Integrand size = 147, antiderivative size = 26 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=2-\frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \]
[Out]
\[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=\int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {-e^x (1-2 x)^2-4 x^2-6 x^2 (-1+2 x) \log \left (5-\frac {5}{2 x}\right )}{(1-2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx \\ & = \int \left (\frac {-1+2 x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2}-\frac {4 x^2 \left (-1+2 \log \left (5-\frac {5}{2 x}\right )-5 x \log \left (5-\frac {5}{2 x}\right )+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}\right ) \, dx \\ & = -\left (4 \int \frac {x^2 \left (-1+2 \log \left (5-\frac {5}{2 x}\right )-5 x \log \left (5-\frac {5}{2 x}\right )+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx\right )+\int \frac {-1+2 x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx \\ & = -\left (4 \int \frac {x^2 \left (1-\left (2-5 x+2 x^2\right ) \log \left (5-\frac {5}{2 x}\right )\right )}{(1-2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx\right )+\int \left (-\frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2}+\frac {2 x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2}\right ) \, dx \\ & = 2 \int \frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx-4 \int \left (\frac {-1+2 \log \left (5-\frac {5}{2 x}\right )-5 x \log \left (5-\frac {5}{2 x}\right )+2 x^2 \log \left (5-\frac {5}{2 x}\right )}{4 \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {x \left (-1+2 \log \left (5-\frac {5}{2 x}\right )-5 x \log \left (5-\frac {5}{2 x}\right )+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )}{2 \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {-1+2 \log \left (5-\frac {5}{2 x}\right )-5 x \log \left (5-\frac {5}{2 x}\right )+2 x^2 \log \left (5-\frac {5}{2 x}\right )}{4 (-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}\right ) \, dx-\int \frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx \\ & = 2 \int \frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx-2 \int \frac {x \left (-1+2 \log \left (5-\frac {5}{2 x}\right )-5 x \log \left (5-\frac {5}{2 x}\right )+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-\int \frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx-\int \frac {-1+2 \log \left (5-\frac {5}{2 x}\right )-5 x \log \left (5-\frac {5}{2 x}\right )+2 x^2 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-\int \frac {-1+2 \log \left (5-\frac {5}{2 x}\right )-5 x \log \left (5-\frac {5}{2 x}\right )+2 x^2 \log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx \\ & = 2 \int \frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx-2 \int \frac {x \left (-1+\left (2-5 x+2 x^2\right ) \log \left (5-\frac {5}{2 x}\right )\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-\int \frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx-\int \frac {1-\left (2-5 x+2 x^2\right ) \log \left (5-\frac {5}{2 x}\right )}{(1-2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-\int \frac {-1+\left (2-5 x+2 x^2\right ) \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx \\ & = 2 \int \frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx-2 \int \left (-\frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {2 x \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}-\frac {5 x^2 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {2 x^3 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}\right ) \, dx-\int \frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx-\int \left (-\frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {2 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}-\frac {5 x \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {2 x^2 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}\right ) \, dx-\int \left (-\frac {1}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {2 \log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}-\frac {5 x \log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {2 x^2 \log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}\right ) \, dx \\ & = 2 \int \frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-2 \int \frac {\log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-2 \int \frac {x^2 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-2 \int \frac {\log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-2 \int \frac {x^2 \log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+2 \int \frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx-4 \int \frac {x \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-4 \int \frac {x^3 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+5 \int \frac {x \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+5 \int \frac {x \log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+10 \int \frac {x^2 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+\int \frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+\int \frac {1}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-\int \frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx \\ & = 2 \int \frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-2 \int \frac {\log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-2 \int \frac {x^2 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-2 \int \frac {\log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+2 \int \frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx-2 \int \left (\frac {\log \left (5-\frac {5}{2 x}\right )}{4 \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {x \log \left (5-\frac {5}{2 x}\right )}{2 \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {\log \left (5-\frac {5}{2 x}\right )}{4 (-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}\right ) \, dx-4 \int \frac {x \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-4 \int \frac {x^3 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+5 \int \frac {x \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+5 \int \left (\frac {\log \left (5-\frac {5}{2 x}\right )}{2 \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}+\frac {\log \left (5-\frac {5}{2 x}\right )}{2 (-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3}\right ) \, dx+10 \int \frac {x^2 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+\int \frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+\int \frac {1}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-\int \frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx \\ & = -\left (\frac {1}{2} \int \frac {\log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx\right )-\frac {1}{2} \int \frac {\log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+2 \int \frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-2 \int \frac {\log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-2 \int \frac {x^2 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-2 \int \frac {\log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+2 \int \frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx+\frac {5}{2} \int \frac {\log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+\frac {5}{2} \int \frac {\log \left (5-\frac {5}{2 x}\right )}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-4 \int \frac {x \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-4 \int \frac {x^3 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+5 \int \frac {x \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+10 \int \frac {x^2 \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+\int \frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx+\int \frac {1}{(-1+2 x) \left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-\int \frac {x \log \left (5-\frac {5}{2 x}\right )}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^3} \, dx-\int \frac {1}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \, dx \\ \end{align*}
Time = 0.77 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=-\frac {x}{\left (e^x+2 x^2 \log \left (5-\frac {5}{2 x}\right )\right )^2} \]
[In]
[Out]
Time = 3.83 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81
method | result | size |
parallelrisch | \(-\frac {x}{4 x^{4} \ln \left (\frac {-\frac {5}{2}+5 x}{x}\right )^{2}+4 \,{\mathrm e}^{x} x^{2} \ln \left (\frac {-\frac {5}{2}+5 x}{x}\right )+{\mathrm e}^{2 x}}\) | \(47\) |
risch | \(\frac {x}{\left (\pi \,x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{2}\right )}{x}\right )-\pi \,x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{2}\right )}{x}\right )^{2}-\pi \,x^{2} \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{2}\right )}{x}\right )^{2}+\pi \,x^{2} \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{2}\right )}{x}\right )^{3}+2 i \ln \left (5\right ) x^{2}-2 i x^{2} \ln \left (x \right )+2 i x^{2} \ln \left (x -\frac {1}{2}\right )+i {\mathrm e}^{x}\right )^{2}}\) | \(133\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=-\frac {x}{4 \, x^{4} \log \left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )^{2} + 4 \, x^{2} e^{x} \log \left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right ) + e^{\left (2 \, x\right )}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=- \frac {x}{4 x^{4} \log {\left (\frac {5 x - \frac {5}{2}}{x} \right )}^{2} + 4 x^{2} e^{x} \log {\left (\frac {5 x - \frac {5}{2}}{x} \right )} + e^{2 x}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (23) = 46\).
Time = 0.36 (sec) , antiderivative size = 123, normalized size of antiderivative = 4.73 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=-\frac {x}{4 \, x^{4} \log \left (2 \, x - 1\right )^{2} - 8 \, x^{4} {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (x\right ) + 4 \, x^{4} \log \left (x\right )^{2} + 4 \, {\left (\log \left (5\right )^{2} - 2 \, \log \left (5\right ) \log \left (2\right ) + \log \left (2\right )^{2}\right )} x^{4} + 4 \, {\left (x^{2} {\left (\log \left (5\right ) - \log \left (2\right )\right )} - x^{2} \log \left (x\right )\right )} e^{x} + 4 \, {\left (2 \, x^{4} {\left (\log \left (5\right ) - \log \left (2\right )\right )} - 2 \, x^{4} \log \left (x\right ) + x^{2} e^{x}\right )} \log \left (2 \, x - 1\right ) + e^{\left (2 \, x\right )}} \]
[In]
[Out]
none
Time = 0.44 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=-\frac {x}{4 \, x^{4} \log \left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )^{2} + 4 \, x^{2} e^{x} \log \left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right ) + e^{\left (2 \, x\right )}} \]
[In]
[Out]
Time = 9.47 (sec) , antiderivative size = 116, normalized size of antiderivative = 4.46 \[ \int \frac {4 x^2+e^x \left (1-4 x+4 x^2\right )+\left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )}{e^{3 x} (-1+2 x)+e^{2 x} \left (-6 x^2+12 x^3\right ) \log \left (\frac {-5+10 x}{2 x}\right )+e^x \left (-12 x^4+24 x^5\right ) \log ^2\left (\frac {-5+10 x}{2 x}\right )+\left (-8 x^6+16 x^7\right ) \log ^3\left (\frac {-5+10 x}{2 x}\right )} \, dx=\frac {{\mathrm {e}}^x\,\left (-4\,x^5+12\,x^4-9\,x^3+2\,x^2\right )+2\,x^4-4\,x^5}{\left ({\mathrm {e}}^{2\,x}+4\,x^4\,{\ln \left (\frac {5\,x-\frac {5}{2}}{x}\right )}^2+4\,x^2\,{\mathrm {e}}^x\,\ln \left (\frac {5\,x-\frac {5}{2}}{x}\right )\right )\,\left (9\,x^2\,{\mathrm {e}}^x-12\,x^3\,{\mathrm {e}}^x+4\,x^4\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^x-2\,x^3+4\,x^4\right )} \]
[In]
[Out]