Integrand size = 51, antiderivative size = 27 \[ \int \frac {-e^{3+2 x}-e^3 x+e^{2 x} \left (e^{2 x}+x\right ) \left (12+48 e^{2 x}+24 x\right )}{e^{2 x}+x} \, dx=1+x \left (-e^3+\frac {12 e^{2 x} \left (e^{2 x}+x\right )}{x}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6820, 2225, 2207} \[ \int \frac {-e^{3+2 x}-e^3 x+e^{2 x} \left (e^{2 x}+x\right ) \left (12+48 e^{2 x}+24 x\right )}{e^{2 x}+x} \, dx=-e^3 x-6 e^{2 x}+12 e^{4 x}+6 e^{2 x} (2 x+1) \]
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Rule 2207
Rule 2225
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \left (-e^3+48 e^{4 x}+12 e^{2 x} (1+2 x)\right ) \, dx \\ & = -e^3 x+12 \int e^{2 x} (1+2 x) \, dx+48 \int e^{4 x} \, dx \\ & = 12 e^{4 x}-e^3 x+6 e^{2 x} (1+2 x)-12 \int e^{2 x} \, dx \\ & = -6 e^{2 x}+12 e^{4 x}-e^3 x+6 e^{2 x} (1+2 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-e^{3+2 x}-e^3 x+e^{2 x} \left (e^{2 x}+x\right ) \left (12+48 e^{2 x}+24 x\right )}{e^{2 x}+x} \, dx=12 e^{4 x}-e^3 x+12 e^{2 x} x \]
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Time = 0.71 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74
method | result | size |
norman | \(12 \,{\mathrm e}^{4 x}-x \,{\mathrm e}^{3}+12 x \,{\mathrm e}^{2 x}\) | \(20\) |
risch | \(12 \,{\mathrm e}^{4 x}-x \,{\mathrm e}^{3}+12 x \,{\mathrm e}^{2 x}\) | \(20\) |
parallelrisch | \(-x \,{\mathrm e}^{3}+12 \,{\mathrm e}^{\ln \left ({\mathrm e}^{2 x}+x \right )+2 x}\) | \(21\) |
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-e^{3+2 x}-e^3 x+e^{2 x} \left (e^{2 x}+x\right ) \left (12+48 e^{2 x}+24 x\right )}{e^{2 x}+x} \, dx=-{\left (x e^{9} - 12 \, x e^{\left (2 \, x + 6\right )} - 12 \, e^{\left (4 \, x + 6\right )}\right )} e^{\left (-6\right )} \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-e^{3+2 x}-e^3 x+e^{2 x} \left (e^{2 x}+x\right ) \left (12+48 e^{2 x}+24 x\right )}{e^{2 x}+x} \, dx=12 x e^{2 x} - x e^{3} + 12 e^{4 x} \]
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Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-e^{3+2 x}-e^3 x+e^{2 x} \left (e^{2 x}+x\right ) \left (12+48 e^{2 x}+24 x\right )}{e^{2 x}+x} \, dx=-x e^{3} + 12 \, x e^{\left (2 \, x\right )} + 12 \, e^{\left (4 \, x\right )} \]
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Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-e^{3+2 x}-e^3 x+e^{2 x} \left (e^{2 x}+x\right ) \left (12+48 e^{2 x}+24 x\right )}{e^{2 x}+x} \, dx=-x e^{3} + 12 \, x e^{\left (2 \, x\right )} + 12 \, e^{\left (4 \, x\right )} \]
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Time = 12.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-e^{3+2 x}-e^3 x+e^{2 x} \left (e^{2 x}+x\right ) \left (12+48 e^{2 x}+24 x\right )}{e^{2 x}+x} \, dx=12\,{\mathrm {e}}^{4\,x}+12\,x\,{\mathrm {e}}^{2\,x}-x\,{\mathrm {e}}^3 \]
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