Integrand size = 26, antiderivative size = 18 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=e^x-\frac {5 (-16+x) (-2+x)}{e^5 x^3} \]
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Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {12, 14, 2225} \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=-\frac {160}{e^5 x^3}+\frac {90}{e^5 x^2}+e^x-\frac {5}{e^5 x} \]
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Rule 12
Rule 14
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {480-180 x+5 x^2+e^{5+x} x^4}{x^4} \, dx}{e^5} \\ & = \frac {\int \left (e^{5+x}+\frac {5 \left (96-36 x+x^2\right )}{x^4}\right ) \, dx}{e^5} \\ & = \frac {\int e^{5+x} \, dx}{e^5}+\frac {5 \int \frac {96-36 x+x^2}{x^4} \, dx}{e^5} \\ & = e^x+\frac {5 \int \left (\frac {96}{x^4}-\frac {36}{x^3}+\frac {1}{x^2}\right ) \, dx}{e^5} \\ & = e^x-\frac {160}{e^5 x^3}+\frac {90}{e^5 x^2}-\frac {5}{e^5 x} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=\frac {e^{5+x}-\frac {160}{x^3}+\frac {90}{x^2}-\frac {5}{x}}{e^5} \]
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Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11
method | result | size |
risch | \(\frac {{\mathrm e}^{-5} \left (-5 x^{2}+90 x -160\right )}{x^{3}}+{\mathrm e}^{x}\) | \(20\) |
parts | \(5 \,{\mathrm e}^{-5} \left (-\frac {1}{x}+\frac {18}{x^{2}}-\frac {32}{x^{3}}\right )+{\mathrm e}^{x}\) | \(26\) |
default | \({\mathrm e}^{-5} \left ({\mathrm e}^{5} {\mathrm e}^{x}-\frac {160}{x^{3}}+\frac {90}{x^{2}}-\frac {5}{x}\right )\) | \(27\) |
parallelrisch | \(\frac {{\mathrm e}^{-5} \left (-160+90 x -5 x^{2}+x^{3} {\mathrm e}^{5} {\mathrm e}^{x}\right )}{x^{3}}\) | \(27\) |
norman | \(\frac {{\mathrm e}^{x} x^{3}-5 x^{2} {\mathrm e}^{-5}+90 \,{\mathrm e}^{-5} x -160 \,{\mathrm e}^{-5}}{x^{3}}\) | \(34\) |
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Time = 0.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=\frac {{\left (x^{3} e^{\left (x + 5\right )} - 5 \, x^{2} + 90 \, x - 160\right )} e^{\left (-5\right )}}{x^{3}} \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=e^{x} + \frac {- 5 x^{2} + 90 x - 160}{x^{3} e^{5}} \]
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Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=-{\left (\frac {5}{x} - \frac {90}{x^{2}} + \frac {160}{x^{3}} - e^{\left (x + 5\right )}\right )} e^{\left (-5\right )} \]
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Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=\frac {{\left (x^{3} e^{\left (x + 5\right )} - 5 \, x^{2} + 90 \, x - 160\right )} e^{\left (-5\right )}}{x^{3}} \]
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Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx={\mathrm {e}}^x-\frac {5\,{\mathrm {e}}^{-5}\,x^2-90\,{\mathrm {e}}^{-5}\,x+160\,{\mathrm {e}}^{-5}}{x^3} \]
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