\(\int \frac {e^{-x+\frac {e^{-x} (2 e^x+5 x)}{\log (-3+2 x)}} (100 e^x+250 x+(375-625 x+250 x^2) \log (-3+2 x))}{(-3+2 x) \log ^2(-3+2 x)} \, dx\) [6237]

Optimal result

Integrand size = 71, antiderivative size = 27 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=25 \left (4-e^{\frac {2+5 e^{-x} x}{\log (-3+2 x)}}\right ) \]

[Out]

100-25*exp((2+5*x/exp(x))/ln(-3+2*x))

Rubi [F]

\[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=\int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx \]

[In]

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]

[Out]

125*Defer[Int][E^(-x + (2*E^x + 5*x)/(E^x*Log[-3 + 2*x]))/Log[-3 + 2*x]^2, x] + 375*Defer[Int][E^(-x + (2*E^x
+ 5*x)/(E^x*Log[-3 + 2*x]))/((-3 + 2*x)*Log[-3 + 2*x]^2), x] + 100*Defer[Int][E^((2*E^x + 5*x)/(E^x*Log[-3 + 2
*x]))/((-3 + 2*x)*Log[-3 + 2*x]^2), x] + (125*Defer[Int][E^(-x + (2*E^x + 5*x)/(E^x*Log[-3 + 2*x]))/Log[-3 + 2
*x], x])/2 + (125*Defer[Int][(E^(-x + (2*E^x + 5*x)/(E^x*Log[-3 + 2*x]))*(-3 + 2*x))/Log[-3 + 2*x], x])/2

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {100 e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)}+\frac {125 \exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) \left (2 x+3 \log (-3+2 x)-5 x \log (-3+2 x)+2 x^2 \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)}\right ) \, dx \\ & = 100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) \left (2 x+3 \log (-3+2 x)-5 x \log (-3+2 x)+2 x^2 \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx \\ & = 100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \left (\frac {2 \exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) x}{(-3+2 x) \log ^2(-3+2 x)}+\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-1+x)}{\log (-3+2 x)}\right ) \, dx \\ & = 100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-1+x)}{\log (-3+2 x)} \, dx+250 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) x}{(-3+2 x) \log ^2(-3+2 x)} \, dx \\ & = 100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \left (\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{2 \log (-3+2 x)}+\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-3+2 x)}{2 \log (-3+2 x)}\right ) \, dx+250 \int \left (\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{2 \log ^2(-3+2 x)}+\frac {3 \exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{2 (-3+2 x) \log ^2(-3+2 x)}\right ) \, dx \\ & = \frac {125}{2} \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{\log (-3+2 x)} \, dx+\frac {125}{2} \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-3+2 x)}{\log (-3+2 x)} \, dx+100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{\log ^2(-3+2 x)} \, dx+375 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.77 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=-25 e^{\frac {2+5 e^{-x} x}{\log (-3+2 x)}} \]

[In]

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]

[Out]

-25*E^((2 + (5*x)/E^x)/Log[-3 + 2*x])

Maple [A] (verified)

Time = 1.00 (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	\(-25 \,{\mathrm e}^{\frac {\left (2 \,{\mathrm e}^{x}+5 x \right ) {\mathrm e}^{-x}}{\ln \left (-3+2 x \right )}}\)	\(25\)

[In]

int(((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,method=_RETURNVERBOSE)

[Out]

-25*exp((2*exp(x)+5*x)*exp(-x)/ln(-3+2*x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=-25 \, e^{\left (x - \frac {{\left (x e^{x} \log \left (2 \, x - 3\right ) - 5 \, x - 2 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (2 \, x - 3\right )}\right )} \]

[In]

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)/e
xp(x)/log(-3+2*x)^2,x, algorithm="fricas")

[Out]

-25*e^(x - (x*e^x*log(2*x - 3) - 5*x - 2*e^x)*e^(-x)/log(2*x - 3))

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=- 25 e^{\frac {\left (5 x + 2 e^{x}\right ) e^{- x}}{\log {\left (2 x - 3 \right )}}} \]

[In]

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)/ex
p(x)/ln(-3+2*x)**2,x)

[Out]

-25*exp((5*x + 2*exp(x))*exp(-x)/log(2*x - 3))

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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)/e
xp(x)/log(-3+2*x)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

Giac [F]

\[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=\int { \frac {25 \, {\left (5 \, {\left (2 \, x^{2} - 5 \, x + 3\right )} \log \left (2 \, x - 3\right ) + 10 \, x + 4 \, e^{x}\right )} e^{\left (-x + \frac {{\left (5 \, x + 2 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (2 \, x - 3\right )}\right )}}{{\left (2 \, x - 3\right )} \log \left (2 \, x - 3\right )^{2}} \,d x } \]

[In]

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)/e
xp(x)/log(-3+2*x)^2,x, algorithm="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 12.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=-25\,{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{-x}+2}{\ln \left (2\,x-3\right )}} \]

[In]

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)

[Out]

-25*exp((5*x*exp(-x) + 2)/log(2*x - 3))