Integrand size = 122, antiderivative size = 29 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {2 \log (\log (4))}{x-\frac {e^{\frac {x^3}{2+x}}}{4 \log (\log (2))}} \]
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Time = 0.41 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {12, 6820, 6818} \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=-\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{x+2}}-4 x \log (\log (2))} \]
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Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \log (\log (4)) \int \frac {e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx \\ & = \log (\log (4)) \int \frac {16 \log (\log (2)) \left (e^{\frac {x^3}{2+x}} x^2 (3+x)-2 (2+x)^2 \log (\log (2))\right )}{(2+x)^2 \left (e^{\frac {x^3}{2+x}}-4 x \log (\log (2))\right )^2} \, dx \\ & = (16 \log (\log (2)) \log (\log (4))) \int \frac {e^{\frac {x^3}{2+x}} x^2 (3+x)-2 (2+x)^2 \log (\log (2))}{(2+x)^2 \left (e^{\frac {x^3}{2+x}}-4 x \log (\log (2))\right )^2} \, dx \\ & = -\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{2+x}}-4 x \log (\log (2))} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=-\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{2+x}}-4 x \log (\log (2))} \]
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Time = 0.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10
method | result | size |
parallelrisch | \(\frac {8 \ln \left (2 \ln \left (2\right )\right ) \ln \left (\ln \left (2\right )\right )}{4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}}\) | \(32\) |
risch | \(\frac {8 \ln \left (\ln \left (2\right )\right )^{2}}{4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}}+\frac {8 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )}{4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}}\) | \(58\) |
norman | \(\frac {\left (8 \ln \left (\ln \left (2\right )\right )^{2}+8 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )\right ) x +16 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )+16 \ln \left (\ln \left (2\right )\right )^{2}}{\left (2+x \right ) \left (4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}\right )}\) | \(60\) |
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Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {8 \, {\left (\log \left (2\right ) \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2}\right )}}{4 \, x \log \left (\log \left (2\right )\right ) - e^{\left (\frac {x^{3}}{x + 2}\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {- 8 \log {\left (\log {\left (2 \right )} \right )}^{2} - 8 \log {\left (2 \right )} \log {\left (\log {\left (2 \right )} \right )}}{- 4 x \log {\left (\log {\left (2 \right )} \right )} + e^{\frac {x^{3}}{x + 2}}} \]
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Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {8 \, e^{\left (2 \, x + \frac {8}{x + 2}\right )} \log \left (2 \, \log \left (2\right )\right ) \log \left (\log \left (2\right )\right )}{4 \, x e^{\left (2 \, x + \frac {8}{x + 2}\right )} \log \left (\log \left (2\right )\right ) - e^{\left (x^{2} + 4\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {8 \, \log \left (2 \, \log \left (2\right )\right ) \log \left (\log \left (2\right )\right )}{4 \, x \log \left (\log \left (2\right )\right ) - e^{\left (\frac {x^{3}}{x + 2}\right )}} \]
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Time = 8.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=-\frac {8\,\ln \left (2\,\ln \left (2\right )\right )\,\ln \left (\ln \left (2\right )\right )}{{\mathrm {e}}^{\frac {x^3}{x+2}}-4\,x\,\ln \left (\ln \left (2\right )\right )} \]
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