Integrand size = 75, antiderivative size = 27 \[ \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \left (-5+e^{\frac {1}{5} \left (5 x+80 x^2+x \log ^2\left (\frac {5}{x}\right )\right )}-160 x+2 \log \left (\frac {5}{x}\right )-\log ^2\left (\frac {5}{x}\right )\right ) \, dx=1+5 e^{-x \left (1+16 x+\frac {1}{5} \log ^2\left (\frac {5}{x}\right )\right )}+x \]
Time = 2.71 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \left (-5+e^{\frac {1}{5} \left (5 x+80 x^2+x \log ^2\left (\frac {5}{x}\right )\right )}-160 x+2 \log \left (\frac {5}{x}\right )-\log ^2\left (\frac {5}{x}\right )\right ) \, dx=5 e^{-\frac {1}{5} x \left (5+80 x+\log ^2\left (\frac {5}{x}\right )\right )}+x \]
Integrate[E^((-5*x - 80*x^2 - x*Log[5/x]^2)/5)*(-5 + E^((5*x + 80*x^2 + x* Log[5/x]^2)/5) - 160*x + 2*Log[5/x] - Log[5/x]^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.
\(\displaystyle \int e^{\frac {1}{5} \left (-80 x^2-5 x-x \log ^2\left (\frac {5}{x}\right )\right )} \left (e^{\frac {1}{5} \left (80 x^2+5 x+x \log ^2\left (\frac {5}{x}\right )\right )}-160 x-\log ^2\left (\frac {5}{x}\right )+2 \log \left (\frac {5}{x}\right )-5\right ) \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\exp \left (\frac {1}{5} \left (-80 x^2-5 x-x \log ^2\left (\frac {5}{x}\right )\right )+\frac {1}{5} x \left (80 x+\log ^2\left (\frac {5}{x}\right )+5\right )\right )-e^{\frac {1}{5} \left (-80 x^2-5 x-x \log ^2\left (\frac {5}{x}\right )\right )} \log ^2\left (\frac {5}{x}\right )+2 e^{\frac {1}{5} \left (-80 x^2-5 x-x \log ^2\left (\frac {5}{x}\right )\right )} \log \left (\frac {5}{x}\right )-5 e^{\frac {1}{5} \left (-80 x^2-5 x-x \log ^2\left (\frac {5}{x}\right )\right )}-160 x e^{\frac {1}{5} \left (-80 x^2-5 x-x \log ^2\left (\frac {5}{x}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \int e^{\frac {1}{5} \left (-80 x^2-\log ^2\left (\frac {5}{x}\right ) x-5 x\right )}dx-160 \int e^{\frac {1}{5} \left (-80 x^2-\log ^2\left (\frac {5}{x}\right ) x-5 x\right )} xdx+2 \int e^{\frac {1}{5} \left (-80 x^2-\log ^2\left (\frac {5}{x}\right ) x-5 x\right )} \log \left (\frac {5}{x}\right )dx-\int e^{\frac {1}{5} \left (-80 x^2-\log ^2\left (\frac {5}{x}\right ) x-5 x\right )} \log ^2\left (\frac {5}{x}\right )dx+x\) |
Int[E^((-5*x - 80*x^2 - x*Log[5/x]^2)/5)*(-5 + E^((5*x + 80*x^2 + x*Log[5/ x]^2)/5) - 160*x + 2*Log[5/x] - Log[5/x]^2),x]
3.22.69.3.1 Defintions of rubi rules used
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81
method | result | size |
risch | \(x +5 \,{\mathrm e}^{-\frac {x \left (\ln \left (\frac {5}{x}\right )^{2}+80 x +5\right )}{5}}\) | \(22\) |
default | \(x +5 \,{\mathrm e}^{-\frac {x \ln \left (\frac {5}{x}\right )^{2}}{5}-16 x^{2}-x}\) | \(26\) |
parts | \(x +5 \,{\mathrm e}^{-\frac {x \ln \left (\frac {5}{x}\right )^{2}}{5}-16 x^{2}-x}\) | \(26\) |
parallelrisch | \(\left (5+{\mathrm e}^{\frac {x \left (\ln \left (\frac {5}{x}\right )^{2}+80 x +5\right )}{5}} x \right ) {\mathrm e}^{-\frac {x \left (\ln \left (\frac {5}{x}\right )^{2}+80 x +5\right )}{5}}\) | \(42\) |
norman | \(\left (5+x \,{\mathrm e}^{\frac {x \ln \left (\frac {5}{x}\right )^{2}}{5}+16 x^{2}+x}\right ) {\mathrm e}^{-\frac {x \ln \left (\frac {5}{x}\right )^{2}}{5}-16 x^{2}-x}\) | \(46\) |
int((exp(1/5*x*ln(5/x)^2+16*x^2+x)-ln(5/x)^2+2*ln(5/x)-160*x-5)/exp(1/5*x* ln(5/x)^2+16*x^2+x),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67 \[ \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \left (-5+e^{\frac {1}{5} \left (5 x+80 x^2+x \log ^2\left (\frac {5}{x}\right )\right )}-160 x+2 \log \left (\frac {5}{x}\right )-\log ^2\left (\frac {5}{x}\right )\right ) \, dx={\left (x e^{\left (\frac {1}{5} \, x \log \left (\frac {5}{x}\right )^{2} + 16 \, x^{2} + x\right )} + 5\right )} e^{\left (-\frac {1}{5} \, x \log \left (\frac {5}{x}\right )^{2} - 16 \, x^{2} - x\right )} \]
integrate((exp(1/5*x*log(5/x)^2+16*x^2+x)-log(5/x)^2+2*log(5/x)-160*x-5)/e xp(1/5*x*log(5/x)^2+16*x^2+x),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \left (-5+e^{\frac {1}{5} \left (5 x+80 x^2+x \log ^2\left (\frac {5}{x}\right )\right )}-160 x+2 \log \left (\frac {5}{x}\right )-\log ^2\left (\frac {5}{x}\right )\right ) \, dx=x + 5 e^{- 16 x^{2} - \frac {x \log {\left (\frac {5}{x} \right )}^{2}}{5} - x} \]
integrate((exp(1/5*x*ln(5/x)**2+16*x**2+x)-ln(5/x)**2+2*ln(5/x)-160*x-5)/e xp(1/5*x*ln(5/x)**2+16*x**2+x),x)
Time = 0.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \left (-5+e^{\frac {1}{5} \left (5 x+80 x^2+x \log ^2\left (\frac {5}{x}\right )\right )}-160 x+2 \log \left (\frac {5}{x}\right )-\log ^2\left (\frac {5}{x}\right )\right ) \, dx=x + 5 \, e^{\left (-\frac {1}{5} \, x \log \left (5\right )^{2} + \frac {2}{5} \, x \log \left (5\right ) \log \left (x\right ) - \frac {1}{5} \, x \log \left (x\right )^{2} - 16 \, x^{2} - x\right )} \]
integrate((exp(1/5*x*log(5/x)^2+16*x^2+x)-log(5/x)^2+2*log(5/x)-160*x-5)/e xp(1/5*x*log(5/x)^2+16*x^2+x),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \left (-5+e^{\frac {1}{5} \left (5 x+80 x^2+x \log ^2\left (\frac {5}{x}\right )\right )}-160 x+2 \log \left (\frac {5}{x}\right )-\log ^2\left (\frac {5}{x}\right )\right ) \, dx=x + 5 \, e^{\left (-\frac {1}{5} \, x \log \left (\frac {5}{x}\right )^{2} - 16 \, x^{2} - x\right )} \]
integrate((exp(1/5*x*log(5/x)^2+16*x^2+x)-log(5/x)^2+2*log(5/x)-160*x-5)/e xp(1/5*x*log(5/x)^2+16*x^2+x),x, algorithm=\
Time = 10.88 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \left (-5+e^{\frac {1}{5} \left (5 x+80 x^2+x \log ^2\left (\frac {5}{x}\right )\right )}-160 x+2 \log \left (\frac {5}{x}\right )-\log ^2\left (\frac {5}{x}\right )\right ) \, dx=x+\frac {5\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-16\,x^2}\,{\mathrm {e}}^{-\frac {x\,{\ln \left (\frac {1}{x}\right )}^2}{5}}\,{\mathrm {e}}^{-\frac {2\,x\,\ln \left (\frac {1}{x}\right )\,\ln \left (5\right )}{5}}}{{\left ({\mathrm {e}}^{x\,{\ln \left (5\right )}^2}\right )}^{1/5}} \]
int(-exp(- x - 16*x^2 - (x*log(5/x)^2)/5)*(160*x - exp(x + 16*x^2 + (x*log (5/x)^2)/5) - 2*log(5/x) + log(5/x)^2 + 5),x)