Optimal. Leaf size=24 \[ x+\frac {\log (\log (4))}{2 x+\frac {x}{1+2 x}+\log (\log (x))} \]
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Rubi [F] time = 1.55, 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 {\left (9 x^3+24 x^4+16 x^5\right ) \log (x)+\left (-1-4 x-4 x^2+\left (-3 x-8 x^2-8 x^3\right ) \log (x)\right ) \log (\log (4))+\left (6 x^2+20 x^3+16 x^4\right ) \log (x) \log (\log (x))+\left (x+4 x^2+4 x^3\right ) \log (x) \log ^2(\log (x))}{\left (9 x^3+24 x^4+16 x^5\right ) \log (x)+\left (6 x^2+20 x^3+16 x^4\right ) \log (x) \log (\log (x))+\left (x+4 x^2+4 x^3\right ) \log (x) \log ^2(\log (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 {-(1+2 x)^2 \log (\log (4))+x \log (x) \left (24 x^3+16 x^4+x^2 (9-8 \log (\log (4)))-3 \log (\log (4))-8 x \log (\log (4))+2 x \left (3+10 x+8 x^2\right ) \log (\log (x))+(1+2 x)^2 \log ^2(\log (x))\right )}{x \log (x) (x (3+4 x)+(1+2 x) \log (\log (x)))^2} \, dx\\ &=\int \left (1-\frac {\left (1+4 x+4 x^2+3 x \log (x)+8 x^2 \log (x)+8 x^3 \log (x)\right ) \log (\log (4))}{x \log (x) \left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2}\right ) \, dx\\ &=x-\log (\log (4)) \int \frac {1+4 x+4 x^2+3 x \log (x)+8 x^2 \log (x)+8 x^3 \log (x)}{x \log (x) \left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2} \, dx\\ &=x-\log (\log (4)) \int \left (\frac {3}{\left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2}+\frac {8 x}{\left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2}+\frac {8 x^2}{\left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2}+\frac {4}{\log (x) \left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2}+\frac {1}{x \log (x) \left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2}+\frac {4 x}{\log (x) \left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2}\right ) \, dx\\ &=x-\log (\log (4)) \int \frac {1}{x \log (x) \left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2} \, dx-(3 \log (\log (4))) \int \frac {1}{\left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2} \, dx-(4 \log (\log (4))) \int \frac {1}{\log (x) \left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2} \, dx-(4 \log (\log (4))) \int \frac {x}{\log (x) \left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2} \, dx-(8 \log (\log (4))) \int \frac {x}{\left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2} \, dx-(8 \log (\log (4))) \int \frac {x^2}{\left (3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 31, normalized size = 1.29 \begin {gather*} x+\frac {(1+2 x) \log (\log (4))}{3 x+4 x^2+\log (\log (x))+2 x \log (\log (x))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.69, size = 54, normalized size = 2.25 \begin {gather*} \frac {4 \, x^{3} + 3 \, x^{2} + {\left (2 \, x + 1\right )} \log \left (2 \, \log \relax (2)\right ) + {\left (2 \, x^{2} + x\right )} \log \left (\log \relax (x)\right )}{4 \, x^{2} + {\left (2 \, x + 1\right )} \log \left (\log \relax (x)\right ) + 3 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.28, size = 40, normalized size = 1.67 \begin {gather*} x + \frac {2 \, x \log \relax (2) + 2 \, x \log \left (\log \relax (2)\right ) + \log \relax (2) + \log \left (\log \relax (2)\right )}{4 \, x^{2} + 2 \, x \log \left (\log \relax (x)\right ) + 3 \, x + \log \left (\log \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 35, normalized size = 1.46
method | result | size |
risch | \(x +\frac {\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) \left (2 x +1\right )}{2 x \ln \left (\ln \relax (x )\right )+4 x^{2}+\ln \left (\ln \relax (x )\right )+3 x}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.50, size = 57, normalized size = 2.38 \begin {gather*} \frac {4 \, x^{3} + 3 \, x^{2} + 2 \, x {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} + {\left (2 \, x^{2} + x\right )} \log \left (\log \relax (x)\right ) + \log \relax (2) + \log \left (\log \relax (2)\right )}{4 \, x^{2} + {\left (2 \, x + 1\right )} \log \left (\log \relax (x)\right ) + 3 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \relax (x)\,\left (4\,x^3+4\,x^2+x\right )\,{\ln \left (\ln \relax (x)\right )}^2+\ln \relax (x)\,\left (16\,x^4+20\,x^3+6\,x^2\right )\,\ln \left (\ln \relax (x)\right )+\ln \relax (x)\,\left (16\,x^5+24\,x^4+9\,x^3\right )-\ln \left (2\,\ln \relax (2)\right )\,\left (4\,x+4\,x^2+\ln \relax (x)\,\left (8\,x^3+8\,x^2+3\,x\right )+1\right )}{\ln \relax (x)\,\left (4\,x^3+4\,x^2+x\right )\,{\ln \left (\ln \relax (x)\right )}^2+\ln \relax (x)\,\left (16\,x^4+20\,x^3+6\,x^2\right )\,\ln \left (\ln \relax (x)\right )+\ln \relax (x)\,\left (16\,x^5+24\,x^4+9\,x^3\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.45, size = 42, normalized size = 1.75 \begin {gather*} x + \frac {2 x \log {\left (\log {\relax (2 )} \right )} + 2 x \log {\relax (2 )} + \log {\left (\log {\relax (2 )} \right )} + \log {\relax (2 )}}{4 x^{2} + 3 x + \left (2 x + 1\right ) \log {\left (\log {\relax (x )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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