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

Optimal. Leaf size=25 \[ \log \left (\log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {12, 6684} \begin {gather*} \log \left (\log \left (\frac {1}{e^3 \log \left (\frac {2-x}{x}\right )-e^3+2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 6684

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 23, normalized size = 0.92 \begin {gather*} \log \left (\log \left (\frac {1}{2-e^3+e^3 \log \left (-1+\frac {2}{x}\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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)]]

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 22, normalized size = 0.88 \begin {gather*} \log \left (\log \left (\frac {1}{e^{3} \log \left (-\frac {x - 2}{x}\right ) - e^{3} + 2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.32, size = 24, normalized size = 0.96

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

Verification of antiderivative is not currently implemented for this CAS.

[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)))

________________________________________________________________________________________

maxima [A]  time = 0.65, size = 23, normalized size = 0.92 \begin {gather*} \log \left (\log \left (-e^{3} \log \relax (x) + e^{3} \log \left (-x + 2\right ) - e^{3} + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

mupad [B]  time = 24.68, size = 22, normalized size = 0.88 \begin {gather*} \ln \left (\ln \left (\frac {1}{{\mathrm {e}}^3\,\ln \left (-\frac {x-2}{x}\right )-{\mathrm {e}}^3+2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

________________________________________________________________________________________

sympy [A]  time = 0.51, size = 19, normalized size = 0.76 \begin {gather*} \log {\left (\log {\left (\frac {1}{e^{3} \log {\left (\frac {2 - x}{x} \right )} - e^{3} + 2} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

________________________________________________________________________________________