Integrand size = 60, antiderivative size = 15 \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=e^{e^{\frac {x}{-4-x}} x} \]
[Out]
\[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=\int \frac {\exp \left (e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}\right ) \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}\right ) \left (16+4 x+x^2\right )}{x \left (16+8 x+x^2\right )} \, dx \\ & = \int \frac {\exp \left (e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}\right ) \left (16+4 x+x^2\right )}{x (4+x)^2} \, dx \\ & = \int \frac {e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}} \left (16+4 x+x^2\right )}{(4+x)^2} \, dx \\ & = \int \left (e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}}+\frac {16 e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}}}{(4+x)^2}-\frac {4 e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}}}{4+x}\right ) \, dx \\ & = -\left (4 \int \frac {e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}}}{4+x} \, dx\right )+16 \int \frac {e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}}}{(4+x)^2} \, dx+\int e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}} \, dx \\ \end{align*}
Time = 0.42 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=e^{e^{-1+\frac {4}{4+x}} x} \]
[In]
[Out]
Time = 0.66 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27
method | result | size |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\frac {\left (4+x \right ) \ln \left (x \right )-x}{4+x}}}\) | \(19\) |
risch | \({\mathrm e}^{{\mathrm e}^{\frac {x \ln \left (x \right )+4 \ln \left (x \right )-x}{4+x}}}\) | \(21\) |
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (13) = 26\).
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 3.73 \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=e^{\left (\frac {{\left (x + 4\right )} e^{\left (\frac {{\left (x + 4\right )} \log \left (x\right ) - x}{x + 4}\right )} + {\left (x + 4\right )} \log \left (x\right ) - x}{x + 4} - \frac {{\left (x + 4\right )} \log \left (x\right ) - x}{x + 4}\right )} \]
[In]
[Out]
Timed out. \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=\text {Timed out} \]
[In]
[Out]
none
Time = 0.40 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=e^{\left (x e^{\left (\frac {4}{x + 4} - 1\right )}\right )} \]
[In]
[Out]
\[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=\int { \frac {{\left (x^{2} + 4 \, x + 16\right )} e^{\left (\frac {{\left (x + 4\right )} \log \left (x\right ) - x}{x + 4} + e^{\left (\frac {{\left (x + 4\right )} \log \left (x\right ) - x}{x + 4}\right )}\right )}}{x^{3} + 8 \, x^{2} + 16 \, x} \,d x } \]
[In]
[Out]
Time = 11.89 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx={\mathrm {e}}^{x\,{\mathrm {e}}^{-\frac {x}{x+4}}} \]
[In]
[Out]