Integrand size = 49, antiderivative size = 20 \[ \int \frac {e^{-4+x^2} \left (2 x-3 x^2+4 x^3-3 x^4+2 x^5-2 x^6\right )}{1+2 x^2+x^4} \, dx=e^{-4+x^2} \left (1-x+\frac {1}{\frac {1}{x}+x}\right ) \]
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Result contains complex when optimal does not.
Time = 0.51 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.80, number of steps used = 18, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {28, 6873, 6874, 2235, 2240, 2243, 2252, 6857} \[ \int \frac {e^{-4+x^2} \left (2 x-3 x^2+4 x^3-3 x^4+2 x^5-2 x^6\right )}{1+2 x^2+x^4} \, dx=-e^{x^2-4} x+e^{x^2-4}-\frac {e^{x^2-4}}{2 (-x+i)}+\frac {e^{x^2-4}}{2 (x+i)} \]
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Rule 28
Rule 2235
Rule 2240
Rule 2243
Rule 2252
Rule 6857
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-4+x^2} \left (2 x-3 x^2+4 x^3-3 x^4+2 x^5-2 x^6\right )}{\left (1+x^2\right )^2} \, dx \\ & = \int \frac {e^{-4+x^2} x \left (2-3 x+4 x^2-3 x^3+2 x^4-2 x^5\right )}{\left (1+x^2\right )^2} \, dx \\ & = \int \left (e^{-4+x^2}+2 e^{-4+x^2} x-2 e^{-4+x^2} x^2+\frac {2 e^{-4+x^2}}{\left (1+x^2\right )^2}-\frac {3 e^{-4+x^2}}{1+x^2}\right ) \, dx \\ & = 2 \int e^{-4+x^2} x \, dx-2 \int e^{-4+x^2} x^2 \, dx+2 \int \frac {e^{-4+x^2}}{\left (1+x^2\right )^2} \, dx-3 \int \frac {e^{-4+x^2}}{1+x^2} \, dx+\int e^{-4+x^2} \, dx \\ & = e^{-4+x^2}-e^{-4+x^2} x+\frac {\sqrt {\pi } \text {erfi}(x)}{2 e^4}+2 \int \left (-\frac {e^{-4+x^2}}{4 (i-x)^2}-\frac {e^{-4+x^2}}{4 (i+x)^2}-\frac {e^{-4+x^2}}{2 \left (-1-x^2\right )}\right ) \, dx-3 \int \left (\frac {i e^{-4+x^2}}{2 (i-x)}+\frac {i e^{-4+x^2}}{2 (i+x)}\right ) \, dx+\int e^{-4+x^2} \, dx \\ & = e^{-4+x^2}-e^{-4+x^2} x+\frac {\sqrt {\pi } \text {erfi}(x)}{e^4}-\frac {3}{2} i \int \frac {e^{-4+x^2}}{i-x} \, dx-\frac {3}{2} i \int \frac {e^{-4+x^2}}{i+x} \, dx-\frac {1}{2} \int \frac {e^{-4+x^2}}{(i-x)^2} \, dx-\frac {1}{2} \int \frac {e^{-4+x^2}}{(i+x)^2} \, dx-\int \frac {e^{-4+x^2}}{-1-x^2} \, dx \\ & = e^{-4+x^2}-\frac {e^{-4+x^2}}{2 (i-x)}-e^{-4+x^2} x+\frac {e^{-4+x^2}}{2 (i+x)}+\frac {\sqrt {\pi } \text {erfi}(x)}{e^4}+i \int \frac {e^{-4+x^2}}{i-x} \, dx+i \int \frac {e^{-4+x^2}}{i+x} \, dx-\frac {3}{2} i \int \frac {e^{-4+x^2}}{i-x} \, dx-\frac {3}{2} i \int \frac {e^{-4+x^2}}{i+x} \, dx-2 \int e^{-4+x^2} \, dx-\int \left (-\frac {i e^{-4+x^2}}{2 (i-x)}-\frac {i e^{-4+x^2}}{2 (i+x)}\right ) \, dx \\ & = e^{-4+x^2}-\frac {e^{-4+x^2}}{2 (i-x)}-e^{-4+x^2} x+\frac {e^{-4+x^2}}{2 (i+x)}+\frac {1}{2} i \int \frac {e^{-4+x^2}}{i-x} \, dx+\frac {1}{2} i \int \frac {e^{-4+x^2}}{i+x} \, dx+i \int \frac {e^{-4+x^2}}{i-x} \, dx+i \int \frac {e^{-4+x^2}}{i+x} \, dx-\frac {3}{2} i \int \frac {e^{-4+x^2}}{i-x} \, dx-\frac {3}{2} i \int \frac {e^{-4+x^2}}{i+x} \, dx \\ \end{align*}
Time = 1.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {e^{-4+x^2} \left (2 x-3 x^2+4 x^3-3 x^4+2 x^5-2 x^6\right )}{1+2 x^2+x^4} \, dx=-\frac {e^{-4+x^2} \left (-1-x^2+x^3\right )}{1+x^2} \]
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Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30
method | result | size |
gosper | \(-\frac {\left (x^{3}-x^{2}-1\right ) {\mathrm e}^{x^{2}-4}}{x^{2}+1}\) | \(26\) |
risch | \(-\frac {\left (x^{3}-x^{2}-1\right ) {\mathrm e}^{\left (-2+x \right ) \left (2+x \right )}}{x^{2}+1}\) | \(28\) |
norman | \(\frac {{\mathrm e}^{x^{2}-4} x^{2}-x^{3} {\mathrm e}^{x^{2}-4}+{\mathrm e}^{x^{2}-4}}{x^{2}+1}\) | \(37\) |
parallelrisch | \(-\frac {x^{3} {\mathrm e}^{x^{2}-4}-{\mathrm e}^{x^{2}-4} x^{2}-{\mathrm e}^{x^{2}-4}}{x^{2}+1}\) | \(40\) |
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Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-4+x^2} \left (2 x-3 x^2+4 x^3-3 x^4+2 x^5-2 x^6\right )}{1+2 x^2+x^4} \, dx=-\frac {{\left (x^{3} - x^{2} - 1\right )} e^{\left (x^{2} - 4\right )}}{x^{2} + 1} \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-4+x^2} \left (2 x-3 x^2+4 x^3-3 x^4+2 x^5-2 x^6\right )}{1+2 x^2+x^4} \, dx=\frac {\left (- x^{3} + x^{2} + 1\right ) e^{x^{2} - 4}}{x^{2} + 1} \]
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\[ \int \frac {e^{-4+x^2} \left (2 x-3 x^2+4 x^3-3 x^4+2 x^5-2 x^6\right )}{1+2 x^2+x^4} \, dx=\int { -\frac {{\left (2 \, x^{6} - 2 \, x^{5} + 3 \, x^{4} - 4 \, x^{3} + 3 \, x^{2} - 2 \, x\right )} e^{\left (x^{2} - 4\right )}}{x^{4} + 2 \, x^{2} + 1} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.85 \[ \int \frac {e^{-4+x^2} \left (2 x-3 x^2+4 x^3-3 x^4+2 x^5-2 x^6\right )}{1+2 x^2+x^4} \, dx=-\frac {x^{3} e^{\left (x^{2}\right )} - x^{2} e^{\left (x^{2}\right )} - e^{\left (x^{2}\right )}}{x^{2} e^{4} + e^{4}} \]
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Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {e^{-4+x^2} \left (2 x-3 x^2+4 x^3-3 x^4+2 x^5-2 x^6\right )}{1+2 x^2+x^4} \, dx=\frac {{\mathrm {e}}^{x^2-4}\,\left (-x^3+x^2+1\right )}{x^2+1} \]
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