Integrand size = 39, antiderivative size = 18 \[ \int \left (1-255 x^4+e^{2 x} \left (-153 x^2-102 x^3\right )+e^x \left (408 x^3+102 x^4\right )\right ) \, dx=x-51 x \left (e^x x-x^2\right )^2 \]
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Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 23, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1607, 2227, 2207, 2225} \[ \int \left (1-255 x^4+e^{2 x} \left (-153 x^2-102 x^3\right )+e^x \left (408 x^3+102 x^4\right )\right ) \, dx=-51 x^5+102 e^x x^4-51 e^{2 x} x^3+x \]
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Rule 1607
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = x-51 x^5+\int e^{2 x} \left (-153 x^2-102 x^3\right ) \, dx+\int e^x \left (408 x^3+102 x^4\right ) \, dx \\ & = x-51 x^5+\int e^{2 x} (-153-102 x) x^2 \, dx+\int e^x x^3 (408+102 x) \, dx \\ & = x-51 x^5+\int \left (-153 e^{2 x} x^2-102 e^{2 x} x^3\right ) \, dx+\int \left (408 e^x x^3+102 e^x x^4\right ) \, dx \\ & = x-51 x^5-102 \int e^{2 x} x^3 \, dx+102 \int e^x x^4 \, dx-153 \int e^{2 x} x^2 \, dx+408 \int e^x x^3 \, dx \\ & = x-\frac {153}{2} e^{2 x} x^2+408 e^x x^3-51 e^{2 x} x^3+102 e^x x^4-51 x^5+153 \int e^{2 x} x \, dx+153 \int e^{2 x} x^2 \, dx-408 \int e^x x^3 \, dx-1224 \int e^x x^2 \, dx \\ & = x+\frac {153}{2} e^{2 x} x-1224 e^x x^2-51 e^{2 x} x^3+102 e^x x^4-51 x^5-\frac {153}{2} \int e^{2 x} \, dx-153 \int e^{2 x} x \, dx+1224 \int e^x x^2 \, dx+2448 \int e^x x \, dx \\ & = -\frac {153 e^{2 x}}{4}+x+2448 e^x x-51 e^{2 x} x^3+102 e^x x^4-51 x^5+\frac {153}{2} \int e^{2 x} \, dx-2448 \int e^x \, dx-2448 \int e^x x \, dx \\ & = -2448 e^x+x-51 e^{2 x} x^3+102 e^x x^4-51 x^5+2448 \int e^x \, dx \\ & = x-51 e^{2 x} x^3+102 e^x x^4-51 x^5 \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \left (1-255 x^4+e^{2 x} \left (-153 x^2-102 x^3\right )+e^x \left (408 x^3+102 x^4\right )\right ) \, dx=x-51 e^{2 x} x^3+102 e^x x^4-51 x^5 \]
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Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33
method | result | size |
default | \(x -51 \,{\mathrm e}^{2 x} x^{3}+102 \,{\mathrm e}^{x} x^{4}-51 x^{5}\) | \(24\) |
norman | \(x -51 \,{\mathrm e}^{2 x} x^{3}+102 \,{\mathrm e}^{x} x^{4}-51 x^{5}\) | \(24\) |
risch | \(x -51 \,{\mathrm e}^{2 x} x^{3}+102 \,{\mathrm e}^{x} x^{4}-51 x^{5}\) | \(24\) |
parallelrisch | \(x -51 \,{\mathrm e}^{2 x} x^{3}+102 \,{\mathrm e}^{x} x^{4}-51 x^{5}\) | \(24\) |
parts | \(x -51 \,{\mathrm e}^{2 x} x^{3}+102 \,{\mathrm e}^{x} x^{4}-51 x^{5}\) | \(24\) |
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \left (1-255 x^4+e^{2 x} \left (-153 x^2-102 x^3\right )+e^x \left (408 x^3+102 x^4\right )\right ) \, dx=-51 \, x^{5} + 102 \, x^{4} e^{x} - 51 \, x^{3} e^{\left (2 \, x\right )} + x \]
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Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \left (1-255 x^4+e^{2 x} \left (-153 x^2-102 x^3\right )+e^x \left (408 x^3+102 x^4\right )\right ) \, dx=- 51 x^{5} + 102 x^{4} e^{x} - 51 x^{3} e^{2 x} + x \]
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Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \left (1-255 x^4+e^{2 x} \left (-153 x^2-102 x^3\right )+e^x \left (408 x^3+102 x^4\right )\right ) \, dx=-51 \, x^{5} + 102 \, x^{4} e^{x} - 51 \, x^{3} e^{\left (2 \, x\right )} + x \]
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Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \left (1-255 x^4+e^{2 x} \left (-153 x^2-102 x^3\right )+e^x \left (408 x^3+102 x^4\right )\right ) \, dx=-51 \, x^{5} + 102 \, x^{4} e^{x} - 51 \, x^{3} e^{\left (2 \, x\right )} + x \]
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Time = 12.84 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \left (1-255 x^4+e^{2 x} \left (-153 x^2-102 x^3\right )+e^x \left (408 x^3+102 x^4\right )\right ) \, dx=x+102\,x^4\,{\mathrm {e}}^x-51\,x^3\,{\mathrm {e}}^{2\,x}-51\,x^5 \]
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