Integrand size = 109, antiderivative size = 30 \[ \int \frac {5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-5+160 x^2-60 x^3\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )}{e^{32 x^2-8 x^3}+e^{16 x^2-4 x^3} (-8-2 x) \log \left (\frac {1}{x}\right )+\left (16+8 x+x^2\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=-3+\frac {5 x}{4+x-\frac {e^{4 (4-x) x^2}}{\log \left (\frac {1}{x}\right )}} \]
Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-5+160 x^2-60 x^3\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )}{e^{32 x^2-8 x^3}+e^{16 x^2-4 x^3} (-8-2 x) \log \left (\frac {1}{x}\right )+\left (16+8 x+x^2\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {5 \left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )\right )}{e^{16 x^2}-e^{4 x^3} (4+x) \log \left (\frac {1}{x}\right )} \]
Integrate[(5*E^(16*x^2 - 4*x^3) + E^(16*x^2 - 4*x^3)*(-5 + 160*x^2 - 60*x^ 3)*Log[x^(-1)] + 20*Log[x^(-1)]^2)/(E^(32*x^2 - 8*x^3) + E^(16*x^2 - 4*x^3 )*(-8 - 2*x)*Log[x^(-1)] + (16 + 8*x + x^2)*Log[x^(-1)]^2),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-60 x^3+160 x^2-5\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )}{\left (x^2+8 x+16\right ) \log ^2\left (\frac {1}{x}\right )+e^{32 x^2-8 x^3}+e^{16 x^2-4 x^3} (-2 x-8) \log \left (\frac {1}{x}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{8 x^3} \left (5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-60 x^3+160 x^2-5\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )\right )}{\left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-5 e^{8 x^3-4 x^2 (x+4)} \left (12 x^3 \log \left (\frac {1}{x}\right )-32 x^2 \log \left (\frac {1}{x}\right )+\log \left (\frac {1}{x}\right )-1\right )+\frac {5 e^{8 x^3} (x+4) \log \left (\frac {1}{x}\right ) \left (12 x^3 \log \left (\frac {1}{x}\right )-32 x^2 \log \left (\frac {1}{x}\right )+\log \left (\frac {1}{x}\right )-1\right )}{-e^{32 x^2}+4 e^{4 x^2 (x+4)} \log \left (\frac {1}{x}\right )+e^{4 x^2 (x+4)} x \log \left (\frac {1}{x}\right )}-\frac {5 e^{8 x^3} \log \left (\frac {1}{x}\right ) \left (12 x^4 \log \left (\frac {1}{x}\right )+16 x^3 \log \left (\frac {1}{x}\right )-128 x^2 \log \left (\frac {1}{x}\right )-x+x \log \left (\frac {1}{x}\right )-4\right )}{\left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {5 e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )+e^{16 x^2}-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {4 e^{4 x^3} \log \left (\frac {1}{x}\right )}{(x+4) \left (4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )-e^{16 x^2}\right )}-\frac {e^{4 x^3+16 x^2} \left (12 \log \left (\frac {1}{x}\right ) x^4+16 \log \left (\frac {1}{x}\right ) x^3-128 \log \left (\frac {1}{x}\right ) x^2+\log \left (\frac {1}{x}\right ) x-x-4\right )}{(x+4) \left (-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 5 \int \frac {e^{4 x^3} \left (4 e^{4 x^3} \log ^2\left (\frac {1}{x}\right )-e^{16 x^2} \left (12 x^3-32 x^2+1\right ) \log \left (\frac {1}{x}\right )+e^{16 x^2}\right )}{\left (e^{16 x^2}-e^{4 x^3} (x+4) \log \left (\frac {1}{x}\right )\right )^2}dx\) |
Int[(5*E^(16*x^2 - 4*x^3) + E^(16*x^2 - 4*x^3)*(-5 + 160*x^2 - 60*x^3)*Log [x^(-1)] + 20*Log[x^(-1)]^2)/(E^(32*x^2 - 8*x^3) + E^(16*x^2 - 4*x^3)*(-8 - 2*x)*Log[x^(-1)] + (16 + 8*x + x^2)*Log[x^(-1)]^2),x]
3.7.32.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23
| method | result | size |
| parallelrisch | \(\frac {5 x \ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )+4 \ln \left (\frac {1}{x}\right )-{\mathrm e}^{-4 x^{3}+16 x^{2}}}\) | \(37\) |
| risch | \(-\frac {20}{4+x}-\frac {5 x \,{\mathrm e}^{-4 \left (x -4\right ) x^{2}}}{\left (4+x \right ) \left (x \ln \left (x \right )+{\mathrm e}^{-4 \left (x -4\right ) x^{2}}+4 \ln \left (x \right )\right )}\) | \(46\) |
int((20*ln(1/x)^2+(-60*x^3+160*x^2-5)*exp(-4*x^3+16*x^2)*ln(1/x)+5*exp(-4* x^3+16*x^2))/((x^2+8*x+16)*ln(1/x)^2+(-2*x-8)*exp(-4*x^3+16*x^2)*ln(1/x)+e xp(-4*x^3+16*x^2)^2),x,method=_RETURNVERBOSE)
Time = 0.39 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-5+160 x^2-60 x^3\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )}{e^{32 x^2-8 x^3}+e^{16 x^2-4 x^3} (-8-2 x) \log \left (\frac {1}{x}\right )+\left (16+8 x+x^2\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {5 \, {\left (e^{\left (-4 \, x^{3} + 16 \, x^{2}\right )} - 4 \, \log \left (\frac {1}{x}\right )\right )}}{{\left (x + 4\right )} \log \left (\frac {1}{x}\right ) - e^{\left (-4 \, x^{3} + 16 \, x^{2}\right )}} \]
integrate((20*log(1/x)^2+(-60*x^3+160*x^2-5)*exp(-4*x^3+16*x^2)*log(1/x)+5 *exp(-4*x^3+16*x^2))/((x^2+8*x+16)*log(1/x)^2+(-2*x-8)*exp(-4*x^3+16*x^2)* log(1/x)+exp(-4*x^3+16*x^2)^2),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-5+160 x^2-60 x^3\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )}{e^{32 x^2-8 x^3}+e^{16 x^2-4 x^3} (-8-2 x) \log \left (\frac {1}{x}\right )+\left (16+8 x+x^2\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=- \frac {5 x \log {\left (\frac {1}{x} \right )}}{- x \log {\left (\frac {1}{x} \right )} + e^{- 4 x^{3} + 16 x^{2}} - 4 \log {\left (\frac {1}{x} \right )}} \]
integrate((20*ln(1/x)**2+(-60*x**3+160*x**2-5)*exp(-4*x**3+16*x**2)*ln(1/x )+5*exp(-4*x**3+16*x**2))/((x**2+8*x+16)*ln(1/x)**2+(-2*x-8)*exp(-4*x**3+1 6*x**2)*ln(1/x)+exp(-4*x**3+16*x**2)**2),x)
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-5+160 x^2-60 x^3\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )}{e^{32 x^2-8 x^3}+e^{16 x^2-4 x^3} (-8-2 x) \log \left (\frac {1}{x}\right )+\left (16+8 x+x^2\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=-\frac {5 \, {\left (4 \, e^{\left (4 \, x^{3}\right )} \log \left (x\right ) + e^{\left (16 \, x^{2}\right )}\right )}}{{\left (x + 4\right )} e^{\left (4 \, x^{3}\right )} \log \left (x\right ) + e^{\left (16 \, x^{2}\right )}} \]
integrate((20*log(1/x)^2+(-60*x^3+160*x^2-5)*exp(-4*x^3+16*x^2)*log(1/x)+5 *exp(-4*x^3+16*x^2))/((x^2+8*x+16)*log(1/x)^2+(-2*x-8)*exp(-4*x^3+16*x^2)* log(1/x)+exp(-4*x^3+16*x^2)^2),x, algorithm=\
Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-5+160 x^2-60 x^3\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )}{e^{32 x^2-8 x^3}+e^{16 x^2-4 x^3} (-8-2 x) \log \left (\frac {1}{x}\right )+\left (16+8 x+x^2\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {5 \, x \log \left (x\right )}{x \log \left (x\right ) + e^{\left (-4 \, x^{3} + 16 \, x^{2}\right )} + 4 \, \log \left (x\right )} \]
integrate((20*log(1/x)^2+(-60*x^3+160*x^2-5)*exp(-4*x^3+16*x^2)*log(1/x)+5 *exp(-4*x^3+16*x^2))/((x^2+8*x+16)*log(1/x)^2+(-2*x-8)*exp(-4*x^3+16*x^2)* log(1/x)+exp(-4*x^3+16*x^2)^2),x, algorithm=\
Timed out. \[ \int \frac {5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-5+160 x^2-60 x^3\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )}{e^{32 x^2-8 x^3}+e^{16 x^2-4 x^3} (-8-2 x) \log \left (\frac {1}{x}\right )+\left (16+8 x+x^2\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\int \frac {20\,{\ln \left (\frac {1}{x}\right )}^2-{\mathrm {e}}^{16\,x^2-4\,x^3}\,\left (60\,x^3-160\,x^2+5\right )\,\ln \left (\frac {1}{x}\right )+5\,{\mathrm {e}}^{16\,x^2-4\,x^3}}{\left (x^2+8\,x+16\right )\,{\ln \left (\frac {1}{x}\right )}^2-{\mathrm {e}}^{16\,x^2-4\,x^3}\,\left (2\,x+8\right )\,\ln \left (\frac {1}{x}\right )+{\mathrm {e}}^{32\,x^2-8\,x^3}} \,d x \]
int((5*exp(16*x^2 - 4*x^3) + 20*log(1/x)^2 - log(1/x)*exp(16*x^2 - 4*x^3)* (60*x^3 - 160*x^2 + 5))/(exp(32*x^2 - 8*x^3) + log(1/x)^2*(8*x + x^2 + 16) - log(1/x)*exp(16*x^2 - 4*x^3)*(2*x + 8)),x)