Integrand size = 70, antiderivative size = 26 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=\frac {25 x}{6 \left (1-e^{x+\frac {1}{5} \left (-7-2 x^3\right )}\right )} \]
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\[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=\int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {14}{5}+\frac {4 x^3}{5}} \left (25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )\right )}{6 \left (e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}\right )^2} \, dx \\ & = \frac {1}{6} \int \frac {e^{\frac {14}{5}+\frac {4 x^3}{5}} \left (25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )\right )}{\left (e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}\right )^2} \, dx \\ & = \frac {1}{6} \int \left (-\frac {5 e^{\frac {14}{5}+\frac {4 x^3}{5}} x \left (-5+6 x^2\right )}{\left (e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}\right )^2}-5 e^{\frac {7}{5}-x+\frac {2 x^3}{5}} \left (5-5 x+6 x^3\right )-\frac {5 e^{\frac {14}{5}-x+\frac {4 x^3}{5}} \left (5-5 x+6 x^3\right )}{e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}}\right ) \, dx \\ & = -\left (\frac {5}{6} \int \frac {e^{\frac {14}{5}+\frac {4 x^3}{5}} x \left (-5+6 x^2\right )}{\left (e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}\right )^2} \, dx\right )-\frac {5}{6} \int e^{\frac {7}{5}-x+\frac {2 x^3}{5}} \left (5-5 x+6 x^3\right ) \, dx-\frac {5}{6} \int \frac {e^{\frac {14}{5}-x+\frac {4 x^3}{5}} \left (5-5 x+6 x^3\right )}{e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}} \, dx \\ & = -\frac {25 e^{\frac {7}{5}-x+\frac {2 x^3}{5}} \left (5 x-6 x^3\right )}{6 \left (5-6 x^2\right )}-\frac {5}{6} \int \left (-\frac {5 e^{\frac {14}{5}+\frac {4 x^3}{5}} x}{\left (e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}\right )^2}+\frac {6 e^{\frac {14}{5}+\frac {4 x^3}{5}} x^3}{\left (e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}\right )^2}\right ) \, dx-\frac {5}{6} \int \left (\frac {5 e^{\frac {14}{5}-x+\frac {4 x^3}{5}}}{e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}}-\frac {5 e^{\frac {14}{5}-x+\frac {4 x^3}{5}} x}{e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}}+\frac {6 e^{\frac {14}{5}-x+\frac {4 x^3}{5}} x^3}{e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}}\right ) \, dx \\ & = -\frac {25 e^{\frac {7}{5}-x+\frac {2 x^3}{5}} \left (5 x-6 x^3\right )}{6 \left (5-6 x^2\right )}-\frac {25}{6} \int \frac {e^{\frac {14}{5}-x+\frac {4 x^3}{5}}}{e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}} \, dx+\frac {25}{6} \int \frac {e^{\frac {14}{5}+\frac {4 x^3}{5}} x}{\left (e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}\right )^2} \, dx+\frac {25}{6} \int \frac {e^{\frac {14}{5}-x+\frac {4 x^3}{5}} x}{e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}} \, dx-5 \int \frac {e^{\frac {14}{5}+\frac {4 x^3}{5}} x^3}{\left (e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}\right )^2} \, dx-5 \int \frac {e^{\frac {14}{5}-x+\frac {4 x^3}{5}} x^3}{e^x-e^{\frac {7}{5}+\frac {2 x^3}{5}}} \, dx \\ \end{align*}
Time = 3.93 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=\frac {25 e^{\frac {7}{5}+\frac {2 x^3}{5}} x}{6 \left (-e^x+e^{\frac {7}{5}+\frac {2 x^3}{5}}\right )} \]
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Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65
method | result | size |
norman | \(-\frac {25 x}{6 \left ({\mathrm e}^{-\frac {2}{5} x^{3}+x -\frac {7}{5}}-1\right )}\) | \(17\) |
risch | \(-\frac {25 x}{6 \left ({\mathrm e}^{-\frac {2}{5} x^{3}+x -\frac {7}{5}}-1\right )}\) | \(17\) |
parallelrisch | \(-\frac {25 x}{6 \left ({\mathrm e}^{-\frac {2}{5} x^{3}+x -\frac {7}{5}}-1\right )}\) | \(17\) |
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Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=-\frac {25 \, x}{6 \, {\left (e^{\left (-\frac {2}{5} \, x^{3} + x - \frac {7}{5}\right )} - 1\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=- \frac {25 x}{6 e^{- \frac {2 x^{3}}{5} + x - \frac {7}{5}} - 6} \]
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Exception generated. \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=-\frac {25 \, x}{6 \, {\left (e^{\left (-\frac {2}{5} \, x^{3} + x - \frac {7}{5}\right )} - 1\right )}} \]
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Time = 11.65 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=-\frac {25\,x}{6\,\left ({\mathrm {e}}^{-\frac {2\,x^3}{5}+x-\frac {7}{5}}-1\right )} \]
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