Integrand size = 107, antiderivative size = 26 \[ \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{\left (4+4 x-63 x^2-64 x^3+236 x^4+254 x^5+96 x^6+16 x^7+x^8\right ) \log (2)} \, dx=\frac {e^{-\frac {5}{-2-x+x^2 (4+x)^2}} x}{\log (2)} \]
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\[ \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{\left (4+4 x-63 x^2-64 x^3+236 x^4+254 x^5+96 x^6+16 x^7+x^8\right ) \log (2)} \, dx=\int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{\left (4+4 x-63 x^2-64 x^3+236 x^4+254 x^5+96 x^6+16 x^7+x^8\right ) \log (2)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{4+4 x-63 x^2-64 x^3+236 x^4+254 x^5+96 x^6+16 x^7+x^8} \, dx}{\log (2)} \\ & = \frac {\int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{\left (2+x-16 x^2-8 x^3-x^4\right )^2} \, dx}{\log (2)} \\ & = \frac {\int \left (e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}}-\frac {5 e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (-8-3 x+32 x^2+8 x^3\right )}{\left (-2-x+16 x^2+8 x^3+x^4\right )^2}+\frac {20 e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}}}{-2-x+16 x^2+8 x^3+x^4}\right ) \, dx}{\log (2)} \\ & = \frac {\int e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \, dx}{\log (2)}-\frac {5 \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (-8-3 x+32 x^2+8 x^3\right )}{\left (-2-x+16 x^2+8 x^3+x^4\right )^2} \, dx}{\log (2)}+\frac {20 \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}}}{-2-x+16 x^2+8 x^3+x^4} \, dx}{\log (2)} \\ & = \frac {\int e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \, dx}{\log (2)}-\frac {5 \int \left (-\frac {8 e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}}}{\left (-2-x+16 x^2+8 x^3+x^4\right )^2}-\frac {3 e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} x}{\left (-2-x+16 x^2+8 x^3+x^4\right )^2}+\frac {32 e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} x^2}{\left (-2-x+16 x^2+8 x^3+x^4\right )^2}+\frac {8 e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} x^3}{\left (-2-x+16 x^2+8 x^3+x^4\right )^2}\right ) \, dx}{\log (2)}+\frac {20 \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}}}{-2-x+16 x^2+8 x^3+x^4} \, dx}{\log (2)} \\ & = \frac {\int e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \, dx}{\log (2)}+\frac {15 \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} x}{\left (-2-x+16 x^2+8 x^3+x^4\right )^2} \, dx}{\log (2)}+\frac {20 \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}}}{-2-x+16 x^2+8 x^3+x^4} \, dx}{\log (2)}+\frac {40 \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}}}{\left (-2-x+16 x^2+8 x^3+x^4\right )^2} \, dx}{\log (2)}-\frac {40 \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} x^3}{\left (-2-x+16 x^2+8 x^3+x^4\right )^2} \, dx}{\log (2)}-\frac {160 \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} x^2}{\left (-2-x+16 x^2+8 x^3+x^4\right )^2} \, dx}{\log (2)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{\left (4+4 x-63 x^2-64 x^3+236 x^4+254 x^5+96 x^6+16 x^7+x^8\right ) \log (2)} \, dx=\frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} x}{\log (2)} \]
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Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15
method | result | size |
risch | \(\frac {x \,{\mathrm e}^{-\frac {5}{x^{4}+8 x^{3}+16 x^{2}-x -2}}}{\ln \left (2\right )}\) | \(30\) |
gosper | \(\frac {x \,{\mathrm e}^{-\frac {5}{x^{4}+8 x^{3}+16 x^{2}-x -2}}}{\ln \left (2\right )}\) | \(32\) |
parallelrisch | \(\frac {\left (11224 x^{5}+89792 x^{4}+179584 x^{3}-11224 x^{2}-22448 x \right ) {\mathrm e}^{-\frac {5}{x^{4}+8 x^{3}+16 x^{2}-x -2}}}{11224 \ln \left (2\right ) \left (x^{4}+8 x^{3}+16 x^{2}-x -2\right )}\) | \(76\) |
norman | \(\frac {\left (\frac {x^{5}}{\ln \left (2\right )}-\frac {2 x}{\ln \left (2\right )}-\frac {x^{2}}{\ln \left (2\right )}+\frac {16 x^{3}}{\ln \left (2\right )}+\frac {8 x^{4}}{\ln \left (2\right )}\right ) {\mathrm e}^{-\frac {5}{x^{4}+8 x^{3}+16 x^{2}-x -2}}}{x^{4}+8 x^{3}+16 x^{2}-x -2}\) | \(90\) |
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{\left (4+4 x-63 x^2-64 x^3+236 x^4+254 x^5+96 x^6+16 x^7+x^8\right ) \log (2)} \, dx=\frac {x e^{\left (-\frac {5}{x^{4} + 8 \, x^{3} + 16 \, x^{2} - x - 2}\right )}}{\log \left (2\right )} \]
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Time = 0.70 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{\left (4+4 x-63 x^2-64 x^3+236 x^4+254 x^5+96 x^6+16 x^7+x^8\right ) \log (2)} \, dx=\frac {x e^{- \frac {5}{x^{4} + 8 x^{3} + 16 x^{2} - x - 2}}}{\log {\left (2 \right )}} \]
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Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{\left (4+4 x-63 x^2-64 x^3+236 x^4+254 x^5+96 x^6+16 x^7+x^8\right ) \log (2)} \, dx=\frac {x e^{\left (-\frac {5}{x^{4} + 8 \, x^{3} + 16 \, x^{2} - x - 2}\right )}}{\log \left (2\right )} \]
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Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{\left (4+4 x-63 x^2-64 x^3+236 x^4+254 x^5+96 x^6+16 x^7+x^8\right ) \log (2)} \, dx=\frac {x e^{\left (-\frac {5 \, {\left (x^{4} + 8 \, x^{3} + 16 \, x^{2} - x\right )}}{2 \, {\left (x^{4} + 8 \, x^{3} + 16 \, x^{2} - x - 2\right )}} + \frac {5}{2}\right )}}{\log \left (2\right )} \]
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Time = 9.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-\frac {5}{-2-x+16 x^2+8 x^3+x^4}} \left (4-x+97 x^2+56 x^3+256 x^4+254 x^5+96 x^6+16 x^7+x^8\right )}{\left (4+4 x-63 x^2-64 x^3+236 x^4+254 x^5+96 x^6+16 x^7+x^8\right ) \log (2)} \, dx=\frac {x\,{\mathrm {e}}^{-\frac {5}{x^4+8\,x^3+16\,x^2-x-2}}}{\ln \left (2\right )} \]
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