Integrand size = 71, antiderivative size = 23 \[ \int \frac {e^{-3-e^x+e^{\frac {-3-e^3 x}{e^3 x}}-x} \left (3 e^{\frac {-3-e^3 x}{e^3 x}}-e^3 x^2-e^{3+x} x^2\right )}{x^2} \, dx=e^{e^{-1-\frac {3}{e^3 x}}-e^x-x} \]
[Out]
Time = 0.43 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6838} \[ \int \frac {e^{-3-e^x+e^{\frac {-3-e^3 x}{e^3 x}}-x} \left (3 e^{\frac {-3-e^3 x}{e^3 x}}-e^3 x^2-e^{3+x} x^2\right )}{x^2} \, dx=e^{-x-e^x+e^{-\frac {e^3 x+3}{e^3 x}}} \]
[In]
[Out]
Rule 6838
Rubi steps \begin{align*} \text {integral}& = e^{-e^x+e^{-\frac {3+e^3 x}{e^3 x}}-x} \\ \end{align*}
Time = 1.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3-e^x+e^{\frac {-3-e^3 x}{e^3 x}}-x} \left (3 e^{\frac {-3-e^3 x}{e^3 x}}-e^3 x^2-e^{3+x} x^2\right )}{x^2} \, dx=e^{e^{-1-\frac {3}{e^3 x}}-e^x-x} \]
[In]
[Out]
Time = 7.67 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
risch | \({\mathrm e}^{-{\mathrm e}^{x}+{\mathrm e}^{-\frac {\left (x \,{\mathrm e}^{3}+3\right ) {\mathrm e}^{-3}}{x}}-x}\) | \(24\) |
norman | \({\mathrm e}^{-{\mathrm e}^{x}+{\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{3}-3\right ) {\mathrm e}^{-3}}{x}}-x}\) | \(26\) |
parallelrisch | \({\mathrm e}^{-{\mathrm e}^{x}+{\mathrm e}^{-\frac {\left (x \,{\mathrm e}^{3}+3\right ) {\mathrm e}^{-3}}{x}}-x}\) | \(26\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {e^{-3-e^x+e^{\frac {-3-e^3 x}{e^3 x}}-x} \left (3 e^{\frac {-3-e^3 x}{e^3 x}}-e^3 x^2-e^{3+x} x^2\right )}{x^2} \, dx=e^{\left (-{\left ({\left (x + 3\right )} e^{3} + e^{\left (x + 3\right )} - e^{\left (-\frac {{\left (x e^{3} + 3\right )} e^{\left (-3\right )}}{x} + 3\right )}\right )} e^{\left (-3\right )} + 3\right )} \]
[In]
[Out]
Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-3-e^x+e^{\frac {-3-e^3 x}{e^3 x}}-x} \left (3 e^{\frac {-3-e^3 x}{e^3 x}}-e^3 x^2-e^{3+x} x^2\right )}{x^2} \, dx=e^{- x - e^{x} + e^{\frac {- x e^{3} - 3}{x e^{3}}}} \]
[In]
[Out]
\[ \int \frac {e^{-3-e^x+e^{\frac {-3-e^3 x}{e^3 x}}-x} \left (3 e^{\frac {-3-e^3 x}{e^3 x}}-e^3 x^2-e^{3+x} x^2\right )}{x^2} \, dx=\int { -\frac {{\left (x^{2} e^{3} + x^{2} e^{\left (x + 3\right )} - 3 \, e^{\left (-\frac {{\left (x e^{3} + 3\right )} e^{\left (-3\right )}}{x}\right )}\right )} e^{\left (-x - e^{x} + e^{\left (-\frac {{\left (x e^{3} + 3\right )} e^{\left (-3\right )}}{x}\right )} - 3\right )}}{x^{2}} \,d x } \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-3-e^x+e^{\frac {-3-e^3 x}{e^3 x}}-x} \left (3 e^{\frac {-3-e^3 x}{e^3 x}}-e^3 x^2-e^{3+x} x^2\right )}{x^2} \, dx=e^{\left (-x - e^{x} + e^{\left (-\frac {3 \, e^{\left (-3\right )}}{x} - 1\right )}\right )} \]
[In]
[Out]
Time = 10.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-3-e^x+e^{\frac {-3-e^3 x}{e^3 x}}-x} \left (3 e^{\frac {-3-e^3 x}{e^3 x}}-e^3 x^2-e^{3+x} x^2\right )}{x^2} \, dx={\mathrm {e}}^{{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^{-3}}{x}}\,{\mathrm {e}}^{-1}}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-{\mathrm {e}}^x} \]
[In]
[Out]