Integrand size = 33, antiderivative size = 27 \[ \int \left (16-2 x+e^{x+2 x^2+x^3} \left (5+5 x+20 x^2+15 x^3\right )\right ) \, dx=5 \left (2+e^{\frac {\left (x+x^2\right )^2}{x}}+\frac {6-x}{5}\right ) x \]
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Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {2326} \[ \int \left (16-2 x+e^{x+2 x^2+x^3} \left (5+5 x+20 x^2+15 x^3\right )\right ) \, dx=-x^2+\frac {5 e^{x^3+2 x^2+x} \left (3 x^3+4 x^2+x\right )}{3 x^2+4 x+1}+16 x \]
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Rule 2326
Rubi steps \begin{align*} \text {integral}& = 16 x-x^2+\int e^{x+2 x^2+x^3} \left (5+5 x+20 x^2+15 x^3\right ) \, dx \\ & = 16 x-x^2+\frac {5 e^{x+2 x^2+x^3} \left (x+4 x^2+3 x^3\right )}{1+4 x+3 x^2} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \left (16-2 x+e^{x+2 x^2+x^3} \left (5+5 x+20 x^2+15 x^3\right )\right ) \, dx=-x \left (-16-5 e^{x (1+x)^2}+x\right ) \]
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Time = 0.44 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78
method | result | size |
risch | \(16 x +5 \,{\mathrm e}^{x \left (1+x \right )^{2}} x -x^{2}\) | \(21\) |
default | \(16 x +5 \,{\mathrm e}^{x^{3}+2 x^{2}+x} x -x^{2}\) | \(24\) |
norman | \(16 x +5 \,{\mathrm e}^{x^{3}+2 x^{2}+x} x -x^{2}\) | \(24\) |
parallelrisch | \(16 x +5 \,{\mathrm e}^{x^{3}+2 x^{2}+x} x -x^{2}\) | \(24\) |
parts | \(16 x +5 \,{\mathrm e}^{x^{3}+2 x^{2}+x} x -x^{2}\) | \(24\) |
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none
Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \left (16-2 x+e^{x+2 x^2+x^3} \left (5+5 x+20 x^2+15 x^3\right )\right ) \, dx=-x^{2} + 5 \, x e^{\left (x^{3} + 2 \, x^{2} + x\right )} + 16 \, x \]
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Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \left (16-2 x+e^{x+2 x^2+x^3} \left (5+5 x+20 x^2+15 x^3\right )\right ) \, dx=- x^{2} + 5 x e^{x^{3} + 2 x^{2} + x} + 16 x \]
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none
Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \left (16-2 x+e^{x+2 x^2+x^3} \left (5+5 x+20 x^2+15 x^3\right )\right ) \, dx=-x^{2} + 5 \, x e^{\left (x^{3} + 2 \, x^{2} + x\right )} + 16 \, x \]
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none
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \left (16-2 x+e^{x+2 x^2+x^3} \left (5+5 x+20 x^2+15 x^3\right )\right ) \, dx=-x^{2} + 5 \, x e^{\left (x^{3} + 2 \, x^{2} + x\right )} + 16 \, x \]
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Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \left (16-2 x+e^{x+2 x^2+x^3} \left (5+5 x+20 x^2+15 x^3\right )\right ) \, dx=16\,x-x^2+5\,x\,{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^x \]
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