\(\int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log (9 e^{-x})+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2(9 e^{-x})-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log (9 e^{-x}) (2 x^3-2 x^3 \log (2 x))} \, dx\) [717]

Optimal result

Integrand size = 108, antiderivative size = 30 \[ \int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log \left (9 e^{-x}\right )+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2\left (9 e^{-x}\right )-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log \left (9 e^{-x}\right ) \left (2 x^3-2 x^3 \log (2 x)\right )} \, dx=\frac {e^x \log (3)}{x \left (x+x \log \left (9 e^{-x}\right )-x \log (2 x)\right )} \]

[Out]

exp(x)/(x+x*ln(9/exp(x))-x*ln(2*x))/x*ln(3)

Rubi [F]

\[ \int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log \left (9 e^{-x}\right )+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2\left (9 e^{-x}\right )-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log \left (9 e^{-x}\right ) \left (2 x^3-2 x^3 \log (2 x)\right )} \, dx=\int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log \left (9 e^{-x}\right )+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2\left (9 e^{-x}\right )-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log \left (9 e^{-x}\right ) \left (2 x^3-2 x^3 \log (2 x)\right )} \, dx \]

[In]

Int[(E^x*(-1 + 2*x)*Log[3] + E^x*(-2 + x)*Log[3]*Log[9/E^x] + E^x*(2 - x)*Log[3]*Log[2*x])/(x^3 + x^3*Log[9/E^
x]^2 - 2*x^3*Log[2*x] + x^3*Log[2*x]^2 + Log[9/E^x]*(2*x^3 - 2*x^3*Log[2*x])),x]

[Out]

Log[3]*Defer[Int][E^x/(x^3*(1 + Log[9/E^x] - Log[2*x])^2), x] + Log[3]*Defer[Int][E^x/(x^2*(1 + Log[9/E^x] - L
og[2*x])^2), x] - 2*Log[3]*Defer[Int][E^x/(x^3*(1 + Log[9/E^x] - Log[2*x])), x] + Log[3]*Defer[Int][E^x/(x^2*(
1 + Log[9/E^x] - Log[2*x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \log (3) \left (-1+2 x+(-2+x) \log \left (9 e^{-x}\right )-(-2+x) \log (2 x)\right )}{x^3 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )^2} \, dx \\ & = \log (3) \int \frac {e^x \left (-1+2 x+(-2+x) \log \left (9 e^{-x}\right )-(-2+x) \log (2 x)\right )}{x^3 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )^2} \, dx \\ & = \log (3) \int \left (\frac {e^x (1+x)}{x^3 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )^2}+\frac {e^x (-2+x)}{x^3 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )}\right ) \, dx \\ & = \log (3) \int \frac {e^x (1+x)}{x^3 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )^2} \, dx+\log (3) \int \frac {e^x (-2+x)}{x^3 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )} \, dx \\ & = \log (3) \int \left (\frac {e^x}{x^3 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )^2}+\frac {e^x}{x^2 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )^2}\right ) \, dx+\log (3) \int \left (-\frac {2 e^x}{x^3 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )}+\frac {e^x}{x^2 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )}\right ) \, dx \\ & = \log (3) \int \frac {e^x}{x^3 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )^2} \, dx+\log (3) \int \frac {e^x}{x^2 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )^2} \, dx+\log (3) \int \frac {e^x}{x^2 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )} \, dx-(2 \log (3)) \int \frac {e^x}{x^3 \left (1+\log \left (9 e^{-x}\right )-\log (2 x)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log \left (9 e^{-x}\right )+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2\left (9 e^{-x}\right )-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log \left (9 e^{-x}\right ) \left (2 x^3-2 x^3 \log (2 x)\right )} \, dx=-\frac {e^x \log (3)}{x^2 \left (-1-\log \left (9 e^{-x}\right )+\log (2 x)\right )} \]

[In]

Integrate[(E^x*(-1 + 2*x)*Log[3] + E^x*(-2 + x)*Log[3]*Log[9/E^x] + E^x*(2 - x)*Log[3]*Log[2*x])/(x^3 + x^3*Lo
g[9/E^x]^2 - 2*x^3*Log[2*x] + x^3*Log[2*x]^2 + Log[9/E^x]*(2*x^3 - 2*x^3*Log[2*x])),x]

[Out]

-((E^x*Log[3])/(x^2*(-1 - Log[9/E^x] + Log[2*x])))

Maple [A] (verified)

Time = 1.65 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(-\frac {\ln \left (3\right ) {\mathrm e}^{x}}{x^{2} \left (-1+\ln \left (2 x \right )-\ln \left (9 \,{\mathrm e}^{-x}\right )\right )}\)	\(27\)
risch	\(\frac {2 \ln \left (3\right ) {\mathrm e}^{x}}{x^{2} \left (2+4 \ln \left (3\right )-2 \ln \left (2 x \right )-2 \ln \left ({\mathrm e}^{x}\right )\right )}\)	\(29\)

[In]

int(((-2+x)*ln(3)*exp(x)*ln(9/exp(x))+(2-x)*ln(3)*exp(x)*ln(2*x)+(-1+2*x)*ln(3)*exp(x))/(x^3*ln(9/exp(x))^2+(-
2*x^3*ln(2*x)+2*x^3)*ln(9/exp(x))+x^3*ln(2*x)^2-2*x^3*ln(2*x)+x^3),x,method=_RETURNVERBOSE)

[Out]

-1/x^2*ln(3)*exp(x)/(-1+ln(2*x)-ln(9/exp(x)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log \left (9 e^{-x}\right )+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2\left (9 e^{-x}\right )-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log \left (9 e^{-x}\right ) \left (2 x^3-2 x^3 \log (2 x)\right )} \, dx=-\frac {e^{x} \log \left (3\right )}{x^{3} - 2 \, x^{2} \log \left (3\right ) + x^{2} \log \left (2 \, x\right ) - x^{2}} \]

[In]

integrate(((-2+x)*log(3)*exp(x)*log(9/exp(x))+(2-x)*log(3)*exp(x)*log(2*x)+(-1+2*x)*log(3)*exp(x))/(x^3*log(9/
exp(x))^2+(-2*x^3*log(2*x)+2*x^3)*log(9/exp(x))+x^3*log(2*x)^2-2*x^3*log(2*x)+x^3),x, algorithm="fricas")

[Out]

-e^x*log(3)/(x^3 - 2*x^2*log(3) + x^2*log(2*x) - x^2)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log \left (9 e^{-x}\right )+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2\left (9 e^{-x}\right )-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log \left (9 e^{-x}\right ) \left (2 x^3-2 x^3 \log (2 x)\right )} \, dx=- \frac {e^{x} \log {\left (3 \right )}}{x^{3} + x^{2} \log {\left (2 x \right )} - 2 x^{2} \log {\left (3 \right )} - x^{2}} \]

[In]

integrate(((-2+x)*ln(3)*exp(x)*ln(9/exp(x))+(2-x)*ln(3)*exp(x)*ln(2*x)+(-1+2*x)*ln(3)*exp(x))/(x**3*ln(9/exp(x
))**2+(-2*x**3*ln(2*x)+2*x**3)*ln(9/exp(x))+x**3*ln(2*x)**2-2*x**3*ln(2*x)+x**3),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log \left (9 e^{-x}\right )+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2\left (9 e^{-x}\right )-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log \left (9 e^{-x}\right ) \left (2 x^3-2 x^3 \log (2 x)\right )} \, dx=-\frac {e^{x} \log \left (3\right )}{x^{3} - x^{2} {\left (2 \, \log \left (3\right ) - \log \left (2\right ) + 1\right )} + x^{2} \log \left (x\right )} \]

[In]

integrate(((-2+x)*log(3)*exp(x)*log(9/exp(x))+(2-x)*log(3)*exp(x)*log(2*x)+(-1+2*x)*log(3)*exp(x))/(x^3*log(9/
exp(x))^2+(-2*x^3*log(2*x)+2*x^3)*log(9/exp(x))+x^3*log(2*x)^2-2*x^3*log(2*x)+x^3),x, algorithm="maxima")

[Out]

-e^x*log(3)/(x^3 - x^2*(2*log(3) - log(2) + 1) + x^2*log(x))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log \left (9 e^{-x}\right )+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2\left (9 e^{-x}\right )-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log \left (9 e^{-x}\right ) \left (2 x^3-2 x^3 \log (2 x)\right )} \, dx=-\frac {e^{x} \log \left (3\right )}{x^{3} - 2 \, x^{2} \log \left (3\right ) + x^{2} \log \left (2\right ) + x^{2} \log \left (x\right ) - x^{2}} \]

[In]

integrate(((-2+x)*log(3)*exp(x)*log(9/exp(x))+(2-x)*log(3)*exp(x)*log(2*x)+(-1+2*x)*log(3)*exp(x))/(x^3*log(9/
exp(x))^2+(-2*x^3*log(2*x)+2*x^3)*log(9/exp(x))+x^3*log(2*x)^2-2*x^3*log(2*x)+x^3),x, algorithm="giac")

[Out]

-e^x*log(3)/(x^3 - 2*x^2*log(3) + x^2*log(2) + x^2*log(x) - x^2)

Mupad [B] (verification not implemented)

Time = 8.47 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log \left (9 e^{-x}\right )+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2\left (9 e^{-x}\right )-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log \left (9 e^{-x}\right ) \left (2 x^3-2 x^3 \log (2 x)\right )} \, dx=\frac {{\mathrm {e}}^x\,\left (\ln \left (3\right )-2\,\ln \left (2\,x\right )\,\ln \left (3\right )+\ln \left (2\,x\right )\,\ln \left (9\right )+x\,\ln \left (3\right )\right )}{x^2\,\left (x+1\right )\,\left (\ln \left (9\,{\mathrm {e}}^{-x}\right )-\ln \left (2\,x\right )+1\right )} \]

[In]

int((exp(x)*log(3)*(2*x - 1) + log(9*exp(-x))*exp(x)*log(3)*(x - 2) - log(2*x)*exp(x)*log(3)*(x - 2))/(x^3*log
(9*exp(-x))^2 - 2*x^3*log(2*x) - log(9*exp(-x))*(2*x^3*log(2*x) - 2*x^3) + x^3 + x^3*log(2*x)^2),x)

[Out]

(exp(x)*(log(3) - 2*log(2*x)*log(3) + log(2*x)*log(9) + x*log(3)))/(x^2*(x + 1)*(log(9*exp(-x)) - log(2*x) + 1
))