Integrand size = 50, antiderivative size = 23 \[ \int \frac {e^{\frac {2 x}{-8+4 x}} \left (-4+5 x-x^2\right )+\left (60-60 x+15 x^2\right ) \log (4)}{20-20 x+5 x^2} \, dx=-\frac {1}{5} e^{\frac {x}{2 (-2+x)}} x+3 x \log (4) \] Output:
6*x*ln(2)-1/5*x*exp(x/(4*x-8))^2
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {2 x}{-8+4 x}} \left (-4+5 x-x^2\right )+\left (60-60 x+15 x^2\right ) \log (4)}{20-20 x+5 x^2} \, dx=-\frac {1}{5} e^{\frac {1}{2} \left (1+\frac {2}{-2+x}\right )} x+x \log (64) \] Input:
Integrate[(E^((2*x)/(-8 + 4*x))*(-4 + 5*x - x^2) + (60 - 60*x + 15*x^2)*Lo g[4])/(20 - 20*x + 5*x^2),x]
Output:
-1/5*(E^((1 + 2/(-2 + x))/2)*x) + x*Log[64]
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.
\(\displaystyle \int \frac {e^{\frac {2 x}{4 x-8}} \left (-x^2+5 x-4\right )+\left (15 x^2-60 x+60\right ) \log (4)}{5 x^2-20 x+20} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 20 \int -\frac {e^{-\frac {x}{2 (2-x)}} \left (x^2-5 x+4\right )-15 \left (x^2-4 x+4\right ) \log (4)}{100 (2-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{5} \int \frac {e^{-\frac {x}{2 (2-x)}} \left (x^2-5 x+4\right )-15 \left (x^2-4 x+4\right ) \log (4)}{(2-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{5} \int \left (\frac {e^{\frac {x}{2 (x-2)}} (x-4) (x-1)}{(x-2)^2}-15 \log (4)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-\int e^{\frac {x}{2 (x-2)}}dx-\sqrt {e} \operatorname {ExpIntegralEi}\left (\frac {1}{x-2}\right )-2 e^{\frac {1}{x-2}+\frac {1}{2}}+15 x \log (4)\right )\) |
Input:
Int[(E^((2*x)/(-8 + 4*x))*(-4 + 5*x - x^2) + (60 - 60*x + 15*x^2)*Log[4])/ (20 - 20*x + 5*x^2),x]
Output:
$Aborted
Time = 0.63 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83
method | result | size |
risch | \(6 x \ln \left (2\right )-\frac {x \,{\mathrm e}^{\frac {x}{2 x -4}}}{5}\) | \(19\) |
parallelrisch | \(-\frac {x \,{\mathrm e}^{\frac {x}{2 x -4}}}{5}+6 x \ln \left (2\right )+48 \ln \left (2\right )\) | \(25\) |
parts | \(-\frac {{\mathrm e}^{\frac {1}{2}+\frac {1}{-2+x}} \left (-2+x \right )}{5}-\frac {2 \,{\mathrm e}^{\frac {1}{2}+\frac {1}{-2+x}}}{5}+6 x \ln \left (2\right )\) | \(38\) |
derivativedivides | \(-\frac {{\mathrm e}^{\frac {1}{2}+\frac {1}{-2+x}} \left (-2+x \right )}{5}+6 \left (-2+x \right ) \ln \left (2\right )-\frac {2 \,{\mathrm e}^{\frac {1}{2}+\frac {1}{-2+x}}}{5}\) | \(40\) |
default | \(-\frac {{\mathrm e}^{\frac {1}{2}+\frac {1}{-2+x}} \left (-2+x \right )}{5}+6 \left (-2+x \right ) \ln \left (2\right )-\frac {2 \,{\mathrm e}^{\frac {1}{2}+\frac {1}{-2+x}}}{5}\) | \(40\) |
norman | \(\frac {\frac {2 x \,{\mathrm e}^{\frac {2 x}{4 x -8}}}{5}-\frac {x^{2} {\mathrm e}^{\frac {2 x}{4 x -8}}}{5}+6 x^{2} \ln \left (2\right )-24 \ln \left (2\right )}{-2+x}\) | \(51\) |
orering | \(\frac {\left (1+x \right ) \left (\left (-x^{2}+5 x -4\right ) {\mathrm e}^{\frac {2 x}{4 x -8}}+2 \left (15 x^{2}-60 x +60\right ) \ln \left (2\right )\right )}{5 x^{2}-20 x +20}+\frac {\left (5 x -4\right ) \left (-2+x \right )^{2} \left (\frac {\left (5-2 x \right ) {\mathrm e}^{\frac {2 x}{4 x -8}}+2 \left (-x^{2}+5 x -4\right ) {\mathrm e}^{\frac {2 x}{4 x -8}} \left (\frac {1}{4 x -8}-\frac {4 x}{\left (4 x -8\right )^{2}}\right )+2 \left (30 x -60\right ) \ln \left (2\right )}{5 x^{2}-20 x +20}-\frac {\left (\left (-x^{2}+5 x -4\right ) {\mathrm e}^{\frac {2 x}{4 x -8}}+2 \left (15 x^{2}-60 x +60\right ) \ln \left (2\right )\right ) \left (10 x -20\right )}{\left (5 x^{2}-20 x +20\right )^{2}}\right )}{5 x -8}\) | \(215\) |
Input:
int(((-x^2+5*x-4)*exp(x/(4*x-8))^2+2*(15*x^2-60*x+60)*ln(2))/(5*x^2-20*x+2 0),x,method=_RETURNVERBOSE)
Output:
6*x*ln(2)-1/5*x*exp(1/2*x/(-2+x))
Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\frac {2 x}{-8+4 x}} \left (-4+5 x-x^2\right )+\left (60-60 x+15 x^2\right ) \log (4)}{20-20 x+5 x^2} \, dx=-\frac {1}{5} \, x e^{\left (\frac {x}{2 \, {\left (x - 2\right )}}\right )} + 6 \, x \log \left (2\right ) \] Input:
integrate(((-x^2+5*x-4)*exp(x/(4*x-8))^2+2*(15*x^2-60*x+60)*log(2))/(5*x^2 -20*x+20),x, algorithm="fricas")
Output:
-1/5*x*e^(1/2*x/(x - 2)) + 6*x*log(2)
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {2 x}{-8+4 x}} \left (-4+5 x-x^2\right )+\left (60-60 x+15 x^2\right ) \log (4)}{20-20 x+5 x^2} \, dx=- \frac {x e^{\frac {2 x}{4 x - 8}}}{5} + 6 x \log {\left (2 \right )} \] Input:
integrate(((-x**2+5*x-4)*exp(x/(4*x-8))**2+2*(15*x**2-60*x+60)*ln(2))/(5*x **2-20*x+20),x)
Output:
-x*exp(2*x/(4*x - 8))/5 + 6*x*log(2)
\[ \int \frac {e^{\frac {2 x}{-8+4 x}} \left (-4+5 x-x^2\right )+\left (60-60 x+15 x^2\right ) \log (4)}{20-20 x+5 x^2} \, dx=\int { -\frac {{\left (x^{2} - 5 \, x + 4\right )} e^{\left (\frac {x}{2 \, {\left (x - 2\right )}}\right )} - 30 \, {\left (x^{2} - 4 \, x + 4\right )} \log \left (2\right )}{5 \, {\left (x^{2} - 4 \, x + 4\right )}} \,d x } \] Input:
integrate(((-x^2+5*x-4)*exp(x/(4*x-8))^2+2*(15*x^2-60*x+60)*log(2))/(5*x^2 -20*x+20),x, algorithm="maxima")
Output:
-1/5*x*e^(1/(x - 2) + 1/2) + 6*(x - 4/(x - 2) + 4*log(x - 2))*log(2) + 24* (2/(x - 2) - log(x - 2))*log(2) - 24*log(2)/(x - 2) + 4/5*e^(1/(x - 2) + 1 /2) + 4/5*integrate(e^(1/(x - 2) + 1/2)/(x^2 - 4*x + 4), x)
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {2 x}{-8+4 x}} \left (-4+5 x-x^2\right )+\left (60-60 x+15 x^2\right ) \log (4)}{20-20 x+5 x^2} \, dx=-\frac {2 \, {\left (\frac {x e^{\left (\frac {x}{2 \, {\left (x - 2\right )}}\right )}}{x - 2} - 30 \, \log \left (2\right )\right )}}{5 \, {\left (\frac {x}{x - 2} - 1\right )}} \] Input:
integrate(((-x^2+5*x-4)*exp(x/(4*x-8))^2+2*(15*x^2-60*x+60)*log(2))/(5*x^2 -20*x+20),x, algorithm="giac")
Output:
-2/5*(x*e^(1/2*x/(x - 2))/(x - 2) - 30*log(2))/(x/(x - 2) - 1)
Time = 0.57 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\frac {2 x}{-8+4 x}} \left (-4+5 x-x^2\right )+\left (60-60 x+15 x^2\right ) \log (4)}{20-20 x+5 x^2} \, dx=x\,\left (\ln \left (64\right )-\frac {{\mathrm {e}}^{\frac {2\,x}{4\,x-8}}}{5}\right ) \] Input:
int((2*log(2)*(15*x^2 - 60*x + 60) - exp((2*x)/(4*x - 8))*(x^2 - 5*x + 4)) /(5*x^2 - 20*x + 20),x)
Output:
x*(log(64) - exp((2*x)/(4*x - 8))/5)
Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {2 x}{-8+4 x}} \left (-4+5 x-x^2\right )+\left (60-60 x+15 x^2\right ) \log (4)}{20-20 x+5 x^2} \, dx=\frac {x \left (-e^{\frac {x}{2 x -4}}+30 \,\mathrm {log}\left (2\right )\right )}{5} \] Input:
int(((-x^2+5*x-4)*exp(x/(4*x-8))^2+2*(15*x^2-60*x+60)*log(2))/(5*x^2-20*x+ 20),x)
Output:
(x*( - e**(x/(2*x - 4)) + 30*log(2)))/5