Integrand size = 66, antiderivative size = 35 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=x-\frac {3 \left (2+x+\frac {1}{2} (i \pi +\log (-\log (\log (2))))^2\right )}{4+\log \left (4 x^2\right )} \]
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Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6873, 6874, 2337, 2209, 2395, 2334, 2339, 30} \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=-\frac {3 \left (4-\pi ^2+\log ^2(-\log (\log (2)))+2 i \pi \log (-\log (\log (2)))\right )}{2 \left (\log \left (4 x^2\right )+4\right )}-\frac {3 x}{\log \left (4 x^2\right )+4}+x \]
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Rule 30
Rule 2209
Rule 2334
Rule 2337
Rule 2339
Rule 2395
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+12 \left (1+\frac {1}{4} (i \pi +\log (-\log (\log (2))))^2\right )}{x \left (4+\log \left (4 x^2\right )\right )^2} \, dx \\ & = \int \left (1-\frac {3}{4+\log \left (4 x^2\right )}+\frac {3 \left (4-\pi ^2+2 x+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))\right )}{x \left (4+\log \left (4 x^2\right )\right )^2}\right ) \, dx \\ & = x-3 \int \frac {1}{4+\log \left (4 x^2\right )} \, dx+3 \int \frac {4-\pi ^2+2 x+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))}{x \left (4+\log \left (4 x^2\right )\right )^2} \, dx \\ & = x+3 \int \left (\frac {2}{\left (4+\log \left (4 x^2\right )\right )^2}+\frac {4-\pi ^2+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))}{x \left (4+\log \left (4 x^2\right )\right )^2}\right ) \, dx-\frac {(3 x) \text {Subst}\left (\int \frac {e^{x/2}}{4+x} \, dx,x,\log \left (4 x^2\right )\right )}{4 \sqrt {x^2}} \\ & = x-\frac {3 x \operatorname {ExpIntegralEi}\left (\frac {1}{2} \left (4+\log \left (4 x^2\right )\right )\right )}{4 e^2 \sqrt {x^2}}+6 \int \frac {1}{\left (4+\log \left (4 x^2\right )\right )^2} \, dx+\left (3 \left (4-\pi ^2+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))\right )\right ) \int \frac {1}{x \left (4+\log \left (4 x^2\right )\right )^2} \, dx \\ & = x-\frac {3 x \operatorname {ExpIntegralEi}\left (\frac {1}{2} \left (4+\log \left (4 x^2\right )\right )\right )}{4 e^2 \sqrt {x^2}}-\frac {3 x}{4+\log \left (4 x^2\right )}+3 \int \frac {1}{4+\log \left (4 x^2\right )} \, dx+\frac {1}{2} \left (3 \left (4-\pi ^2+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))\right )\right ) \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,4+\log \left (4 x^2\right )\right ) \\ & = x-\frac {3 x \operatorname {ExpIntegralEi}\left (\frac {1}{2} \left (4+\log \left (4 x^2\right )\right )\right )}{4 e^2 \sqrt {x^2}}-\frac {3 x}{4+\log \left (4 x^2\right )}-\frac {3 \left (4-\pi ^2+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))\right )}{2 \left (4+\log \left (4 x^2\right )\right )}+\frac {(3 x) \text {Subst}\left (\int \frac {e^{x/2}}{4+x} \, dx,x,\log \left (4 x^2\right )\right )}{4 \sqrt {x^2}} \\ & = x-\frac {3 x}{4+\log \left (4 x^2\right )}-\frac {3 \left (4-\pi ^2+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))\right )}{2 \left (4+\log \left (4 x^2\right )\right )} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=x+\frac {3 \left (-4+\pi ^2-2 x-2 i \pi \log (-\log (\log (2)))-\log ^2(-\log (\log (2)))\right )}{2 \left (4+\log \left (4 x^2\right )\right )} \]
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Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74
method | result | size |
risch | \(x -\frac {3 \left (\ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}+2 x +4\right )}{2 \left (\ln \left (4 x^{2}\right )+4\right )}\) | \(26\) |
norman | \(\frac {x +x \ln \left (4 x^{2}\right )-6-\frac {3 \ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}}{2}}{\ln \left (4 x^{2}\right )+4}\) | \(31\) |
parallelrisch | \(-\frac {12+3 \ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}-2 x \ln \left (4 x^{2}\right )-2 x}{2 \left (\ln \left (4 x^{2}\right )+4\right )}\) | \(35\) |
default | \(\frac {x \ln \left (x^{2}\right )+\left (1+2 \ln \left (2\right )\right ) x -6}{2 \ln \left (2\right )+\ln \left (x^{2}\right )+4}-\frac {3 \ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}}{2 \left (2 \ln \left (2\right )+\ln \left (x^{2}\right )+4\right )}\) | \(51\) |
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=\frac {2 \, x \log \left (4 \, x^{2}\right ) - 3 \, \log \left (\log \left (\log \left (2\right )\right )\right )^{2} + 2 \, x - 12}{2 \, {\left (\log \left (4 \, x^{2}\right ) + 4\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=x + \frac {- 6 x - 12 - 3 \log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}^{2} + 3 \pi ^{2} - 6 i \pi \log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}}{2 \log {\left (x^{2} \right )} + 4 \log {\left (2 \right )} + 8} \]
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Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (25) = 50\).
Time = 0.30 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.46 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=-\frac {3 \, \log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{4 \, {\left (\log \left (2\right ) + \log \left (x\right ) + 2\right )}} + \frac {x {\left (2 \, \log \left (2\right ) + 1\right )} + 2 \, x \log \left (x\right )}{2 \, {\left (\log \left (2\right ) + \log \left (x\right ) + 2\right )}} - \frac {3}{\log \left (2\right ) + \log \left (x\right ) + 2} \]
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Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=x - \frac {3 \, {\left (\log \left (\log \left (\log \left (2\right )\right )\right )^{2} + 2 \, x + 4\right )}}{2 \, {\left (2 \, \log \left (2\right ) + \log \left (x^{2}\right ) + 4\right )}} \]
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Time = 11.81 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=\frac {x\,\left (\ln \left (4\,x^2\right )+1\right )}{\ln \left (4\,x^2\right )+4}+\frac {\ln \left (4\,x^2\right )\,\left (\frac {3\,{\ln \left (\ln \left (\ln \left (2\right )\right )\right )}^2}{8}+\frac {3}{2}\right )}{\ln \left (4\,x^2\right )+4} \]
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