Optimal. Leaf size=28 \[ \frac {5+\frac {1}{2} \left (x+\log (x)+(1+x)^2 \log \left (\frac {\log (5)}{25}\right )\right )}{\log (x)} \]
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Rubi [F] time = 0.38, 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 {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \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} &=\frac {1}{2} \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{x \log ^2(x)} \, dx\\ &=\frac {1}{2} \int \left (\frac {1+2 \log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)}+\frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {1+2 \log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)} \, dx+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx\\ &=\frac {1}{2} \int \left (\frac {2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)}+\frac {1+2 \log \left (\frac {\log (5)}{25}\right )}{\log (x)}\right ) \, dx+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx\\ &=\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx+\log \left (\frac {\log (5)}{25}\right ) \int \frac {x}{\log (x)} \, dx+\frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \int \frac {1}{\log (x)} \, dx\\ &=\frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \text {li}(x)+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx+\log \left (\frac {\log (5)}{25}\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\text {Ei}(2 \log (x)) \log \left (\frac {\log (5)}{25}\right )+\frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \text {li}(x)+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 39, normalized size = 1.39 \begin {gather*} \frac {10+x+\log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )+x^2 \log \left (\frac {\log (5)}{25}\right )}{2 \log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.92, size = 23, normalized size = 0.82 \begin {gather*} \frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (\frac {1}{25} \, \log \relax (5)\right ) + x + 10}{2 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.74, size = 46, normalized size = 1.64 \begin {gather*} -\frac {2 \, x^{2} \log \relax (5) - x^{2} \log \left (\log \relax (5)\right ) + 4 \, x \log \relax (5) - 2 \, x \log \left (\log \relax (5)\right ) - x + 2 \, \log \relax (5) - \log \left (\log \relax (5)\right ) - 10}{2 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 42, normalized size = 1.50
method | result | size |
norman | \(\frac {\left (-\ln \relax (5)+\frac {\ln \left (\ln \relax (5)\right )}{2}\right ) x^{2}+\left (-2 \ln \relax (5)+\ln \left (\ln \relax (5)\right )+\frac {1}{2}\right ) x +5-\ln \relax (5)+\frac {\ln \left (\ln \relax (5)\right )}{2}}{\ln \relax (x )}\) | \(42\) |
risch | \(-\frac {2 x^{2} \ln \relax (5)-\ln \left (\ln \relax (5)\right ) x^{2}+4 x \ln \relax (5)-2 x \ln \left (\ln \relax (5)\right )+2 \ln \relax (5)-\ln \left (\ln \relax (5)\right )-x -10}{2 \ln \relax (x )}\) | \(47\) |
default | \(2 \ln \relax (5) \expIntegralEi \left (1, -2 \ln \relax (x )\right )-\ln \left (\ln \relax (5)\right ) \expIntegralEi \left (1, -2 \ln \relax (x )\right )+2 \ln \relax (5) \expIntegralEi \left (1, -\ln \relax (x )\right )-\ln \left (\ln \relax (5)\right ) \expIntegralEi \left (1, -\ln \relax (x )\right )+\ln \relax (5) \left (-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )-\frac {\ln \left (\ln \relax (5)\right ) \left (-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )}{2}+2 \ln \relax (5) \left (-\frac {x}{\ln \relax (x )}-\expIntegralEi \left (1, -\ln \relax (x )\right )\right )-\ln \left (\ln \relax (5)\right ) \left (-\frac {x}{\ln \relax (x )}-\expIntegralEi \left (1, -\ln \relax (x )\right )\right )-\frac {\ln \relax (5)}{\ln \relax (x )}+\frac {\ln \left (\ln \relax (5)\right )}{2 \ln \relax (x )}+\frac {x}{2 \ln \relax (x )}+\frac {5}{\ln \relax (x )}\) | \(159\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 1.00, size = 77, normalized size = 2.75 \begin {gather*} {\rm Ei}\left (2 \, \log \relax (x)\right ) \log \left (\frac {1}{25} \, \log \relax (5)\right ) + {\rm Ei}\left (\log \relax (x)\right ) \log \left (\frac {1}{25} \, \log \relax (5)\right ) - \Gamma \left (-1, -\log \relax (x)\right ) \log \left (\frac {1}{25} \, \log \relax (5)\right ) - \Gamma \left (-1, -2 \, \log \relax (x)\right ) \log \left (\frac {1}{25} \, \log \relax (5)\right ) + \frac {\log \left (\frac {1}{25} \, \log \relax (5)\right )}{2 \, \log \relax (x)} + \frac {5}{\log \relax (x)} + \frac {1}{2} \, {\rm Ei}\left (\log \relax (x)\right ) - \frac {1}{2} \, \Gamma \left (-1, -\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.46, size = 43, normalized size = 1.54 \begin {gather*} \frac {\ln \left (\frac {\ln \relax (5)}{25}\right )\,x^4+\left (2\,\ln \left (\frac {\ln \relax (5)}{25}\right )+1\right )\,x^3+\left (\ln \left (\frac {\ln \relax (5)}{25}\right )+10\right )\,x^2}{2\,x^2\,\ln \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.14, size = 49, normalized size = 1.75 \begin {gather*} \frac {- 2 x^{2} \log {\relax (5 )} + x^{2} \log {\left (\log {\relax (5 )} \right )} - 4 x \log {\relax (5 )} + 2 x \log {\left (\log {\relax (5 )} \right )} + x - 2 \log {\relax (5 )} + \log {\left (\log {\relax (5 )} \right )} + 10}{2 \log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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