Integrand size = 71, antiderivative size = 27 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=25 \left (4-e^{\frac {2+5 e^{-x} x}{\log (-3+2 x)}}\right ) \] Output:
100-25*exp((2+5*x/exp(x))/ln(-3+2*x))
Time = 1.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=-25 e^{\frac {2+5 e^{-x} x}{\log (-3+2 x)}} \] Input:
Integrate[(E^(-x + (2*E^x + 5*x)/(E^x*Log[-3 + 2*x]))*(100*E^x + 250*x + ( 375 - 625*x + 250*x^2)*Log[-3 + 2*x]))/((-3 + 2*x)*Log[-3 + 2*x]^2),x]
Output:
-25*E^((2 + (5*x)/E^x)/Log[-3 + 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.
\(\displaystyle \int \frac {e^{\frac {e^{-x} \left (5 x+2 e^x\right )}{\log (2 x-3)}-x} \left (\left (250 x^2-625 x+375\right ) \log (2 x-3)+250 x+100 e^x\right )}{(2 x-3) \log ^2(2 x-3)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {125 e^{\frac {e^{-x} \left (5 x+2 e^x\right )}{\log (2 x-3)}-x} \left (2 x^2 \log (2 x-3)+2 x-5 x \log (2 x-3)+3 \log (2 x-3)\right )}{(2 x-3) \log ^2(2 x-3)}+\frac {100 e^{\frac {e^{-x} \left (5 x+2 e^x\right )}{\log (2 x-3)}}}{(2 x-3) \log ^2(2 x-3)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 125 \int \frac {e^{\frac {e^{-x} \left (5 x+2 e^x\right )}{\log (2 x-3)}-x}}{\log ^2(2 x-3)}dx+375 \int \frac {e^{\frac {e^{-x} \left (5 x+2 e^x\right )}{\log (2 x-3)}-x}}{(2 x-3) \log ^2(2 x-3)}dx+100 \int \frac {e^{\frac {e^{-x} \left (5 x+2 e^x\right )}{\log (2 x-3)}}}{(2 x-3) \log ^2(2 x-3)}dx+\frac {125}{2} \int \frac {e^{\frac {e^{-x} \left (5 x+2 e^x\right )}{\log (2 x-3)}-x}}{\log (2 x-3)}dx+\frac {125}{2} \int \frac {e^{\frac {e^{-x} \left (5 x+2 e^x\right )}{\log (2 x-3)}-x} (2 x-3)}{\log (2 x-3)}dx\) |
Input:
Int[(E^(-x + (2*E^x + 5*x)/(E^x*Log[-3 + 2*x]))*(100*E^x + 250*x + (375 - 625*x + 250*x^2)*Log[-3 + 2*x]))/((-3 + 2*x)*Log[-3 + 2*x]^2),x]
Output:
$Aborted
Time = 0.77 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
risch | \(-25 \,{\mathrm e}^{\frac {\left (2 \,{\mathrm e}^{x}+5 x \right ) {\mathrm e}^{-x}}{\ln \left (-3+2 x \right )}}\) | \(25\) |
parallelrisch | \(-25 \,{\mathrm e}^{\frac {\left (2 \,{\mathrm e}^{x}+5 x \right ) {\mathrm e}^{-x}}{\ln \left (-3+2 x \right )}}\) | \(25\) |
Input:
int(((250*x^2-625*x+375)*ln(-3+2*x)+100*exp(x)+250*x)*exp((2*exp(x)+5*x)/e xp(x)/ln(-3+2*x))/(-3+2*x)/exp(x)/ln(-3+2*x)^2,x,method=_RETURNVERBOSE)
Output:
-25*exp((2*exp(x)+5*x)*exp(-x)/ln(-3+2*x))
Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=-25 \, e^{\left (x - \frac {{\left (x e^{x} \log \left (2 \, x - 3\right ) - 5 \, x - 2 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (2 \, x - 3\right )}\right )} \] Input:
integrate(((250*x^2-625*x+375)*log(-3+2*x)+100*exp(x)+250*x)*exp((2*exp(x) +5*x)/exp(x)/log(-3+2*x))/(-3+2*x)/exp(x)/log(-3+2*x)^2,x, algorithm="fric as")
Output:
-25*e^(x - (x*e^x*log(2*x - 3) - 5*x - 2*e^x)*e^(-x)/log(2*x - 3))
Time = 0.46 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=- 25 e^{\frac {\left (5 x + 2 e^{x}\right ) e^{- x}}{\log {\left (2 x - 3 \right )}}} \] Input:
integrate(((250*x**2-625*x+375)*ln(-3+2*x)+100*exp(x)+250*x)*exp((2*exp(x) +5*x)/exp(x)/ln(-3+2*x))/(-3+2*x)/exp(x)/ln(-3+2*x)**2,x)
Output:
-25*exp((5*x + 2*exp(x))*exp(-x)/log(2*x - 3))
Exception generated. \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(((250*x^2-625*x+375)*log(-3+2*x)+100*exp(x)+250*x)*exp((2*exp(x) +5*x)/exp(x)/log(-3+2*x))/(-3+2*x)/exp(x)/log(-3+2*x)^2,x, algorithm="maxi ma")
Output:
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
\[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=\int { \frac {25 \, {\left (5 \, {\left (2 \, x^{2} - 5 \, x + 3\right )} \log \left (2 \, x - 3\right ) + 10 \, x + 4 \, e^{x}\right )} e^{\left (-x + \frac {{\left (5 \, x + 2 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (2 \, x - 3\right )}\right )}}{{\left (2 \, x - 3\right )} \log \left (2 \, x - 3\right )^{2}} \,d x } \] Input:
integrate(((250*x^2-625*x+375)*log(-3+2*x)+100*exp(x)+250*x)*exp((2*exp(x) +5*x)/exp(x)/log(-3+2*x))/(-3+2*x)/exp(x)/log(-3+2*x)^2,x, algorithm="giac ")
Output:
integrate(25*(5*(2*x^2 - 5*x + 3)*log(2*x - 3) + 10*x + 4*e^x)*e^(-x + (5* x + 2*e^x)*e^(-x)/log(2*x - 3))/((2*x - 3)*log(2*x - 3)^2), x)
Time = 3.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=-25\,{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{-x}+2}{\ln \left (2\,x-3\right )}} \] Input:
int((exp(-x)*exp((exp(-x)*(5*x + 2*exp(x)))/log(2*x - 3))*(250*x + 100*exp (x) + log(2*x - 3)*(250*x^2 - 625*x + 375)))/(log(2*x - 3)^2*(2*x - 3)),x)
Output:
-25*exp((5*x*exp(-x) + 2)/log(2*x - 3))
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=-25 e^{\frac {2 e^{x}+5 x}{e^{x} \mathrm {log}\left (2 x -3\right )}} \] Input:
int(((250*x^2-625*x+375)*log(-3+2*x)+100*exp(x)+250*x)*exp((2*exp(x)+5*x)/ exp(x)/log(-3+2*x))/(-3+2*x)/exp(x)/log(-3+2*x)^2,x)
Output:
- 25*e**((2*e**x + 5*x)/(e**x*log(2*x - 3)))