Integrand size = 64, antiderivative size = 24 \[ \int \frac {-450-1350 x+e^x (-30+120 x)+\left (-450 x+30 e^x x\right ) \log (x)}{225 x+e^{2 x} x-450 x^2+225 x^3+e^x \left (30 x-30 x^2\right )} \, dx=\frac {4+\log (x)}{-\frac {e^x}{30}+\frac {1}{4} (-2+2 x)} \]
[Out]
\[ \int \frac {-450-1350 x+e^x (-30+120 x)+\left (-450 x+30 e^x x\right ) \log (x)}{225 x+e^{2 x} x-450 x^2+225 x^3+e^x \left (30 x-30 x^2\right )} \, dx=\int \frac {-450-1350 x+e^x (-30+120 x)+\left (-450 x+30 e^x x\right ) \log (x)}{225 x+e^{2 x} x-450 x^2+225 x^3+e^x \left (30 x-30 x^2\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {30 \left (-15 (1+3 x)+e^x (-1+4 x)+\left (-15+e^x\right ) x \log (x)\right )}{\left (15+e^x-15 x\right )^2 x} \, dx \\ & = 30 \int \frac {-15 (1+3 x)+e^x (-1+4 x)+\left (-15+e^x\right ) x \log (x)}{\left (15+e^x-15 x\right )^2 x} \, dx \\ & = 30 \int \left (\frac {15 (-2+x) (4+\log (x))}{\left (15+e^x-15 x\right )^2}+\frac {-1+4 x+x \log (x)}{\left (15+e^x-15 x\right ) x}\right ) \, dx \\ & = 30 \int \frac {-1+4 x+x \log (x)}{\left (15+e^x-15 x\right ) x} \, dx+450 \int \frac {(-2+x) (4+\log (x))}{\left (15+e^x-15 x\right )^2} \, dx \\ & = 30 \int \left (\frac {4}{15+e^x-15 x}-\frac {1}{\left (15+e^x-15 x\right ) x}+\frac {\log (x)}{15+e^x-15 x}\right ) \, dx+450 \int \left (-\frac {2 (4+\log (x))}{\left (15+e^x-15 x\right )^2}+\frac {x (4+\log (x))}{\left (15+e^x-15 x\right )^2}\right ) \, dx \\ & = -\left (30 \int \frac {1}{\left (15+e^x-15 x\right ) x} \, dx\right )+30 \int \frac {\log (x)}{15+e^x-15 x} \, dx+120 \int \frac {1}{15+e^x-15 x} \, dx+450 \int \frac {x (4+\log (x))}{\left (15+e^x-15 x\right )^2} \, dx-900 \int \frac {4+\log (x)}{\left (15+e^x-15 x\right )^2} \, dx \\ & = -\left (30 \int \frac {1}{\left (15+e^x-15 x\right ) x} \, dx\right )-30 \int \frac {\int \frac {1}{15+e^x-15 x} \, dx}{x} \, dx+120 \int \frac {1}{15+e^x-15 x} \, dx+450 \int \left (\frac {4 x}{\left (15+e^x-15 x\right )^2}+\frac {x \log (x)}{\left (15+e^x-15 x\right )^2}\right ) \, dx-900 \int \left (\frac {4}{\left (15+e^x-15 x\right )^2}+\frac {\log (x)}{\left (15+e^x-15 x\right )^2}\right ) \, dx+(30 \log (x)) \int \frac {1}{15+e^x-15 x} \, dx \\ & = -\left (30 \int \frac {1}{\left (15+e^x-15 x\right ) x} \, dx\right )-30 \int \frac {\int \frac {1}{15+e^x-15 x} \, dx}{x} \, dx+120 \int \frac {1}{15+e^x-15 x} \, dx+450 \int \frac {x \log (x)}{\left (15+e^x-15 x\right )^2} \, dx-900 \int \frac {\log (x)}{\left (15+e^x-15 x\right )^2} \, dx+1800 \int \frac {x}{\left (15+e^x-15 x\right )^2} \, dx-3600 \int \frac {1}{\left (15+e^x-15 x\right )^2} \, dx+(30 \log (x)) \int \frac {1}{15+e^x-15 x} \, dx \\ & = -\left (30 \int \frac {1}{\left (15+e^x-15 x\right ) x} \, dx\right )-30 \int \frac {\int \frac {1}{15+e^x-15 x} \, dx}{x} \, dx+120 \int \frac {1}{15+e^x-15 x} \, dx-450 \int \frac {\int \frac {x}{\left (15+e^x-15 x\right )^2} \, dx}{x} \, dx+900 \int \frac {\int \frac {1}{\left (15+e^x-15 x\right )^2} \, dx}{x} \, dx+1800 \int \frac {x}{\left (15+e^x-15 x\right )^2} \, dx-3600 \int \frac {1}{\left (15+e^x-15 x\right )^2} \, dx+(30 \log (x)) \int \frac {1}{15+e^x-15 x} \, dx+(450 \log (x)) \int \frac {x}{\left (15+e^x-15 x\right )^2} \, dx-(900 \log (x)) \int \frac {1}{\left (15+e^x-15 x\right )^2} \, dx \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-450-1350 x+e^x (-30+120 x)+\left (-450 x+30 e^x x\right ) \log (x)}{225 x+e^{2 x} x-450 x^2+225 x^3+e^x \left (30 x-30 x^2\right )} \, dx=-\frac {30 (4+\log (x))}{15+e^x-15 x} \]
[In]
[Out]
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79
method | result | size |
parallelrisch | \(\frac {120+30 \ln \left (x \right )}{15 x -{\mathrm e}^{x}-15}\) | \(19\) |
risch | \(\frac {30 \ln \left (x \right )}{15 x -{\mathrm e}^{x}-15}+\frac {120}{15 x -{\mathrm e}^{x}-15}\) | \(30\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-450-1350 x+e^x (-30+120 x)+\left (-450 x+30 e^x x\right ) \log (x)}{225 x+e^{2 x} x-450 x^2+225 x^3+e^x \left (30 x-30 x^2\right )} \, dx=\frac {30 \, {\left (\log \left (x\right ) + 4\right )}}{15 \, x - e^{x} - 15} \]
[In]
[Out]
Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {-450-1350 x+e^x (-30+120 x)+\left (-450 x+30 e^x x\right ) \log (x)}{225 x+e^{2 x} x-450 x^2+225 x^3+e^x \left (30 x-30 x^2\right )} \, dx=\frac {- 30 \log {\left (x \right )} - 120}{- 15 x + e^{x} + 15} \]
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-450-1350 x+e^x (-30+120 x)+\left (-450 x+30 e^x x\right ) \log (x)}{225 x+e^{2 x} x-450 x^2+225 x^3+e^x \left (30 x-30 x^2\right )} \, dx=\frac {30 \, {\left (\log \left (x\right ) + 4\right )}}{15 \, x - e^{x} - 15} \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-450-1350 x+e^x (-30+120 x)+\left (-450 x+30 e^x x\right ) \log (x)}{225 x+e^{2 x} x-450 x^2+225 x^3+e^x \left (30 x-30 x^2\right )} \, dx=\frac {30 \, {\left (\log \left (x\right ) + 4\right )}}{15 \, x - e^{x} - 15} \]
[In]
[Out]
Timed out. \[ \int \frac {-450-1350 x+e^x (-30+120 x)+\left (-450 x+30 e^x x\right ) \log (x)}{225 x+e^{2 x} x-450 x^2+225 x^3+e^x \left (30 x-30 x^2\right )} \, dx=\int -\frac {1350\,x+\ln \left (x\right )\,\left (450\,x-30\,x\,{\mathrm {e}}^x\right )-{\mathrm {e}}^x\,\left (120\,x-30\right )+450}{225\,x+x\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (30\,x-30\,x^2\right )-450\,x^2+225\,x^3} \,d x \]
[In]
[Out]