\(\int \frac {e^2 (27-9 x)-2 \log (\frac {\log (2)}{(-6+2 x) \log (6)})}{e^2 (-27+9 x)} \, dx\) [10114]

Optimal result

Integrand size = 38, antiderivative size = 29 \[ \int \frac {e^2 (27-9 x)-2 \log \left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{e^2 (-27+9 x)} \, dx=-x+\frac {\log ^2\left (\frac {\log (2)}{2 (-3+x) \log (6)}\right )}{9 e^2} \]

[Out]

1/9/exp(1)^2*ln(1/2*ln(2)/(-3+x)/ln(6))^2-x

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {12, 14, 2338} \[ \int \frac {e^2 (27-9 x)-2 \log \left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{e^2 (-27+9 x)} \, dx=\frac {\log ^2\left (-\frac {\log (2)}{(3-x) \log (36)}\right )}{9 e^2}-x \]

[In]

Int[(E^2*(27 - 9*x) - 2*Log[Log[2]/((-6 + 2*x)*Log[6])])/(E^2*(-27 + 9*x)),x]

[Out]

-x + Log[-(Log[2]/((3 - x)*Log[36]))]^2/(9*E^2)

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^2 (27-9 x)-2 \log \left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{-27+9 x} \, dx}{e^2} \\ & = \frac {\text {Subst}\left (\int \frac {-9 e^2 x-2 \log \left (\frac {\log (2)}{2 x \log (6)}\right )}{9 x} \, dx,x,-3+x\right )}{e^2} \\ & = \frac {\text {Subst}\left (\int \frac {-9 e^2 x-2 \log \left (\frac {\log (2)}{2 x \log (6)}\right )}{x} \, dx,x,-3+x\right )}{9 e^2} \\ & = \frac {\text {Subst}\left (\int \left (-9 e^2-\frac {2 \log \left (\frac {\log (2)}{x \log (36)}\right )}{x}\right ) \, dx,x,-3+x\right )}{9 e^2} \\ & = -x-\frac {2 \text {Subst}\left (\int \frac {\log \left (\frac {\log (2)}{x \log (36)}\right )}{x} \, dx,x,-3+x\right )}{9 e^2} \\ & = -x+\frac {\log ^2\left (-\frac {\log (2)}{(3-x) \log (36)}\right )}{9 e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^2 (27-9 x)-2 \log \left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{e^2 (-27+9 x)} \, dx=-x+\frac {\log ^2\left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{9 e^2} \]

[In]

Integrate[(E^2*(27 - 9*x) - 2*Log[Log[2]/((-6 + 2*x)*Log[6])])/(E^2*(-27 + 9*x)),x]

[Out]

-x + Log[Log[2]/((-6 + 2*x)*Log[6])]^2/(9*E^2)

Maple [A] (verified)

Time = 1.56 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97

method	result	size
parts	\(-x +\frac {{\mathrm e}^{-2} \ln \left (\frac {\ln \left (2\right )}{\left (2 x -6\right ) \ln \left (6\right )}\right )^{2}}{9}\)	\(28\)
risch	\(-x +\frac {{\mathrm e}^{-2} \ln \left (\frac {\ln \left (2\right )}{\left (2 x -6\right ) \left (\ln \left (2\right )+\ln \left (3\right )\right )}\right )^{2}}{9}\)	\(29\)
default	\(\frac {{\mathrm e}^{-2} \left (-9 \left (-3+x \right ) {\mathrm e}^{2}+\ln \left (\frac {\ln \left (2\right )}{2 \left (-3+x \right ) \ln \left (6\right )}\right )^{2}\right )}{9}\)	\(33\)
derivativedivides	\(-\frac {{\mathrm e}^{-2} \left (9 \left (-3+x \right ) {\mathrm e}^{2}-\ln \left (\frac {\ln \left (2\right )}{2 \left (-3+x \right ) \ln \left (6\right )}\right )^{2}\right )}{9}\)	\(35\)
norman	\(\left (-x \,{\mathrm e}+\frac {{\mathrm e}^{-1} \ln \left (\frac {\ln \left (2\right )}{\left (2 x -6\right ) \ln \left (6\right )}\right )^{2}}{9}\right ) {\mathrm e}^{-1}\)	\(35\)

[In]

int((-2*ln(ln(2)/(2*x-6)/ln(6))+(-9*x+27)*exp(1)^2)/(9*x-27)/exp(1)^2,x,method=_RETURNVERBOSE)

[Out]

-x+1/9/exp(1)^2*ln(ln(2)/(2*x-6)/ln(6))^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^2 (27-9 x)-2 \log \left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{e^2 (-27+9 x)} \, dx=-\frac {1}{9} \, {\left (9 \, x e^{2} - \log \left (\frac {\log \left (2\right )}{2 \, {\left (x - 3\right )} \log \left (6\right )}\right )^{2}\right )} e^{\left (-2\right )} \]

[In]

integrate((-2*log(log(2)/(2*x-6)/log(6))+(-9*x+27)*exp(1)^2)/(9*x-27)/exp(1)^2,x, algorithm="fricas")

[Out]

-1/9*(9*x*e^2 - log(1/2*log(2)/((x - 3)*log(6)))^2)*e^(-2)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {e^2 (27-9 x)-2 \log \left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{e^2 (-27+9 x)} \, dx=- x + \frac {\log {\left (\frac {\log {\left (2 \right )}}{\left (2 x - 6\right ) \log {\left (6 \right )}} \right )}^{2}}{9 e^{2}} \]

[In]

integrate((-2*ln(ln(2)/(2*x-6)/ln(6))+(-9*x+27)*exp(1)**2)/(9*x-27)/exp(1)**2,x)

[Out]

-x + exp(-2)*log(log(2)/((2*x - 6)*log(6)))**2/9

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (24) = 48\).

Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 3.62 \[ \int \frac {e^2 (27-9 x)-2 \log \left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{e^2 (-27+9 x)} \, dx=-\frac {1}{9} \, {\left (9 \, {\left (x + 3 \, \log \left (x - 3\right )\right )} e^{2} - {\left (\frac {2 \, \log \left (2 \, x \log \left (6\right ) - 6 \, \log \left (6\right )\right ) \log \left (x - 3\right )}{\log \left (6\right )} - \frac {\log \left (x - 3\right )^{2}}{\log \left (3\right ) + \log \left (2\right )}\right )} \log \left (6\right ) - 27 \, e^{2} \log \left (x - 3\right ) + 2 \, \log \left (2 \, x \log \left (6\right ) - 6 \, \log \left (6\right )\right ) \log \left (x - 3\right ) + 2 \, \log \left (x - 3\right ) \log \left (\frac {\log \left (2\right )}{2 \, {\left (x \log \left (6\right ) - 3 \, \log \left (6\right )\right )}}\right )\right )} e^{\left (-2\right )} \]

[In]

integrate((-2*log(log(2)/(2*x-6)/log(6))+(-9*x+27)*exp(1)^2)/(9*x-27)/exp(1)^2,x, algorithm="maxima")

[Out]

-1/9*(9*(x + 3*log(x - 3))*e^2 - (2*log(2*x*log(6) - 6*log(6))*log(x - 3)/log(6) - log(x - 3)^2/(log(3) + log(
2)))*log(6) - 27*e^2*log(x - 3) + 2*log(2*x*log(6) - 6*log(6))*log(x - 3) + 2*log(x - 3)*log(1/2*log(2)/(x*log
(6) - 3*log(6))))*e^(-2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (24) = 48\).

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 5.00 \[ \int \frac {e^2 (27-9 x)-2 \log \left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{e^2 (-27+9 x)} \, dx=\frac {{\left (\frac {\log \left (3\right ) \log \left (2\right )^{2} \log \left (\frac {\log \left (2\right )}{2 \, {\left (x \log \left (6\right ) - 3 \, \log \left (6\right )\right )}}\right )^{2}}{x \log \left (6\right ) - 3 \, \log \left (6\right )} + \frac {\log \left (2\right )^{3} \log \left (\frac {\log \left (2\right )}{2 \, {\left (x \log \left (6\right ) - 3 \, \log \left (6\right )\right )}}\right )^{2}}{x \log \left (6\right ) - 3 \, \log \left (6\right )} - 9 \, e^{2} \log \left (2\right )^{2}\right )} e^{\left (-2\right )} \log \left (6\right )}{9 \, {\left (\frac {\log \left (3\right )^{2} \log \left (2\right )}{x \log \left (6\right ) - 3 \, \log \left (6\right )} + \frac {2 \, \log \left (3\right ) \log \left (2\right )^{2}}{x \log \left (6\right ) - 3 \, \log \left (6\right )} + \frac {\log \left (2\right )^{3}}{x \log \left (6\right ) - 3 \, \log \left (6\right )}\right )} \log \left (2\right )} \]

[In]

integrate((-2*log(log(2)/(2*x-6)/log(6))+(-9*x+27)*exp(1)^2)/(9*x-27)/exp(1)^2,x, algorithm="giac")

[Out]

1/9*(log(3)*log(2)^2*log(1/2*log(2)/(x*log(6) - 3*log(6)))^2/(x*log(6) - 3*log(6)) + log(2)^3*log(1/2*log(2)/(
x*log(6) - 3*log(6)))^2/(x*log(6) - 3*log(6)) - 9*e^2*log(2)^2)*e^(-2)*log(6)/((log(3)^2*log(2)/(x*log(6) - 3*
log(6)) + 2*log(3)*log(2)^2/(x*log(6) - 3*log(6)) + log(2)^3/(x*log(6) - 3*log(6)))*log(2))

Mupad [B] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {e^2 (27-9 x)-2 \log \left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{e^2 (-27+9 x)} \, dx=\frac {{\mathrm {e}}^{-2}\,{\ln \left (\frac {\ln \left (2\right )}{\ln \left (6\right )\,\left (2\,x-6\right )}\right )}^2}{9}-x \]

[In]

int(-(exp(-2)*(2*log(log(2)/(log(6)*(2*x - 6))) + exp(2)*(9*x - 27)))/(9*x - 27),x)

[Out]

(exp(-2)*log(log(2)/(log(6)*(2*x - 6)))^2)/9 - x