Integrand size = 208, antiderivative size = 35 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=x+\frac {x \log (\log (3))}{x-\frac {x^2}{\log \left (\frac {2+\frac {5}{x}+x}{5+e^3}\right )}} \]
[Out]
\[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=\int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x^3+x^4+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )-\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right ) (2 x-\log (\log (3)))-x^2 (-5+\log (\log (3)))+5 \log (\log (3))}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx \\ & = \int \left (\frac {2 x^3}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {x^4}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {\log ^2\left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}-\frac {x^2 (-5+\log (\log (3)))}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {5 \log (\log (3))}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {\log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right ) (-2 x+\log (\log (3)))}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx \\ & = 2 \int \frac {x^3}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+(5-\log (\log (3))) \int \frac {x^2}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+(5 \log (\log (3))) \int \frac {1}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {x^4}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {\log ^2\left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {\log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right ) (-2 x+\log (\log (3)))}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx \\ & = 2 \int \left (-\frac {2}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {x}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {10-x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx+(5-\log (\log (3))) \int \left (\frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {-5-2 x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx+(5 \log (\log (3))) \int \left (\frac {i}{2 ((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {i}{2 ((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx+\int \left (-\frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}-\frac {2 x}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {x^2}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {5+12 x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx+\int \frac {\log ^2\left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \left (-\frac {2 x \log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {\log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right ) \log (\log (3))}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx \\ & = 2 \int \frac {10-x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-2 \int \frac {x \log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-4 \int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+(5-\log (\log (3))) \int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+(5-\log (\log (3))) \int \frac {-5-2 x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\frac {1}{2} (5 i \log (\log (3))) \int \frac {1}{((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\frac {1}{2} (5 i \log (\log (3))) \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\log (\log (3)) \int \frac {\log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-\int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {x^2}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {5+12 x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {\log ^2\left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx \\ & = 2 \int \left (\frac {10}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}-\frac {x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx-2 \int \frac {x \log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-4 \int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+(5-\log (\log (3))) \int \left (-\frac {5}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}-\frac {2 x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx+(5-\log (\log (3))) \int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\frac {1}{2} (5 i \log (\log (3))) \int \frac {1}{((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\frac {1}{2} (5 i \log (\log (3))) \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\log (\log (3)) \int \frac {\log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \left (\frac {5}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {12 x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx-\int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {x^2}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {\log ^2\left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx \\ & = -\left (2 \int \frac {x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx\right )-2 \int \frac {x \log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-4 \int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+5 \int \frac {1}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+12 \int \frac {x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+20 \int \frac {1}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+(5-\log (\log (3))) \int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-(2 (5-\log (\log (3)))) \int \frac {x}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-(5 (5-\log (\log (3)))) \int \frac {1}{\left (5+2 x+x^2\right ) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\frac {1}{2} (5 i \log (\log (3))) \int \frac {1}{((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\frac {1}{2} (5 i \log (\log (3))) \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\log (\log (3)) \int \frac {\log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-\int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {x^2}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {\log ^2\left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx \\ & = -\left (2 \int \left (\frac {1+\frac {i}{2}}{((2-4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {1-\frac {i}{2}}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx\right )-2 \int \frac {x \log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-4 \int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+5 \int \left (\frac {i}{2 ((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {i}{2 ((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx+12 \int \left (\frac {1+\frac {i}{2}}{((2-4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {1-\frac {i}{2}}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx+20 \int \left (\frac {i}{2 ((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {i}{2 ((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx+(5-\log (\log (3))) \int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-(2 (5-\log (\log (3)))) \int \left (\frac {1+\frac {i}{2}}{((2-4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {1-\frac {i}{2}}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx-(5 (5-\log (\log (3)))) \int \left (\frac {i}{2 ((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}+\frac {i}{2 ((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2}\right ) \, dx+\frac {1}{2} (5 i \log (\log (3))) \int \frac {1}{((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\frac {1}{2} (5 i \log (\log (3))) \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\log (\log (3)) \int \frac {\log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-\int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {x^2}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {\log ^2\left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx \\ & = \frac {5}{2} i \int \frac {1}{((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\frac {5}{2} i \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+10 i \int \frac {1}{((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+10 i \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-2 \int \frac {x \log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-(2-i) \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-(2+i) \int \frac {1}{((2-4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-4 \int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+(12-6 i) \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+(12+6 i) \int \frac {1}{((2-4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-\frac {1}{2} (5 i (5-\log (\log (3)))) \int \frac {1}{((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-\frac {1}{2} (5 i (5-\log (\log (3)))) \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+(5-\log (\log (3))) \int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-((2-i) (5-\log (\log (3)))) \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-((2+i) (5-\log (\log (3)))) \int \frac {1}{((2-4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\frac {1}{2} (5 i \log (\log (3))) \int \frac {1}{((-2+4 i)-2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\frac {1}{2} (5 i \log (\log (3))) \int \frac {1}{((2+4 i)+2 x) \left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\log (\log (3)) \int \frac {\log \left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx-\int \frac {1}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {x^2}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx+\int \frac {\log ^2\left (\frac {5+2 x+x^2}{\left (5+e^3\right ) x}\right )}{\left (x+\log \left (5+e^3\right )-\log \left (2+\frac {5}{x}+x\right )\right )^2} \, dx \\ \end{align*}
\[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=\int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx \]
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Time = 1.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00
method | result | size |
risch | \(x -\frac {x \ln \left (\ln \left (3\right )\right )}{x -\ln \left (\frac {x^{2}+2 x +5}{x \,{\mathrm e}^{3}+5 x}\right )}\) | \(35\) |
default | \(x +\frac {-\ln \left (\ln \left (3\right )\right ) \ln \left (\frac {x^{2}+2 x +5}{x}\right )+\ln \left (\ln \left (3\right )\right ) \ln \left ({\mathrm e}^{3}+5\right )}{\ln \left ({\mathrm e}^{3}+5\right )+x -\ln \left (\frac {x^{2}+2 x +5}{x}\right )}\) | \(56\) |
norman | \(\frac {x^{2}-\ln \left (\ln \left (3\right )\right ) \ln \left (\frac {x^{2}+2 x +5}{x \,{\mathrm e}^{3}+5 x}\right )-x \ln \left (\frac {x^{2}+2 x +5}{x \,{\mathrm e}^{3}+5 x}\right )}{x -\ln \left (\frac {x^{2}+2 x +5}{x \,{\mathrm e}^{3}+5 x}\right )}\) | \(80\) |
parallelrisch | \(-\frac {\ln \left (\ln \left (3\right )\right ) x -x^{2}+\ln \left (\frac {x^{2}+2 x +5}{x \left ({\mathrm e}^{3}+5\right )}\right ) x +4 x -4 \ln \left (\frac {x^{2}+2 x +5}{x \left ({\mathrm e}^{3}+5\right )}\right )}{x -\ln \left (\frac {x^{2}+2 x +5}{x \left ({\mathrm e}^{3}+5\right )}\right )}\) | \(84\) |
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Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.71 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=\frac {x^{2} - x \log \left (\frac {x^{2} + 2 \, x + 5}{x e^{3} + 5 \, x}\right ) - x \log \left (\log \left (3\right )\right )}{x - \log \left (\frac {x^{2} + 2 \, x + 5}{x e^{3} + 5 \, x}\right )} \]
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Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=x + \frac {x \log {\left (\log {\left (3 \right )} \right )}}{- x + \log {\left (\frac {x^{2} + 2 x + 5}{5 x + x e^{3}} \right )}} \]
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Time = 0.32 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.60 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=\frac {x^{2} + x {\left (\log \left (e^{3} + 5\right ) - \log \left (\log \left (3\right )\right )\right )} - x \log \left (x^{2} + 2 \, x + 5\right ) + x \log \left (x\right )}{x - \log \left (x^{2} + 2 \, x + 5\right ) + \log \left (x\right ) + \log \left (e^{3} + 5\right )} \]
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Time = 0.42 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.66 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=\frac {x^{2} - x \log \left (x^{2} + 2 \, x + 5\right ) + x \log \left (x e^{3} + 5 \, x\right ) - x \log \left (\log \left (3\right )\right )}{x - \log \left (x^{2} + 2 \, x + 5\right ) + \log \left (x e^{3} + 5 \, x\right )} \]
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Time = 13.49 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.29 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=-\frac {x\,\ln \left (\frac {x^2+2\,x+5}{5\,x+x\,{\mathrm {e}}^3}\right )-x^2+\ln \left (\frac {x^2+2\,x+5}{5\,x+x\,{\mathrm {e}}^3}\right )\,\ln \left (\ln \left (3\right )\right )}{x-\ln \left (\frac {x^2+2\,x+5}{5\,x+x\,{\mathrm {e}}^3}\right )} \]
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