Integrand size = 90, antiderivative size = 27 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=\log \left (\frac {8 \left (e^4+\left (-x+\frac {\log \left (\log ^2(x)\right )}{x}\right )^2\right )^4}{x^2}\right ) \]
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Time = 0.49 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6873, 6874, 6816} \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=4 \log \left (x^4+e^4 x^2-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )-10 \log (x) \]
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Rule 6816
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{x \log (x) \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )} \, dx \\ & = \int \left (-\frac {10}{x}+\frac {8 \left (-2 x^2+e^4 x^2 \log (x)+2 x^4 \log (x)+2 \log \left (\log ^2(x)\right )-2 x^2 \log (x) \log \left (\log ^2(x)\right )\right )}{x \log (x) \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )}\right ) \, dx \\ & = -10 \log (x)+8 \int \frac {-2 x^2+e^4 x^2 \log (x)+2 x^4 \log (x)+2 \log \left (\log ^2(x)\right )-2 x^2 \log (x) \log \left (\log ^2(x)\right )}{x \log (x) \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )} \, dx \\ & = -10 \log (x)+4 \log \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=-2 \left (5 \log (x)-2 \log \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )\right ) \]
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Time = 8.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33
method | result | size |
parallelrisch | \(-10 \ln \left (x \right )+4 \ln \left (x^{4}+x^{2} {\mathrm e}^{4}-2 x^{2} \ln \left (\ln \left (x \right )^{2}\right )+\ln \left (\ln \left (x \right )^{2}\right )^{2}\right )\) | \(36\) |
risch | \(-10 \ln \left (x \right )+4 \ln \left (-\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{4} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}}{16}+\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{3} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}}{4}-\frac {3 \pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{4}}{8}+\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{5}}{4}-\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{6}}{16}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )}{4}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}}{4}+\frac {x^{4}}{4}+\frac {x^{2} {\mathrm e}^{4}}{4}+\left (-\frac {i \pi \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )}{2}+i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}-\frac {i \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}}{2}-x^{2}\right ) \ln \left (\ln \left (x \right )\right )+\ln \left (\ln \left (x \right )\right )^{2}\right )\) | \(259\) |
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Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=4 \, \log \left (x^{4} + x^{2} e^{4} - 2 \, x^{2} \log \left (\log \left (x\right )^{2}\right ) + \log \left (\log \left (x\right )^{2}\right )^{2}\right ) - 10 \, \log \left (x\right ) \]
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Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=- 10 \log {\left (x \right )} + 4 \log {\left (x^{4} - 2 x^{2} \log {\left (\log {\left (x \right )}^{2} \right )} + x^{2} e^{4} + \log {\left (\log {\left (x \right )}^{2} \right )}^{2} \right )} \]
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=4 \, \log \left (\frac {1}{4} \, x^{4} + \frac {1}{4} \, x^{2} e^{4} - x^{2} \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2}\right ) - 10 \, \log \left (x\right ) \]
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Time = 0.36 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=4 \, \log \left (x^{4} + x^{2} e^{4} - 2 \, x^{2} \log \left (\log \left (x\right )^{2}\right ) + \log \left (\log \left (x\right )^{2}\right )^{2}\right ) - 10 \, \log \left (x\right ) \]
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Timed out. \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=\int -\frac {10\,{\ln \left ({\ln \left (x\right )}^2\right )}^2\,\ln \left (x\right )+\ln \left (x\right )\,\left (2\,x^2\,{\mathrm {e}}^4-6\,x^4\right )-\ln \left ({\ln \left (x\right )}^2\right )\,\left (4\,x^2\,\ln \left (x\right )+16\right )+16\,x^2}{\ln \left (x\right )\,\left (x^5+{\mathrm {e}}^4\,x^3\right )+x\,{\ln \left ({\ln \left (x\right )}^2\right )}^2\,\ln \left (x\right )-2\,x^3\,\ln \left ({\ln \left (x\right )}^2\right )\,\ln \left (x\right )} \,d x \]
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