subst}\left (\int \frac {1}{x} \, dx,x,-5+2 \log (x)\right )\\ &=-\log (5-2 \log (x))+\log (\log (x))+\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{x}+\frac {2}{1+x}\right ) \, dx,x,x^2\right )\\ &=\log (x)+\log \left (1+x^2\right )-\log (5-2 \log (x))+\log (\log (x))\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.07, size = 21, normalized size = 0.95 \begin {gather*} \log (x)+\log \left (1+x^2\right )-\log (5-2 \log (x))+\log (\log (x)) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.98, size = 22, normalized size = 1.00
method | result | size |
risch | \(\ln \left (x^{3}+x \right )-\ln \left (\ln \left (x \right )-\frac {5}{2}\right )+\ln \left (\ln \left (x \right )\right )\) | \(18\) |
default | \(\ln \left (x \right )-\ln \left (2 \ln \left (x \right )-5\right )+\ln \left (\ln \left (x \right )\right )+\ln \left (x^{2}+1\right )\) | \(22\) |
norman | \(\ln \left (x \right )-\ln \left (2 \ln \left (x \right )-5\right )+\ln \left (\ln \left (x \right )\right )+\ln \left (x^{2}+1\right )\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.54, size = 19, normalized size = 0.86 \begin {gather*} \log \left (x^{2} + 1\right ) + \log \left (x\right ) - \log \left (\log \left (x\right ) - \frac {5}{2}\right ) + \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 = 0.35, size = 19, normalized size = 0.86 \begin {gather*} \log \left (x^{3} + x\right ) - \log \left (2 \, \log \left (x\right ) - 5\right ) + \log \left (\log \left (x\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.09, size = 19, normalized size = 0.86 \begin {gather*} \log {\left (x^{3} + x \right )} - \log {\left (\log {\left (x \right )} - \frac {5}{2} \right )} + \log {\left (\log {\left (x \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 21, normalized size = 0.95 \begin {gather*} \log \left (x^{2} + 1\right ) + \log \left (x\right ) - \log \left (2 \, \log \left (x\right ) - 5\right ) + \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 = 3.35, size = 21, normalized size = 0.95 \begin {gather*} \ln \left (x\right )-\ln \left (2\,\ln \left (x\right )-5\right )+\ln \left (\ln \left (x\right )\,\left (x^2+1\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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