Integrand size = 28, antiderivative size = 23 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=3 x \left (e^x-x-\left (-4+\frac {4 x^2}{5}\right )^2\right ) \]
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Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 2207, 2225} \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-\frac {48 x^5}{25}+\frac {96 x^3}{5}-3 x^2-48 x-3 e^x+3 e^x (x+1) \]
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Rule 12
Rule 2207
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx \\ & = -48 x-3 x^2+\frac {96 x^3}{5}-\frac {48 x^5}{25}+\frac {1}{5} \int e^x (15+15 x) \, dx \\ & = -48 x-3 x^2+\frac {96 x^3}{5}-\frac {48 x^5}{25}+3 e^x (1+x)-3 \int e^x \, dx \\ & = -3 e^x-48 x-3 x^2+\frac {96 x^3}{5}-\frac {48 x^5}{25}+3 e^x (1+x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-48 x+3 e^x x-3 x^2+\frac {96 x^3}{5}-\frac {48 x^5}{25} \]
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Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
method | result | size |
default | \(-48 x +3 \,{\mathrm e}^{x} x -3 x^{2}+\frac {96 x^{3}}{5}-\frac {48 x^{5}}{25}\) | \(25\) |
norman | \(-48 x +3 \,{\mathrm e}^{x} x -3 x^{2}+\frac {96 x^{3}}{5}-\frac {48 x^{5}}{25}\) | \(25\) |
risch | \(-48 x +3 \,{\mathrm e}^{x} x -3 x^{2}+\frac {96 x^{3}}{5}-\frac {48 x^{5}}{25}\) | \(25\) |
parallelrisch | \(-48 x +3 \,{\mathrm e}^{x} x -3 x^{2}+\frac {96 x^{3}}{5}-\frac {48 x^{5}}{25}\) | \(25\) |
parts | \(-48 x +3 \,{\mathrm e}^{x} x -3 x^{2}+\frac {96 x^{3}}{5}-\frac {48 x^{5}}{25}\) | \(25\) |
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Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-\frac {48}{25} \, x^{5} + \frac {96}{5} \, x^{3} - 3 \, x^{2} + 3 \, x e^{x} - 48 \, x \]
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Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=- \frac {48 x^{5}}{25} + \frac {96 x^{3}}{5} - 3 x^{2} + 3 x e^{x} - 48 x \]
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Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-\frac {48}{25} \, x^{5} + \frac {96}{5} \, x^{3} - 3 \, x^{2} + 3 \, x e^{x} - 48 \, x \]
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Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-\frac {48}{25} \, x^{5} + \frac {96}{5} \, x^{3} - 3 \, x^{2} + 3 \, x e^{x} - 48 \, x \]
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Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-\frac {3\,x\,\left (25\,x-25\,{\mathrm {e}}^x-160\,x^2+16\,x^4+400\right )}{25} \]
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