\(\int \frac {3 e^x+e^x (-3+6 x-3 x^2) \log (\frac {1-x}{x})}{e (-x^2+x^3) \log ^2(\frac {1-x}{x})} \, dx\) [10290]

Optimal result

Integrand size = 57, antiderivative size = 18 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3 e^{-1+x}}{x \log \left (-1+\frac {1}{x}\right )} \]

[Out]

-3*exp(x)/x/ln(1/x-1)/exp(1)

Rubi [F]

\[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=\int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx \]

[In]

Int[(3*E^x + E^x*(-3 + 6*x - 3*x^2)*Log[(1 - x)/x])/(E*(-x^2 + x^3)*Log[(1 - x)/x]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{\left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx}{e} \\ & = \frac {\int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{(-1+x) x^2 \log ^2\left (\frac {1-x}{x}\right )} \, dx}{e} \\ & = \frac {\int \frac {3 e^x \left (-1+(-1+x)^2 \log \left (-1+\frac {1}{x}\right )\right )}{(1-x) x^2 \log ^2\left (-1+\frac {1}{x}\right )} \, dx}{e} \\ & = \frac {3 \int \frac {e^x \left (-1+(-1+x)^2 \log \left (-1+\frac {1}{x}\right )\right )}{(1-x) x^2 \log ^2\left (-1+\frac {1}{x}\right )} \, dx}{e} \\ & = \frac {3 \int \left (\frac {e^x}{(-1+x) x^2 \log ^2\left (-1+\frac {1}{x}\right )}+\frac {e^x (1-x)}{x^2 \log \left (-1+\frac {1}{x}\right )}\right ) \, dx}{e} \\ & = \frac {3 \int \frac {e^x}{(-1+x) x^2 \log ^2\left (-1+\frac {1}{x}\right )} \, dx}{e}+\frac {3 \int \frac {e^x (1-x)}{x^2 \log \left (-1+\frac {1}{x}\right )} \, dx}{e} \\ & = \frac {3 \int \left (\frac {e^x}{(-1+x) \log ^2\left (-1+\frac {1}{x}\right )}-\frac {e^x}{x^2 \log ^2\left (-1+\frac {1}{x}\right )}-\frac {e^x}{x \log ^2\left (-1+\frac {1}{x}\right )}\right ) \, dx}{e}+\frac {3 \int \left (\frac {e^x}{x^2 \log \left (-1+\frac {1}{x}\right )}-\frac {e^x}{x \log \left (-1+\frac {1}{x}\right )}\right ) \, dx}{e} \\ & = \frac {3 \int \frac {e^x}{(-1+x) \log ^2\left (-1+\frac {1}{x}\right )} \, dx}{e}-\frac {3 \int \frac {e^x}{x^2 \log ^2\left (-1+\frac {1}{x}\right )} \, dx}{e}-\frac {3 \int \frac {e^x}{x \log ^2\left (-1+\frac {1}{x}\right )} \, dx}{e}+\frac {3 \int \frac {e^x}{x^2 \log \left (-1+\frac {1}{x}\right )} \, dx}{e}-\frac {3 \int \frac {e^x}{x \log \left (-1+\frac {1}{x}\right )} \, dx}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3 e^{-1+x}}{x \log \left (-1+\frac {1}{x}\right )} \]

[In]

Integrate[(3*E^x + E^x*(-3 + 6*x - 3*x^2)*Log[(1 - x)/x])/(E*(-x^2 + x^3)*Log[(1 - x)/x]^2),x]

[Out]

(-3*E^(-1 + x))/(x*Log[-1 + x^(-1)])

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28

method	result	size
parallelrisch	\(-\frac {3 \,{\mathrm e}^{-1} {\mathrm e}^{x}}{x \ln \left (-\frac {-1+x}{x}\right )}\)	\(23\)
risch	\(-\frac {6 i {\mathrm e}^{-1+x}}{\left (\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}+2 \pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{3}-2 \pi -2 i \ln \left (x \right )+2 i \ln \left (-1+x \right )\right ) x}\)	\(129\)

[In]

int(((-3*x^2+6*x-3)*exp(x)*ln((1-x)/x)+3*exp(x))/(x^3-x^2)/exp(1)/ln((1-x)/x)^2,x,method=_RETURNVERBOSE)

[Out]

-3/exp(1)*exp(x)/x/ln(-(-1+x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3 \, e^{\left (x - 1\right )}}{x \log \left (-\frac {x - 1}{x}\right )} \]

[In]

integrate(((-3*x^2+6*x-3)*exp(x)*log((1-x)/x)+3*exp(x))/(x^3-x^2)/exp(1)/log((1-x)/x)^2,x, algorithm="fricas")

[Out]

-3*e^(x - 1)/(x*log(-(x - 1)/x))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=- \frac {3 e^{x}}{e x \log {\left (\frac {1 - x}{x} \right )}} \]

[In]

integrate(((-3*x**2+6*x-3)*exp(x)*ln((1-x)/x)+3*exp(x))/(x**3-x**2)/exp(1)/ln((1-x)/x)**2,x)

[Out]

-3*exp(-1)*exp(x)/(x*log((1 - x)/x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=\frac {3 \, e^{\left (x - 1\right )}}{x \log \left (x\right ) - x \log \left (-x + 1\right )} \]

[In]

integrate(((-3*x^2+6*x-3)*exp(x)*log((1-x)/x)+3*exp(x))/(x^3-x^2)/exp(1)/log((1-x)/x)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3 \, e^{\left (x - 1\right )}}{x \log \left (-\frac {x - 1}{x}\right )} \]

[In]

integrate(((-3*x^2+6*x-3)*exp(x)*log((1-x)/x)+3*exp(x))/(x^3-x^2)/exp(1)/log((1-x)/x)^2,x, algorithm="giac")

[Out]

-3*e^(x - 1)/(x*log(-(x - 1)/x))

Mupad [B] (verification not implemented)

Time = 17.86 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x}{x\,\ln \left (-\frac {x-1}{x}\right )} \]

[In]

int(-(exp(-1)*(3*exp(x) - exp(x)*log(-(x - 1)/x)*(3*x^2 - 6*x + 3)))/(log(-(x - 1)/x)^2*(x^2 - x^3)),x)

[Out]

-(3*exp(-1)*exp(x))/(x*log(-(x - 1)/x))