Integrand size = 73, antiderivative size = 33 \[ \int \frac {e^{-x} \left (e^{10 e^{-x}} \left (-2 e^x-10 x\right )-e^{e^x+2 x} x^3+e^{5 e^{-x}} \left (4 e^x+10 x\right )+e^x \left (-6-x^3\right )\right )}{x^3} \, dx=2-e^{e^x}+\frac {2+\left (1-e^{5 e^{-x}}\right )^2}{x^2}-x \]
[Out]
\[ \int \frac {e^{-x} \left (e^{10 e^{-x}} \left (-2 e^x-10 x\right )-e^{e^x+2 x} x^3+e^{5 e^{-x}} \left (4 e^x+10 x\right )+e^x \left (-6-x^3\right )\right )}{x^3} \, dx=\int \frac {e^{-x} \left (e^{10 e^{-x}} \left (-2 e^x-10 x\right )-e^{e^x+2 x} x^3+e^{5 e^{-x}} \left (4 e^x+10 x\right )+e^x \left (-6-x^3\right )\right )}{x^3} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \left (-e^{e^x+x}-\frac {10 e^{5 e^{-x}-x} \left (-1+e^{5 e^{-x}}\right )}{x^2}-\frac {6-4 e^{5 e^{-x}}+2 e^{10 e^{-x}}+x^3}{x^3}\right ) \, dx \\ & = -\left (10 \int \frac {e^{5 e^{-x}-x} \left (-1+e^{5 e^{-x}}\right )}{x^2} \, dx\right )-\int e^{e^x+x} \, dx-\int \frac {6-4 e^{5 e^{-x}}+2 e^{10 e^{-x}}+x^3}{x^3} \, dx \\ & = -\left (10 \int \left (-\frac {e^{5 e^{-x}-x}}{x^2}+\frac {e^{10 e^{-x}-x}}{x^2}\right ) \, dx\right )-\int \left (-\frac {4 e^{5 e^{-x}}}{x^3}+\frac {2 e^{10 e^{-x}}}{x^3}+\frac {6+x^3}{x^3}\right ) \, dx-\text {Subst}\left (\int e^x \, dx,x,e^x\right ) \\ & = -e^{e^x}-2 \int \frac {e^{10 e^{-x}}}{x^3} \, dx+4 \int \frac {e^{5 e^{-x}}}{x^3} \, dx+10 \int \frac {e^{5 e^{-x}-x}}{x^2} \, dx-10 \int \frac {e^{10 e^{-x}-x}}{x^2} \, dx-\int \frac {6+x^3}{x^3} \, dx \\ & = -e^{e^x}-2 \int \frac {e^{10 e^{-x}}}{x^3} \, dx+4 \int \frac {e^{5 e^{-x}}}{x^3} \, dx+10 \int \frac {e^{5 e^{-x}-x}}{x^2} \, dx-10 \int \frac {e^{10 e^{-x}-x}}{x^2} \, dx-\int \left (1+\frac {6}{x^3}\right ) \, dx \\ & = -e^{e^x}+\frac {3}{x^2}-x-2 \int \frac {e^{10 e^{-x}}}{x^3} \, dx+4 \int \frac {e^{5 e^{-x}}}{x^3} \, dx+10 \int \frac {e^{5 e^{-x}-x}}{x^2} \, dx-10 \int \frac {e^{10 e^{-x}-x}}{x^2} \, dx \\ \end{align*}
Time = 4.79 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {e^{-x} \left (e^{10 e^{-x}} \left (-2 e^x-10 x\right )-e^{e^x+2 x} x^3+e^{5 e^{-x}} \left (4 e^x+10 x\right )+e^x \left (-6-x^3\right )\right )}{x^3} \, dx=-e^{e^x}+\frac {3}{x^2}-\frac {2 e^{5 e^{-x}}}{x^2}+\frac {e^{10 e^{-x}}}{x^2}-x \]
[In]
[Out]
Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15
method | result | size |
risch | \(-x +\frac {3}{x^{2}}-{\mathrm e}^{{\mathrm e}^{x}}+\frac {{\mathrm e}^{10 \,{\mathrm e}^{-x}}}{x^{2}}-\frac {2 \,{\mathrm e}^{5 \,{\mathrm e}^{-x}}}{x^{2}}\) | \(38\) |
parallelrisch | \(-\frac {-3+\ln \left ({\mathrm e}^{x}\right ) x^{2}+{\mathrm e}^{{\mathrm e}^{x}} x^{2}-{\mathrm e}^{10 \,{\mathrm e}^{-x}}+2 \,{\mathrm e}^{5 \,{\mathrm e}^{-x}}}{x^{2}}\) | \(42\) |
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (27) = 54\).
Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {e^{-x} \left (e^{10 e^{-x}} \left (-2 e^x-10 x\right )-e^{e^x+2 x} x^3+e^{5 e^{-x}} \left (4 e^x+10 x\right )+e^x \left (-6-x^3\right )\right )}{x^3} \, dx=-\frac {{\left (x^{2} e^{\left (2 \, x + e^{x}\right )} + {\left (x^{3} - 3\right )} e^{\left (2 \, x\right )} - e^{\left (2 \, x + 10 \, e^{\left (-x\right )}\right )} + 2 \, e^{\left (2 \, x + 5 \, e^{\left (-x\right )}\right )}\right )} e^{\left (-2 \, x\right )}}{x^{2}} \]
[In]
[Out]
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x} \left (e^{10 e^{-x}} \left (-2 e^x-10 x\right )-e^{e^x+2 x} x^3+e^{5 e^{-x}} \left (4 e^x+10 x\right )+e^x \left (-6-x^3\right )\right )}{x^3} \, dx=- x - e^{e^{x}} + \frac {3}{x^{2}} + \frac {x^{2} e^{10 e^{- x}} - 2 x^{2} e^{5 e^{- x}}}{x^{4}} \]
[In]
[Out]
\[ \int \frac {e^{-x} \left (e^{10 e^{-x}} \left (-2 e^x-10 x\right )-e^{e^x+2 x} x^3+e^{5 e^{-x}} \left (4 e^x+10 x\right )+e^x \left (-6-x^3\right )\right )}{x^3} \, dx=\int { -\frac {{\left (x^{3} e^{\left (2 \, x + e^{x}\right )} + {\left (x^{3} + 6\right )} e^{x} + 2 \, {\left (5 \, x + e^{x}\right )} e^{\left (10 \, e^{\left (-x\right )}\right )} - 2 \, {\left (5 \, x + 2 \, e^{x}\right )} e^{\left (5 \, e^{\left (-x\right )}\right )}\right )} e^{\left (-x\right )}}{x^{3}} \,d x } \]
[In]
[Out]
\[ \int \frac {e^{-x} \left (e^{10 e^{-x}} \left (-2 e^x-10 x\right )-e^{e^x+2 x} x^3+e^{5 e^{-x}} \left (4 e^x+10 x\right )+e^x \left (-6-x^3\right )\right )}{x^3} \, dx=\int { -\frac {{\left (x^{3} e^{\left (2 \, x + e^{x}\right )} + {\left (x^{3} + 6\right )} e^{x} + 2 \, {\left (5 \, x + e^{x}\right )} e^{\left (10 \, e^{\left (-x\right )}\right )} - 2 \, {\left (5 \, x + 2 \, e^{x}\right )} e^{\left (5 \, e^{\left (-x\right )}\right )}\right )} e^{\left (-x\right )}}{x^{3}} \,d x } \]
[In]
[Out]
Time = 13.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-x} \left (e^{10 e^{-x}} \left (-2 e^x-10 x\right )-e^{e^x+2 x} x^3+e^{5 e^{-x}} \left (4 e^x+10 x\right )+e^x \left (-6-x^3\right )\right )}{x^3} \, dx=\frac {{\mathrm {e}}^{10\,{\mathrm {e}}^{-x}}}{x^2}-{\mathrm {e}}^{{\mathrm {e}}^x}-\frac {2\,{\mathrm {e}}^{5\,{\mathrm {e}}^{-x}}}{x^2}-x+\frac {3}{x^2} \]
[In]
[Out]