Integrand size = 29, antiderivative size = 14 \[ \int \frac {e^{12+\frac {2-x^2-x \log (x)}{x}} \left (2+x^2\right )}{x} \, dx=-e^{12+\frac {2}{x}-x} \]
[Out]
Time = 0.16 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {6820, 6838} \[ \int \frac {e^{12+\frac {2-x^2-x \log (x)}{x}} \left (2+x^2\right )}{x} \, dx=-e^{-x+\frac {2}{x}+12} \]
[In]
[Out]
Rule 6820
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{12+\frac {2}{x}-x} \left (2+x^2\right )}{x^2} \, dx \\ & = -e^{12+\frac {2}{x}-x} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{12+\frac {2-x^2-x \log (x)}{x}} \left (2+x^2\right )}{x} \, dx=-e^{12+\frac {2}{x}-x} \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21
method | result | size |
risch | \(-{\mathrm e}^{-\frac {x^{2}-12 x -2}{x}}\) | \(17\) |
gosper | \(-x \,{\mathrm e}^{12} {\mathrm e}^{-\frac {x \ln \left (x \right )+x^{2}-2}{x}}\) | \(23\) |
default | \(-x \,{\mathrm e}^{12} {\mathrm e}^{-\frac {x \ln \left (x \right )+x^{2}-2}{x}}\) | \(23\) |
parallelrisch | \(-x \,{\mathrm e}^{12} {\mathrm e}^{-\frac {x \ln \left (x \right )+x^{2}-2}{x}}\) | \(23\) |
norman | \(-x \,{\mathrm e}^{12} {\mathrm e}^{\frac {-x \ln \left (x \right )-x^{2}+2}{x}}\) | \(25\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.50 \[ \int \frac {e^{12+\frac {2-x^2-x \log (x)}{x}} \left (2+x^2\right )}{x} \, dx=-x e^{\left (-\frac {x^{2} + x \log \left (x\right ) - 12 \, x - 2}{x}\right )} \]
[In]
[Out]
Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {e^{12+\frac {2-x^2-x \log (x)}{x}} \left (2+x^2\right )}{x} \, dx=- x e^{12} e^{\frac {- x^{2} - x \log {\left (x \right )} + 2}{x}} \]
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {e^{12+\frac {2-x^2-x \log (x)}{x}} \left (2+x^2\right )}{x} \, dx=-e^{\left (-x + \frac {2}{x} + 12\right )} \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {e^{12+\frac {2-x^2-x \log (x)}{x}} \left (2+x^2\right )}{x} \, dx=-e^{\left (-\frac {x^{2} - 12 \, x - 2}{x}\right )} \]
[In]
[Out]
Time = 8.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{12+\frac {2-x^2-x \log (x)}{x}} \left (2+x^2\right )}{x} \, dx=-{\mathrm {e}}^{-x}\,{\mathrm {e}}^{12}\,{\mathrm {e}}^{2/x} \]
[In]
[Out]