Integrand size = 14, antiderivative size = 18 \[ \int \frac {1}{3} \left (100+5 e^x-40 x\right ) \, dx=\frac {5}{3} \left (-1+e^x-(-5+2 x)^2\right ) \]
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Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2225} \[ \int \frac {1}{3} \left (100+5 e^x-40 x\right ) \, dx=-\frac {20 x^2}{3}+\frac {100 x}{3}+\frac {5 e^x}{3} \]
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Rule 12
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \left (100+5 e^x-40 x\right ) \, dx \\ & = \frac {100 x}{3}-\frac {20 x^2}{3}+\frac {5 \int e^x \, dx}{3} \\ & = \frac {5 e^x}{3}+\frac {100 x}{3}-\frac {20 x^2}{3} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1}{3} \left (100+5 e^x-40 x\right ) \, dx=\frac {5}{3} \left (e^x+20 x-4 x^2\right ) \]
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Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78
method | result | size |
default | \(-\frac {20 x^{2}}{3}+\frac {100 x}{3}+\frac {5 \,{\mathrm e}^{x}}{3}\) | \(14\) |
norman | \(-\frac {20 x^{2}}{3}+\frac {100 x}{3}+\frac {5 \,{\mathrm e}^{x}}{3}\) | \(14\) |
risch | \(-\frac {20 x^{2}}{3}+\frac {100 x}{3}+\frac {5 \,{\mathrm e}^{x}}{3}\) | \(14\) |
parallelrisch | \(-\frac {20 x^{2}}{3}+\frac {100 x}{3}+\frac {5 \,{\mathrm e}^{x}}{3}\) | \(14\) |
parts | \(-\frac {20 x^{2}}{3}+\frac {100 x}{3}+\frac {5 \,{\mathrm e}^{x}}{3}\) | \(14\) |
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Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{3} \left (100+5 e^x-40 x\right ) \, dx=-\frac {20}{3} \, x^{2} + \frac {100}{3} \, x + \frac {5}{3} \, e^{x} \]
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Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {1}{3} \left (100+5 e^x-40 x\right ) \, dx=- \frac {20 x^{2}}{3} + \frac {100 x}{3} + \frac {5 e^{x}}{3} \]
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Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{3} \left (100+5 e^x-40 x\right ) \, dx=-\frac {20}{3} \, x^{2} + \frac {100}{3} \, x + \frac {5}{3} \, e^{x} \]
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Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{3} \left (100+5 e^x-40 x\right ) \, dx=-\frac {20}{3} \, x^{2} + \frac {100}{3} \, x + \frac {5}{3} \, e^{x} \]
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Time = 11.32 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{3} \left (100+5 e^x-40 x\right ) \, dx=\frac {100\,x}{3}+\frac {5\,{\mathrm {e}}^x}{3}-\frac {20\,x^2}{3} \]
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