Integrand size = 65, antiderivative size = 23 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{3+x \left (-8-e^{\left .\frac {1}{4}\right /x}+\log \left (x^2\right )\right )} \]
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Time = 0.36 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12, 6838} \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{-e^{\left .\frac {1}{4}\right /x} x-8 x+3} \left (x^2\right )^x \]
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Rule 12
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{x} \, dx \\ & = e^{3-8 x-e^{\left .\frac {1}{4}\right /x} x} \left (x^2\right )^x \\ \end{align*}
Time = 0.88 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{3-8 x-e^{\left .\frac {1}{4}\right /x} x} \left (x^2\right )^x \]
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Time = 3.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
method | result | size |
parallelrisch | \({\mathrm e}^{x \left (\ln \left (x^{2}\right )-8-{\mathrm e}^{\frac {1}{4 x}}\right )} {\mathrm e}^{3}\) | \(21\) |
default | \({\mathrm e}^{3} {\mathrm e}^{x \ln \left (x^{2}\right )-x \,{\mathrm e}^{\frac {1}{4 x}}-8 x}\) | \(24\) |
norman | \({\mathrm e}^{3} {\mathrm e}^{x \ln \left (x^{2}\right )-x \,{\mathrm e}^{\frac {1}{4 x}}-8 x}\) | \(24\) |
risch | \(x^{2 x} {\mathrm e}^{3-\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}+i x \operatorname {csgn}\left (i x^{2}\right )^{2} \pi \,\operatorname {csgn}\left (i x \right )-\frac {i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}-x \,{\mathrm e}^{\frac {1}{4 x}}-8 x}\) | \(74\) |
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{\left ({\left (x e^{3} \log \left (x^{2}\right ) - 8 \, x e^{3} - x e^{\left (\frac {12 \, x + 1}{4 \, x}\right )}\right )} e^{\left (-3\right )} + 3\right )} \]
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{3} e^{- x e^{\frac {1}{4 x}} + x \log {\left (x^{2} \right )} - 8 x} \]
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{\left (-x e^{\left (\frac {1}{4 \, x}\right )} + 2 \, x \log \left (x\right ) - 8 \, x + 3\right )} \]
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Time = 0.35 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{\left (-x e^{\left (\frac {1}{4 \, x}\right )} + x \log \left (x^{2}\right ) - 8 \, x + 3\right )} \]
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Time = 14.94 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx={\mathrm {e}}^{-8\,x}\,{\mathrm {e}}^3\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{\frac {1}{4\,x}}}\,{\left (x^2\right )}^x \]
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