Integrand size = 38, antiderivative size = 18 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=e^{x-(5-x)^2 x^3} x \]
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Leaf count is larger than twice the leaf count of optimal. \(56\) vs. \(2(18)=36\).
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 3.11, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2326} \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=\frac {e^{-x^5+10 x^4-25 x^3+x} \left (-5 x^5+40 x^4-75 x^3+x\right )}{-5 x^4+40 x^3-75 x^2+1} \]
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Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {e^{x-25 x^3+10 x^4-x^5} \left (x-75 x^3+40 x^4-5 x^5\right )}{1-75 x^2+40 x^3-5 x^4} \\ \end{align*}
Time = 0.77 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=e^{x-25 x^3+10 x^4-x^5} x \]
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Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17
method | result | size |
gosper | \({\mathrm e}^{\ln \left (x \right )-x^{5}+10 x^{4}-25 x^{3}+x}\) | \(21\) |
default | \({\mathrm e}^{\ln \left (x \right )-x^{5}+10 x^{4}-25 x^{3}+x}\) | \(21\) |
norman | \({\mathrm e}^{\ln \left (x \right )-x^{5}+10 x^{4}-25 x^{3}+x}\) | \(21\) |
parallelrisch | \({\mathrm e}^{\ln \left (x \right )-x^{5}+10 x^{4}-25 x^{3}+x}\) | \(21\) |
risch | \(x \,{\mathrm e}^{-x \left (x^{2}-5 x +1\right ) \left (x^{2}-5 x -1\right )}\) | \(23\) |
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none
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=e^{\left (-x^{5} + 10 \, x^{4} - 25 \, x^{3} + x + \log \left (x\right )\right )} \]
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Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=x e^{- x^{5} + 10 x^{4} - 25 x^{3} + x} \]
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none
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=x e^{\left (-x^{5} + 10 \, x^{4} - 25 \, x^{3} + x\right )} \]
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none
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=e^{\left (-x^{5} + 10 \, x^{4} - 25 \, x^{3} + x + \log \left (x\right )\right )} \]
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Time = 13.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=x\,{\mathrm {e}}^{-x^5}\,{\mathrm {e}}^{10\,x^4}\,{\mathrm {e}}^{-25\,x^3}\,{\mathrm {e}}^x \]
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