Integrand size = 94, antiderivative size = 25 \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\log \left (4+\frac {e^x}{x+\log (3)}+\frac {5+x}{4 \log (\log (4))}\right ) \]
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\[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{(x+\log (3)) \left (x^2+4 e^x \log (\log (4))+5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )+5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )\right )} \, dx \\ & = \int \left (\frac {-1+x+\log (3)}{x+\log (3)}+\frac {5-x^2-\log (81)+16 (1-\log (3)) \log (\log (4))-x (3+\log (3)+16 \log (\log (4)))}{x^2+4 e^x \log (\log (4))+5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )+5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )}\right ) \, dx \\ & = \int \frac {-1+x+\log (3)}{x+\log (3)} \, dx+\int \frac {5-x^2-\log (81)+16 (1-\log (3)) \log (\log (4))-x (3+\log (3)+16 \log (\log (4)))}{x^2+4 e^x \log (\log (4))+5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )+5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )} \, dx \\ & = \int \left (1+\frac {1}{-x-\log (3)}\right ) \, dx+\int \left (\frac {x^2}{-x^2-4 e^x \log (\log (4))-5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )-5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )}+\frac {\log (81) \left (1+\frac {-5+16 (-1+\log (3)) \log (\log (4))}{\log (81)}\right )}{-x^2-4 e^x \log (\log (4))-5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )-5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )}+\frac {x (-3-\log (3)-16 \log (\log (4)))}{x^2+4 e^x \log (\log (4))+5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )+5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )}\right ) \, dx \\ & = x-\log (x+\log (3))+(-3-\log (3)-16 \log (\log (4))) \int \frac {x}{x^2+4 e^x \log (\log (4))+5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )+5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )} \, dx+(-5+\log (81)-16 (1-\log (3)) \log (\log (4))) \int \frac {1}{-x^2-4 e^x \log (\log (4))-5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )-5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )} \, dx+\int \frac {x^2}{-x^2-4 e^x \log (\log (4))-5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )-5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )} \, dx \\ \end{align*}
\[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. \(50\) vs. \(2(24)=48\).
Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04
method | result | size |
norman | \(-\ln \left (\ln \left (3\right )+x \right )+\ln \left (16 \ln \left (3\right ) \ln \left (2 \ln \left (2\right )\right )+x \ln \left (3\right )+16 x \ln \left (2 \ln \left (2\right )\right )+4 \,{\mathrm e}^{x} \ln \left (2 \ln \left (2\right )\right )+x^{2}+5 \ln \left (3\right )+5 x \right )\) | \(51\) |
parallelrisch | \(-\ln \left (\ln \left (3\right )+x \right )+\ln \left (16 \ln \left (3\right ) \ln \left (2 \ln \left (2\right )\right )+x \ln \left (3\right )+16 x \ln \left (2 \ln \left (2\right )\right )+4 \,{\mathrm e}^{x} \ln \left (2 \ln \left (2\right )\right )+x^{2}+5 \ln \left (3\right )+5 x \right )\) | \(51\) |
risch | \(-\ln \left (\ln \left (3\right )+x \right )+\ln \left ({\mathrm e}^{x}+\frac {16 \ln \left (2\right ) \ln \left (3\right )+16 \ln \left (3\right ) \ln \left (\ln \left (2\right )\right )+x \ln \left (3\right )+16 x \ln \left (2\right )+16 x \ln \left (\ln \left (2\right )\right )+x^{2}+5 \ln \left (3\right )+5 x}{4 \ln \left (2\right )+4 \ln \left (\ln \left (2\right )\right )}\right )\) | \(62\) |
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\log \left (x^{2} + {\left (x + 5\right )} \log \left (3\right ) + 4 \, {\left (4 \, x + e^{x} + 4 \, \log \left (3\right )\right )} \log \left (2 \, \log \left (2\right )\right ) + 5 \, x\right ) - \log \left (x + \log \left (3\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (24) = 48\).
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.92 \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=- \log {\left (x + \log {\left (3 \right )} \right )} + \log {\left (\frac {x^{2} + 16 x \log {\left (\log {\left (2 \right )} \right )} + x \log {\left (3 \right )} + 5 x + 16 x \log {\left (2 \right )} + 16 \log {\left (3 \right )} \log {\left (\log {\left (2 \right )} \right )} + 5 \log {\left (3 \right )} + 16 \log {\left (2 \right )} \log {\left (3 \right )}}{4 \log {\left (\log {\left (2 \right )} \right )} + 4 \log {\left (2 \right )}} + e^{x} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (24) = 48\).
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.56 \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=-\log \left (x + \log \left (3\right )\right ) + \log \left (\frac {x^{2} + x {\left (\log \left (3\right ) + 16 \, \log \left (2\right ) + 16 \, \log \left (\log \left (2\right )\right ) + 5\right )} + 4 \, {\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )} e^{x} + {\left (16 \, \log \left (\log \left (2\right )\right ) + 5\right )} \log \left (3\right ) + 16 \, \log \left (3\right ) \log \left (2\right )}{4 \, {\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )}}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (24) = 48\).
Time = 0.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.44 \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\log \left (x^{2} + x \log \left (3\right ) + 16 \, x \log \left (2\right ) + 4 \, e^{x} \log \left (2\right ) + 16 \, \log \left (3\right ) \log \left (2\right ) + 16 \, x \log \left (\log \left (2\right )\right ) + 4 \, e^{x} \log \left (\log \left (2\right )\right ) + 16 \, \log \left (3\right ) \log \left (\log \left (2\right )\right ) + 5 \, x + 5 \, \log \left (3\right )\right ) - \log \left (x + \log \left (3\right )\right ) \]
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Timed out. \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\int \frac {2\,x\,\ln \left (3\right )+{\ln \left (3\right )}^2+x^2+\ln \left (2\,\ln \left (2\right )\right )\,{\mathrm {e}}^x\,\left (4\,x+4\,\ln \left (3\right )-4\right )}{{\ln \left (3\right )}^2\,\left (x+5\right )+\ln \left (3\right )\,\left (2\,x^2+10\,x\right )+5\,x^2+x^3+\ln \left (2\,\ln \left (2\right )\right )\,\left ({\mathrm {e}}^x\,\left (4\,x+4\,\ln \left (3\right )\right )+32\,x\,\ln \left (3\right )+16\,{\ln \left (3\right )}^2+16\,x^2\right )} \,d x \]
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