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

Optimal. Leaf size=18 \[ -\frac {3 e^{-1+x}}{x \log \left (-1+\frac {1}{x}\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 1.60, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 5.03, size = 18, normalized size = 1.00 \begin {gather*} -\frac {3 e^{-1+x}}{x \log \left (-1+\frac {1}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[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)])

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 20, normalized size = 1.11 \begin {gather*} -\frac {3 \, e^{\left (x - 1\right )}}{x \log \left (-\frac {x - 1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*e^(x - 1)/(x*log(-(x - 1)/x))

________________________________________________________________________________________

giac [A]  time = 0.17, size = 20, normalized size = 1.11 \begin {gather*} -\frac {3 \, e^{\left (x - 1\right )}}{x \log \left (-\frac {x - 1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*e^(x - 1)/(x*log(-(x - 1)/x))

________________________________________________________________________________________

maple [C]  time = 0.13, size = 129, normalized size = 7.17

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

-6*I/(Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I/x*(x-1))-Pi*csgn(I/x)*csgn(I/x*(x-1))^2+2*Pi*csgn(I/x*(x-1))^2-Pi*csgn
(I*(x-1))*csgn(I/x*(x-1))^2-Pi*csgn(I/x*(x-1))^3-2*Pi-2*I*ln(x)+2*I*ln(x-1))/x*exp(x-1)

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 22, normalized size = 1.22 \begin {gather*} \frac {3 \, e^{\left (x - 1\right )}}{x \log \relax (x) - x \log \left (-x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 7.51, size = 20, normalized size = 1.11 \begin {gather*} -\frac {3\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x}{x\,\ln \left (-\frac {x-1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [A]  time = 0.52, size = 17, normalized size = 0.94 \begin {gather*} - \frac {3 e^{x}}{e x \log {\left (\frac {1 - x}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*exp(-1)*exp(x)/(x*log((1 - x)/x))

________________________________________________________________________________________