Integrand size = 20, antiderivative size = 17 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=x \left (-5-\frac {4}{e^3}+2 x-25 x^4\right ) \]
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Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {12} \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-25 x^5+2 x^2-\frac {4 x}{e^3}-5 x \]
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Rule 12
Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (-4+e^3 \left (-5+4 x-125 x^4\right )\right ) \, dx}{e^3} \\ & = -\frac {4 x}{e^3}+\int \left (-5+4 x-125 x^4\right ) \, dx \\ & = -5 x-\frac {4 x}{e^3}+2 x^2-25 x^5 \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-5 x-\frac {4 x}{e^3}+2 x^2-25 x^5 \]
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Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18
method | result | size |
risch | \(-25 x^{5}+2 x^{2}-5 x -4 \,{\mathrm e}^{-3} x\) | \(20\) |
norman | \(2 x^{2}-25 x^{5}-{\mathrm e}^{-3} \left (5 \,{\mathrm e}^{3}+4\right ) x\) | \(25\) |
gosper | \(-x \left (25 x^{4} {\mathrm e}^{3}-2 x \,{\mathrm e}^{3}+5 \,{\mathrm e}^{3}+4\right ) {\mathrm e}^{-3}\) | \(26\) |
parallelrisch | \({\mathrm e}^{-3} \left ({\mathrm e}^{3} \left (-25 x^{5}+2 x^{2}-5 x \right )-4 x \right )\) | \(27\) |
default | \({\mathrm e}^{-3} \left (-25 x^{5} {\mathrm e}^{3}+2 x^{2} {\mathrm e}^{3}-5 x \,{\mathrm e}^{3}-4 x \right )\) | \(29\) |
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Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-{\left ({\left (25 \, x^{5} - 2 \, x^{2} + 5 \, x\right )} e^{3} + 4 \, x\right )} e^{\left (-3\right )} \]
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Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=- 25 x^{5} + 2 x^{2} + \frac {x \left (- 5 e^{3} - 4\right )}{e^{3}} \]
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Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-{\left ({\left (25 \, x^{5} - 2 \, x^{2} + 5 \, x\right )} e^{3} + 4 \, x\right )} e^{\left (-3\right )} \]
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Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-{\left ({\left (25 \, x^{5} - 2 \, x^{2} + 5 \, x\right )} e^{3} + 4 \, x\right )} e^{\left (-3\right )} \]
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Time = 12.83 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-25\,x^5+2\,x^2+\left (-4\,{\mathrm {e}}^{-3}-5\right )\,x \]
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