Integrand size = 41, antiderivative size = 20 \[ \int \frac {-64+64 x-16 x^2+e^x \left (-32+64 x-16 x^2\right )}{4 x^2-4 x^3+x^4} \, dx=\frac {16}{x}-\frac {16 e^x}{x+(-3+x) x} \]
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Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {1608, 27, 6874, 2208, 2209} \[ \int \frac {-64+64 x-16 x^2+e^x \left (-32+64 x-16 x^2\right )}{4 x^2-4 x^3+x^4} \, dx=\frac {8 e^x}{x}+\frac {16}{x}+\frac {8 e^x}{2-x} \]
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Rule 27
Rule 1608
Rule 2208
Rule 2209
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-64+64 x-16 x^2+e^x \left (-32+64 x-16 x^2\right )}{x^2 \left (4-4 x+x^2\right )} \, dx \\ & = \int \frac {-64+64 x-16 x^2+e^x \left (-32+64 x-16 x^2\right )}{(-2+x)^2 x^2} \, dx \\ & = \int \left (-\frac {16}{x^2}-\frac {16 e^x \left (2-4 x+x^2\right )}{(-2+x)^2 x^2}\right ) \, dx \\ & = \frac {16}{x}-16 \int \frac {e^x \left (2-4 x+x^2\right )}{(-2+x)^2 x^2} \, dx \\ & = \frac {16}{x}-16 \int \left (-\frac {e^x}{2 (-2+x)^2}+\frac {e^x}{2 (-2+x)}+\frac {e^x}{2 x^2}-\frac {e^x}{2 x}\right ) \, dx \\ & = \frac {16}{x}+8 \int \frac {e^x}{(-2+x)^2} \, dx-8 \int \frac {e^x}{-2+x} \, dx-8 \int \frac {e^x}{x^2} \, dx+8 \int \frac {e^x}{x} \, dx \\ & = \frac {8 e^x}{2-x}+\frac {16}{x}+\frac {8 e^x}{x}-8 e^2 \text {Ei}(-2+x)+8 \text {Ei}(x)+8 \int \frac {e^x}{-2+x} \, dx-8 \int \frac {e^x}{x} \, dx \\ & = \frac {8 e^x}{2-x}+\frac {16}{x}+\frac {8 e^x}{x} \\ \end{align*}
Time = 0.34 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-64+64 x-16 x^2+e^x \left (-32+64 x-16 x^2\right )}{4 x^2-4 x^3+x^4} \, dx=-\frac {16 \left (2+e^x-x\right )}{(-2+x) x} \]
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Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95
method | result | size |
norman | \(\frac {-32+16 x -16 \,{\mathrm e}^{x}}{\left (-2+x \right ) x}\) | \(19\) |
risch | \(\frac {16}{x}-\frac {16 \,{\mathrm e}^{x}}{\left (-2+x \right ) x}\) | \(19\) |
parallelrisch | \(\frac {-32+16 x -16 \,{\mathrm e}^{x}}{\left (-2+x \right ) x}\) | \(19\) |
default | \(\frac {16}{x}+\frac {8 \,{\mathrm e}^{x}}{x}-\frac {8 \,{\mathrm e}^{x}}{-2+x}\) | \(23\) |
parts | \(\frac {16}{x}+\frac {8 \,{\mathrm e}^{x}}{x}-\frac {8 \,{\mathrm e}^{x}}{-2+x}\) | \(23\) |
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-64+64 x-16 x^2+e^x \left (-32+64 x-16 x^2\right )}{4 x^2-4 x^3+x^4} \, dx=\frac {16 \, {\left (x - e^{x} - 2\right )}}{x^{2} - 2 \, x} \]
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Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-64+64 x-16 x^2+e^x \left (-32+64 x-16 x^2\right )}{4 x^2-4 x^3+x^4} \, dx=- \frac {16 e^{x}}{x^{2} - 2 x} + \frac {16}{x} \]
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Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75 \[ \int \frac {-64+64 x-16 x^2+e^x \left (-32+64 x-16 x^2\right )}{4 x^2-4 x^3+x^4} \, dx=\frac {32 \, {\left (x - 1\right )}}{x^{2} - 2 \, x} - \frac {16 \, e^{x}}{x^{2} - 2 \, x} - \frac {16}{x - 2} \]
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Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-64+64 x-16 x^2+e^x \left (-32+64 x-16 x^2\right )}{4 x^2-4 x^3+x^4} \, dx=\frac {16 \, {\left (x - e^{x} - 2\right )}}{x^{2} - 2 \, x} \]
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Time = 14.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-64+64 x-16 x^2+e^x \left (-32+64 x-16 x^2\right )}{4 x^2-4 x^3+x^4} \, dx=-\frac {8\,\left (2\,{\mathrm {e}}^x-x^2+4\right )}{x\,\left (x-2\right )} \]
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