Integrand size = 66, antiderivative size = 28 \[ \int \frac {2-5 e^x x+e^{e^{16 e^{-x}} x} \left (-5 e^x+e^{16 e^{-x}} \left (2-32 e^{-x} x\right )\right )}{e^{e^{16 e^{-x}} x}+x} \, dx=5 \left (25-e^x\right )+\log \left (\left (e^{e^{16 e^{-x}} x}+x\right )^2\right ) \]
[Out]
\[ \int \frac {2-5 e^x x+e^{e^{16 e^{-x}} x} \left (-5 e^x+e^{16 e^{-x}} \left (2-32 e^{-x} x\right )\right )}{e^{e^{16 e^{-x}} x}+x} \, dx=\int \frac {2-5 e^x x+e^{e^{16 e^{-x}} x} \left (-5 e^x+e^{16 e^{-x}} \left (2-32 e^{-x} x\right )\right )}{e^{e^{16 e^{-x}} x}+x} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \left (-e^{-x} \left (5 e^{2 x}-2 e^{16 e^{-x}+x}+32 e^{16 e^{-x}} x\right )-\frac {2 e^{-x} \left (-e^x+e^{16 e^{-x}+x} x-16 e^{16 e^{-x}} x^2\right )}{e^{e^{16 e^{-x}} x}+x}\right ) \, dx \\ & = -\left (2 \int \frac {e^{-x} \left (-e^x+e^{16 e^{-x}+x} x-16 e^{16 e^{-x}} x^2\right )}{e^{e^{16 e^{-x}} x}+x} \, dx\right )-\int e^{-x} \left (5 e^{2 x}-2 e^{16 e^{-x}+x}+32 e^{16 e^{-x}} x\right ) \, dx \\ & = -\left (2 \int \left (-\frac {1}{e^{e^{16 e^{-x}} x}+x}+\frac {e^{16 e^{-x}} x}{e^{e^{16 e^{-x}} x}+x}-\frac {16 e^{16 e^{-x}-x} x^2}{e^{e^{16 e^{-x}} x}+x}\right ) \, dx\right )-\int \left (-2 e^{16 e^{-x}}+5 e^x+32 e^{16 e^{-x}-x} x\right ) \, dx \\ & = 2 \int e^{16 e^{-x}} \, dx+2 \int \frac {1}{e^{e^{16 e^{-x}} x}+x} \, dx-2 \int \frac {e^{16 e^{-x}} x}{e^{e^{16 e^{-x}} x}+x} \, dx-5 \int e^x \, dx-32 \int e^{16 e^{-x}-x} x \, dx+32 \int \frac {e^{16 e^{-x}-x} x^2}{e^{e^{16 e^{-x}} x}+x} \, dx \\ & = -5 e^x+2 \int \frac {1}{e^{e^{16 e^{-x}} x}+x} \, dx-2 \int \frac {e^{16 e^{-x}} x}{e^{e^{16 e^{-x}} x}+x} \, dx-2 \text {Subst}\left (\int \frac {e^{16 x}}{x} \, dx,x,e^{-x}\right )-32 \int e^{16 e^{-x}-x} x \, dx+32 \int \frac {e^{16 e^{-x}-x} x^2}{e^{e^{16 e^{-x}} x}+x} \, dx \\ & = -5 e^x-2 \operatorname {ExpIntegralEi}\left (16 e^{-x}\right )+2 \int \frac {1}{e^{e^{16 e^{-x}} x}+x} \, dx-2 \int \frac {e^{16 e^{-x}} x}{e^{e^{16 e^{-x}} x}+x} \, dx-32 \int e^{16 e^{-x}-x} x \, dx+32 \int \frac {e^{16 e^{-x}-x} x^2}{e^{e^{16 e^{-x}} x}+x} \, dx \\ \end{align*}
Time = 3.94 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {2-5 e^x x+e^{e^{16 e^{-x}} x} \left (-5 e^x+e^{16 e^{-x}} \left (2-32 e^{-x} x\right )\right )}{e^{e^{16 e^{-x}} x}+x} \, dx=-5 e^x+2 \log \left (e^{e^{16 e^{-x}} x}+x\right ) \]
[In]
[Out]
Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75
method | result | size |
norman | \(-5 \,{\mathrm e}^{x}+2 \ln \left ({\mathrm e}^{x \,{\mathrm e}^{16 \,{\mathrm e}^{-x}}}+x \right )\) | \(21\) |
risch | \(-5 \,{\mathrm e}^{x}+2 \ln \left ({\mathrm e}^{x \,{\mathrm e}^{16 \,{\mathrm e}^{-x}}}+x \right )\) | \(21\) |
parallelrisch | \(-5 \,{\mathrm e}^{x}+2 \ln \left ({\mathrm e}^{x \,{\mathrm e}^{{\mathrm e}^{4 \ln \left (2\right )-x}}}+x \right )\) | \(24\) |
parts | \(-5 \,{\mathrm e}^{x}+2 \ln \left ({\mathrm e}^{x \,{\mathrm e}^{{\mathrm e}^{4 \ln \left (2\right )-x}}}+x \right )\) | \(24\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {2-5 e^x x+e^{e^{16 e^{-x}} x} \left (-5 e^x+e^{16 e^{-x}} \left (2-32 e^{-x} x\right )\right )}{e^{e^{16 e^{-x}} x}+x} \, dx=2 \, {\left (e^{\left (-x + 4 \, \log \left (2\right )\right )} \log \left (x + e^{\left (x e^{\left (e^{\left (-x + 4 \, \log \left (2\right )\right )}\right )}\right )}\right ) - 40\right )} e^{\left (x - 4 \, \log \left (2\right )\right )} \]
[In]
[Out]
Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {2-5 e^x x+e^{e^{16 e^{-x}} x} \left (-5 e^x+e^{16 e^{-x}} \left (2-32 e^{-x} x\right )\right )}{e^{e^{16 e^{-x}} x}+x} \, dx=- 5 e^{x} + 2 \log {\left (x + e^{x e^{16 e^{- x}}} \right )} \]
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {2-5 e^x x+e^{e^{16 e^{-x}} x} \left (-5 e^x+e^{16 e^{-x}} \left (2-32 e^{-x} x\right )\right )}{e^{e^{16 e^{-x}} x}+x} \, dx=-5 \, e^{x} + 2 \, \log \left (x + e^{\left (x e^{\left (16 \, e^{\left (-x\right )}\right )}\right )}\right ) \]
[In]
[Out]
\[ \int \frac {2-5 e^x x+e^{e^{16 e^{-x}} x} \left (-5 e^x+e^{16 e^{-x}} \left (2-32 e^{-x} x\right )\right )}{e^{e^{16 e^{-x}} x}+x} \, dx=\int { -\frac {{\left (2 \, {\left (x e^{\left (-x + 4 \, \log \left (2\right )\right )} - 1\right )} e^{\left (e^{\left (-x + 4 \, \log \left (2\right )\right )}\right )} + 5 \, e^{x}\right )} e^{\left (x e^{\left (e^{\left (-x + 4 \, \log \left (2\right )\right )}\right )}\right )} + 5 \, x e^{x} - 2}{x + e^{\left (x e^{\left (e^{\left (-x + 4 \, \log \left (2\right )\right )}\right )}\right )}} \,d x } \]
[In]
[Out]
Time = 13.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {2-5 e^x x+e^{e^{16 e^{-x}} x} \left (-5 e^x+e^{16 e^{-x}} \left (2-32 e^{-x} x\right )\right )}{e^{e^{16 e^{-x}} x}+x} \, dx=2\,\ln \left (x+{\mathrm {e}}^{x\,{\mathrm {e}}^{16\,{\mathrm {e}}^{-x}}}\right )-5\,{\mathrm {e}}^x \]
[In]
[Out]