Integrand size = 31, antiderivative size = 18 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {e^{16-4 x^2} x^4}{(1+x)^4} \]
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Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1608, 2326} \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {e^{16-4 x^2} x^2 \left (x^3+x^2\right )}{(x+1)^5} \]
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Rule 1608
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{16-4 x^2} x^3 \left (4-8 x^2-8 x^3\right )}{(1+x)^5} \, dx \\ & = \frac {e^{16-4 x^2} x^2 \left (x^2+x^3\right )}{(1+x)^5} \\ \end{align*}
Time = 2.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {e^{16-4 x^2} x^4}{(1+x)^4} \]
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Time = 0.47 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06
method | result | size |
risch | \(\frac {x^{4} {\mathrm e}^{-4 \left (-2+x \right ) \left (2+x \right )}}{\left (1+x \right )^{4}}\) | \(19\) |
gosper | \(\frac {x^{4} {\mathrm e}^{-4 x^{2}+16}}{\left (1+x \right )^{4}}\) | \(21\) |
default | \(\frac {x^{4} {\mathrm e}^{-4 x^{2}+16}}{\left (1+x \right )^{4}}\) | \(21\) |
parallelrisch | \(\frac {x^{4} {\mathrm e}^{-4 x^{2}+16}}{\left (1+x \right )^{4}}\) | \(21\) |
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none
Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=x^{4} e^{\left (-4 \, x^{2} - 4 \, \log \left (x + 1\right ) + 16\right )} \]
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Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {x^{4} e^{16 - 4 x^{2}}}{x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1} \]
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none
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {x^{4} e^{\left (-4 \, x^{2} + 16\right )}}{x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1} \]
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none
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {x^{4} e^{\left (-4 \, x^{2} + 16\right )}}{x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1} \]
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Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {x^4\,{\mathrm {e}}^{16}\,{\mathrm {e}}^{-4\,x^2}}{{\left (x+1\right )}^4} \]
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