Integrand size = 64, antiderivative size = 31 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=5+\frac {e^{\frac {e^{-x} x}{15}} \left (-e^{3+3 x}+\log (x)\right )}{\log (2)} \]
[Out]
\[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{x} \, dx}{15 \log (2)} \\ & = \frac {\int \left (-45 e^{3+3 x+\frac {e^{-x} x}{15}}+e^{3+2 x+\frac {e^{-x} x}{15}} (-1+x)+\frac {15 e^{\frac {e^{-x} x}{15}}}{x}-e^{-x+\frac {e^{-x} x}{15}} (-1+x) \log (x)\right ) \, dx}{15 \log (2)} \\ & = \frac {\int e^{3+2 x+\frac {e^{-x} x}{15}} (-1+x) \, dx}{15 \log (2)}-\frac {\int e^{-x+\frac {e^{-x} x}{15}} (-1+x) \log (x) \, dx}{15 \log (2)}+\frac {\int \frac {e^{\frac {e^{-x} x}{15}}}{x} \, dx}{\log (2)}-\frac {3 \int e^{3+3 x+\frac {e^{-x} x}{15}} \, dx}{\log (2)} \\ & = \frac {\int \left (-e^{3+2 x+\frac {e^{-x} x}{15}}+e^{3+2 x+\frac {e^{-x} x}{15}} x\right ) \, dx}{15 \log (2)}+\frac {\int \frac {-\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} \, dx+\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} x \, dx}{x} \, dx}{15 \log (2)}+\frac {\int \frac {e^{\frac {e^{-x} x}{15}}}{x} \, dx}{\log (2)}-\frac {3 \int e^{3+3 x+\frac {e^{-x} x}{15}} \, dx}{\log (2)}+\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} \, dx}{15 \log (2)}-\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} x \, dx}{15 \log (2)} \\ & = -\frac {\int e^{3+2 x+\frac {e^{-x} x}{15}} \, dx}{15 \log (2)}+\frac {\int e^{3+2 x+\frac {e^{-x} x}{15}} x \, dx}{15 \log (2)}+\frac {\int \left (-\frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} \, dx}{x}+\frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} x \, dx}{x}\right ) \, dx}{15 \log (2)}+\frac {\int \frac {e^{\frac {e^{-x} x}{15}}}{x} \, dx}{\log (2)}-\frac {3 \int e^{3+3 x+\frac {e^{-x} x}{15}} \, dx}{\log (2)}+\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} \, dx}{15 \log (2)}-\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} x \, dx}{15 \log (2)} \\ & = -\frac {\int e^{3+2 x+\frac {e^{-x} x}{15}} \, dx}{15 \log (2)}+\frac {\int e^{3+2 x+\frac {e^{-x} x}{15}} x \, dx}{15 \log (2)}-\frac {\int \frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} \, dx}{x} \, dx}{15 \log (2)}+\frac {\int \frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} x \, dx}{x} \, dx}{15 \log (2)}+\frac {\int \frac {e^{\frac {e^{-x} x}{15}}}{x} \, dx}{\log (2)}-\frac {3 \int e^{3+3 x+\frac {e^{-x} x}{15}} \, dx}{\log (2)}+\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} \, dx}{15 \log (2)}-\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} x \, dx}{15 \log (2)} \\ \end{align*}
Time = 2.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\frac {e^{\frac {e^{-x} x}{15}} \left (-15 e^{3+3 x}+15 \log (x)\right )}{15 \log (2)} \]
[In]
[Out]
Time = 0.42 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\frac {\left (-15 \,{\mathrm e}^{3 x +3}+15 \ln \left (x \right )\right ) {\mathrm e}^{\frac {x \,{\mathrm e}^{-x}}{15}}}{15 \ln \left (2\right )}\) | \(28\) |
parallelrisch | \(\frac {15 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-x}}{15}} \ln \left (x \right )-15 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-x}}{15}} {\mathrm e}^{3 x +3}}{15 \ln \left (2\right )}\) | \(36\) |
[In]
[Out]
none
Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\frac {{\left (e^{x} \log \left (x\right ) - e^{\left (4 \, x + 3\right )}\right )} e^{\left (-\frac {1}{15} \, {\left (15 \, x e^{x} - x\right )} e^{\left (-x\right )}\right )}}{\log \left (2\right )} \]
[In]
[Out]
Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\frac {\left (- e^{3} e^{3 x} + \log {\left (x \right )}\right ) e^{\frac {x e^{- x}}{15}}}{\log {\left (2 \right )}} \]
[In]
[Out]
\[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\int { \frac {{\left ({\left (x^{2} - 45 \, x e^{x} - x\right )} e^{\left (3 \, x + 3\right )} - {\left (x^{2} - x\right )} \log \left (x\right ) + 15 \, e^{x}\right )} e^{\left (\frac {1}{15} \, x e^{\left (-x\right )} - x\right )}}{15 \, x \log \left (2\right )} \,d x } \]
[In]
[Out]
\[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\int { \frac {{\left ({\left (x^{2} - 45 \, x e^{x} - x\right )} e^{\left (3 \, x + 3\right )} - {\left (x^{2} - x\right )} \log \left (x\right ) + 15 \, e^{x}\right )} e^{\left (\frac {1}{15} \, x e^{\left (-x\right )} - x\right )}}{15 \, x \log \left (2\right )} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\int \frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-x}}{15}-x}\,\left (15\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (x-x^2\right )-{\mathrm {e}}^{3\,x+3}\,\left (x+45\,x\,{\mathrm {e}}^x-x^2\right )\right )}{15\,x\,\ln \left (2\right )} \,d x \]
[In]
[Out]