Integrand size = 71, antiderivative size = 21 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{\frac {5 x}{3-4 e^{8-2 x} x^4}} \]
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Time = 0.58 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6838} \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{\frac {5 x}{3-4 e^{8-2 x} x^4}} \]
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Rule 6838
Rubi steps \begin{align*} \text {integral}& = e^{\frac {5 x}{3-4 e^{8-2 x} x^4}} \\ \end{align*}
Time = 0.77 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \]
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Time = 38.55 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95
method | result | size |
risch | \({\mathrm e}^{-\frac {5 x}{4 x^{4} {\mathrm e}^{-2 x +8}-3}}\) | \(20\) |
parallelrisch | \({\mathrm e}^{-\frac {5 x}{4 x^{4} {\mathrm e}^{-2 x +8}-3}}\) | \(22\) |
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none
Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{\left (-\frac {5 \, x}{4 \, x^{4} e^{\left (-2 \, x + 8\right )} - 3}\right )} \]
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Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{- \frac {5 x}{4 x^{4} e^{8 - 2 x} - 3}} \]
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none
Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{\left (-\frac {5 \, x e^{\left (2 \, x\right )}}{4 \, x^{4} e^{8} - 3 \, e^{\left (2 \, x\right )}}\right )} \]
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Timed out. \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=\text {Timed out} \]
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Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx={\mathrm {e}}^{-\frac {5\,x}{4\,x^4\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^8-3}} \]
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