Integrand size = 26, antiderivative size = 18 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=-4+e^{\frac {1}{5} e^{5-3 x}}-2 x \]
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Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {12, 2320, 2225} \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=e^{\frac {1}{5} e^{5-3 x}}-2 x \]
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Rule 12
Rule 2225
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx \\ & = -2 x-\frac {3}{5} \int e^{5+\frac {1}{5} e^{5-3 x}-3 x} \, dx \\ & = -2 x+\frac {1}{5} \text {Subst}\left (\int e^{5+\frac {e^5 x}{5}} \, dx,x,e^{-3 x}\right ) \\ & = e^{\frac {1}{5} e^{5-3 x}}-2 x \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=e^{\frac {1}{5} e^{5-3 x}}-2 x \]
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Time = 0.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78
method | result | size |
default | \(-2 x +{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\) | \(14\) |
norman | \(-2 x +{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\) | \(14\) |
risch | \(-2 x +{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\) | \(14\) |
parallelrisch | \(-2 x +{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\) | \(14\) |
parts | \(-2 x +{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\) | \(14\) |
derivativedivides | \(\frac {2 \ln \left (\frac {{\mathrm e}^{-3 x +5}}{5}\right )}{3}+{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\) | \(22\) |
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Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (14) = 28\).
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.89 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=-{\left (2 \, x e^{\left (-3 \, x + 5\right )} - e^{\left (-3 \, x + \frac {1}{5} \, e^{\left (-3 \, x + 5\right )} + 5\right )}\right )} e^{\left (3 \, x - 5\right )} \]
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Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=- 2 x + e^{\frac {e^{5 - 3 x}}{5}} \]
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none
Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=-2 \, x + e^{\left (\frac {1}{5} \, e^{\left (-3 \, x + 5\right )}\right )} \]
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none
Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=-2 \, x + e^{\left (\frac {1}{5} \, e^{\left (-3 \, x + 5\right )}\right )} \]
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Time = 0.10 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{5-3\,x}}{5}}-2\,x \]
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