\(\int -\frac {2 e^3}{(-4 x+2 x^2+e^3 (2 x-x^2)+e^3 (-2 x+x^2) \log (\frac {2-x}{x})) \log (\frac {1}{2-e^3+e^3 \log (\frac {2-x}{x})})} \, dx\) [2244]

Optimal result

Integrand size = 76, antiderivative size = 25 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log \left (\log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {12, 6816} \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log \left (\log \left (\frac {1}{e^3 \log \left (\frac {2-x}{x}\right )-e^3+2}\right )\right ) \]

[In]

Int[(-2*E^3)/((-4*x + 2*x^2 + E^3*(2*x - x^2) + E^3*(-2*x + x^2)*Log[(2 - x)/x])*Log[(2 - E^3 + E^3*Log[(2 - x
)/x])^(-1)]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps \begin{align*} \text {integral}& = -\left (\left (2 e^3\right ) \int \frac {1}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx\right ) \\ & = \log \left (\log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log \left (\log \left (\frac {1}{2-e^3+e^3 \log \left (-1+\frac {2}{x}\right )}\right )\right ) \]

[In]

Integrate[(-2*E^3)/((-4*x + 2*x^2 + E^3*(2*x - x^2) + E^3*(-2*x + x^2)*Log[(2 - x)/x])*Log[(2 - E^3 + E^3*Log[
(2 - x)/x])^(-1)]),x]

[Out]

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

Maple [A] (verified)

Time = 7.46 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

method	result	size
parallelrisch	\(\ln \left (\ln \left (\frac {1}{{\mathrm e}^{3} \ln \left (-\frac {-2+x}{x}\right )-{\mathrm e}^{3}+2}\right )\right )\)	\(23\)
norman	\(\ln \left (\ln \left (\frac {1}{{\mathrm e}^{3} \ln \left (\frac {2-x}{x}\right )-{\mathrm e}^{3}+2}\right )\right )\)	\(24\)

[In]

int(-2*exp(3)/((x^2-2*x)*exp(3)*ln((2-x)/x)+(-x^2+2*x)*exp(3)+2*x^2-4*x)/ln(1/(exp(3)*ln((2-x)/x)-exp(3)+2)),x
,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log \left (\log \left (\frac {1}{e^{3} \log \left (-\frac {x - 2}{x}\right ) - e^{3} + 2}\right )\right ) \]

[In]

integrate(-2*exp(3)/((x^2-2*x)*exp(3)*log((2-x)/x)+(-x^2+2*x)*exp(3)+2*x^2-4*x)/log(1/(exp(3)*log((2-x)/x)-exp
(3)+2)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log {\left (\log {\left (\frac {1}{e^{3} \log {\left (\frac {2 - x}{x} \right )} - e^{3} + 2} \right )} \right )} \]

[In]

integrate(-2*exp(3)/((x**2-2*x)*exp(3)*ln((2-x)/x)+(-x**2+2*x)*exp(3)+2*x**2-4*x)/ln(1/(exp(3)*ln((2-x)/x)-exp
(3)+2)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log \left (\log \left (-e^{3} \log \left (x\right ) + e^{3} \log \left (-x + 2\right ) - e^{3} + 2\right )\right ) \]

[In]

integrate(-2*exp(3)/((x^2-2*x)*exp(3)*log((2-x)/x)+(-x^2+2*x)*exp(3)+2*x^2-4*x)/log(1/(exp(3)*log((2-x)/x)-exp
(3)+2)),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\int { -\frac {2 \, e^{3}}{{\left ({\left (x^{2} - 2 \, x\right )} e^{3} \log \left (-\frac {x - 2}{x}\right ) + 2 \, x^{2} - {\left (x^{2} - 2 \, x\right )} e^{3} - 4 \, x\right )} \log \left (\frac {1}{e^{3} \log \left (-\frac {x - 2}{x}\right ) - e^{3} + 2}\right )} \,d x } \]

[In]

integrate(-2*exp(3)/((x^2-2*x)*exp(3)*log((2-x)/x)+(-x^2+2*x)*exp(3)+2*x^2-4*x)/log(1/(exp(3)*log((2-x)/x)-exp
(3)+2)),x, algorithm="giac")

[Out]

integrate(-2*e^3/(((x^2 - 2*x)*e^3*log(-(x - 2)/x) + 2*x^2 - (x^2 - 2*x)*e^3 - 4*x)*log(1/(e^3*log(-(x - 2)/x)
- e^3 + 2))), x)

Mupad [B] (verification not implemented)

Time = 32.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\ln \left (\ln \left (\frac {1}{{\mathrm {e}}^3\,\ln \left (-\frac {x-2}{x}\right )-{\mathrm {e}}^3+2}\right )\right ) \]

[In]

int((2*exp(3))/(log(1/(exp(3)*log(-(x - 2)/x) - exp(3) + 2))*(4*x - exp(3)*(2*x - x^2) - 2*x^2 + exp(3)*log(-(
x - 2)/x)*(2*x - x^2))),x)

[Out]

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