Integrand size = 102, antiderivative size = 31 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=e^{-2+\frac {-1+e^2-e^4-\log (3)}{5+e^{4 x}-x}} \]
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\[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=\int \frac {\exp \left (\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}\right ) \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) \left (1-4 e^{4 x}\right ) \left (-1+e^2-e^4-\log (3)\right )}{\left (5+e^{4 x}-x\right )^2} \, dx \\ & = \left (-1+e^2-e^4-\log (3)\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) \left (1-4 e^{4 x}\right )}{\left (5+e^{4 x}-x\right )^2} \, dx \\ & = \left (-1+e^2-e^4-\log (3)\right ) \int \left (-\frac {4 \exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{5+e^{4 x}-x}-\frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) (-21+4 x)}{\left (-5-e^{4 x}+x\right )^2}\right ) \, dx \\ & = \left (1-e^2+e^4+\log (3)\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) (-21+4 x)}{\left (-5-e^{4 x}+x\right )^2} \, dx+\left (4 \left (1-e^2+e^4+\log (3)\right )\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{5+e^{4 x}-x} \, dx \\ & = \left (1-e^2+e^4+\log (3)\right ) \int \left (-\frac {21 \exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{\left (5+e^{4 x}-x\right )^2}+\frac {4 \exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) x}{\left (5+e^{4 x}-x\right )^2}\right ) \, dx+\left (4 \left (1-e^2+e^4+\log (3)\right )\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{5+e^{4 x}-x} \, dx \\ & = \left (4 \left (1-e^2+e^4+\log (3)\right )\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{5+e^{4 x}-x} \, dx+\left (4 \left (1-e^2+e^4+\log (3)\right )\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) x}{\left (5+e^{4 x}-x\right )^2} \, dx-\left (21 \left (1-e^2+e^4+\log (3)\right )\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{\left (5+e^{4 x}-x\right )^2} \, dx \\ \end{align*}
Time = 3.80 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=3^{-\frac {1}{5+e^{4 x}-x}} e^{-2+\frac {-1+e^2-e^4}{5+e^{4 x}-x}} \]
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Time = 0.68 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10
method | result | size |
parallelrisch | \({\mathrm e}^{-\frac {2 \,{\mathrm e}^{4 x}+\ln \left (3\right )+{\mathrm e}^{4}-{\mathrm e}^{2}-2 x +11}{{\mathrm e}^{4 x}+5-x}}\) | \(34\) |
risch | \({\mathrm e}^{-\frac {-2 \,{\mathrm e}^{4 x}-\ln \left (3\right )-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{-{\mathrm e}^{4 x}-5+x}}\) | \(36\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {-2 \,{\mathrm e}^{4 x}-\ln \left (3\right )-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{{\mathrm e}^{4 x}+5-x}}-{\mathrm e}^{4 x} {\mathrm e}^{\frac {-2 \,{\mathrm e}^{4 x}-\ln \left (3\right )-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{{\mathrm e}^{4 x}+5-x}}-5 \,{\mathrm e}^{\frac {-2 \,{\mathrm e}^{4 x}-\ln \left (3\right )-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{{\mathrm e}^{4 x}+5-x}}}{-{\mathrm e}^{4 x}-5+x}\) | \(126\) |
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=e^{\left (-\frac {2 \, x - e^{4} + e^{2} - 2 \, e^{\left (4 \, x\right )} - \log \left (3\right ) - 11}{x - e^{\left (4 \, x\right )} - 5}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=e^{\frac {2 x - 2 e^{4 x} - e^{4} - 11 - \log {\left (3 \right )} + e^{2}}{- x + e^{4 x} + 5}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).
Time = 0.45 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.84 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=e^{\left (\frac {e^{4}}{x - e^{\left (4 \, x\right )} - 5} - \frac {e^{2}}{x - e^{\left (4 \, x\right )} - 5} + \frac {\log \left (3\right )}{x - e^{\left (4 \, x\right )} - 5} + \frac {1}{x - e^{\left (4 \, x\right )} - 5} - 2\right )} \]
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\[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=\int { \frac {{\left (4 \, {\left (e^{4} - e^{2} + \log \left (3\right ) + 1\right )} e^{\left (4 \, x\right )} - e^{4} + e^{2} - \log \left (3\right ) - 1\right )} e^{\left (-\frac {2 \, x - e^{4} + e^{2} - 2 \, e^{\left (4 \, x\right )} - \log \left (3\right ) - 11}{x - e^{\left (4 \, x\right )} - 5}\right )}}{x^{2} - 2 \, {\left (x - 5\right )} e^{\left (4 \, x\right )} - 10 \, x + e^{\left (8 \, x\right )} + 25} \,d x } \]
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Time = 13.64 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.03 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=\frac {{\mathrm {e}}^{\frac {2\,x}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{-\frac {11}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{4\,x}}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^4}{{\mathrm {e}}^{4\,x}-x+5}}}{3^{\frac {1}{{\mathrm {e}}^{4\,x}-x+5}}} \]
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