Integrand size = 39, antiderivative size = 19 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 e^{-1+x}}{25 x \left (3+x^2\right )} \]
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Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1608, 28, 2326} \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 e^{x-1} \left (x^3+3 x\right )}{25 x^2 \left (x^2+3\right )^2} \]
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Rule 28
Rule 1608
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{x^2 \left (225+150 x^2+25 x^4\right )} \, dx \\ & = 25 \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{x^2 \left (75+25 x^2\right )^2} \, dx \\ & = \frac {168 e^{-1+x} \left (3 x+x^3\right )}{25 x^2 \left (3+x^2\right )^2} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 e^{-1+x}}{25 \left (3 x+x^3\right )} \]
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Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89
method | result | size |
risch | \(\frac {168 \,{\mathrm e}^{-1+x}}{25 \left (x^{2}+3\right ) x}\) | \(17\) |
gosper | \(\frac {168 \,{\mathrm e}^{-1+x}}{25 \left (x^{2}+3\right ) x}\) | \(21\) |
norman | \(\frac {168 \,{\mathrm e}^{-1+x}}{25 \left (x^{2}+3\right ) x}\) | \(21\) |
parallelrisch | \(\frac {168 \,{\mathrm e}^{-1+x}}{25 \left (x^{2}+3\right ) x}\) | \(21\) |
derivativedivides | \(-\frac {56 \,{\mathrm e}^{-1+x} \left (\left (1-x \right )^{2}+1+2 x \right )}{25 \left (\left (1-x \right )^{3}-3 \left (1-x \right )^{2}+2-6 x \right )}+\frac {56 \,{\mathrm e}^{-1+x} \left (1-x \right ) \left (-1-x \right )}{25 \left (\left (1-x \right )^{3}-3 \left (1-x \right )^{2}+2-6 x \right )}\) | \(82\) |
default | \(-\frac {56 \,{\mathrm e}^{-1+x} \left (\left (1-x \right )^{2}+1+2 x \right )}{25 \left (\left (1-x \right )^{3}-3 \left (1-x \right )^{2}+2-6 x \right )}+\frac {56 \,{\mathrm e}^{-1+x} \left (1-x \right ) \left (-1-x \right )}{25 \left (\left (1-x \right )^{3}-3 \left (1-x \right )^{2}+2-6 x \right )}\) | \(82\) |
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Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 \, e^{\left (x - 1\right )}}{25 \, {\left (x^{3} + 3 \, x\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 e^{x - 1}}{25 x^{3} + 75 x} \]
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Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 \, e^{x}}{25 \, {\left (x^{3} e + 3 \, x e\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 \, e^{x}}{25 \, {\left (x^{3} e + 3 \, x e\right )}} \]
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Time = 13.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x}{25\,\left (x^3+3\,x\right )} \]
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