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

Optimal. Leaf size=29 \[ -x+\frac {\log ^2\left (\frac {\log (2)}{2 (-3+x) \log (6)}\right )}{9 e^2} \]

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Rubi [A]  time = 0.08, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {12, 14, 2301} \begin {gather*} \frac {\log ^2\left (-\frac {\log (2)}{(3-x) \log (36)}\right )}{9 e^2}-x \end {gather*}

Antiderivative was successfully verified.

[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 2301

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.97 \begin {gather*} -x+\frac {\log ^2\left (\frac {\log (2)}{(-6+2 x) \log (6)}\right )}{9 e^2} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.61, size = 28, normalized size = 0.97 \begin {gather*} -\frac {1}{9} \, {\left (9 \, x e^{2} - \log \left (\frac {\log \relax (2)}{2 \, {\left (x - 3\right )} \log \relax (6)}\right )^{2}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.14, size = 145, normalized size = 5.00 \begin {gather*} \frac {{\left (\frac {\log \relax (3) \log \relax (2)^{2} \log \left (\frac {\log \relax (2)}{2 \, {\left (x \log \relax (6) - 3 \, \log \relax (6)\right )}}\right )^{2}}{x \log \relax (6) - 3 \, \log \relax (6)} + \frac {\log \relax (2)^{3} \log \left (\frac {\log \relax (2)}{2 \, {\left (x \log \relax (6) - 3 \, \log \relax (6)\right )}}\right )^{2}}{x \log \relax (6) - 3 \, \log \relax (6)} - 9 \, e^{2} \log \relax (2)^{2}\right )} e^{\left (-2\right )} \log \relax (6)}{9 \, {\left (\frac {\log \relax (3)^{2} \log \relax (2)}{x \log \relax (6) - 3 \, \log \relax (6)} + \frac {2 \, \log \relax (3) \log \relax (2)^{2}}{x \log \relax (6) - 3 \, \log \relax (6)} + \frac {\log \relax (2)^{3}}{x \log \relax (6) - 3 \, \log \relax (6)}\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.15, size = 28, normalized size = 0.97

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

Verification of antiderivative is not currently implemented for this CAS.

[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(-2)*ln(1/2*ln(2)/(x-3)/(ln(2)+ln(3)))^2

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maxima [B]  time = 0.44, size = 105, normalized size = 3.62 \begin {gather*} -\frac {1}{9} \, {\left (9 \, {\left (x + 3 \, \log \left (x - 3\right )\right )} e^{2} - {\left (\frac {2 \, \log \left (2 \, x \log \relax (6) - 6 \, \log \relax (6)\right ) \log \left (x - 3\right )}{\log \relax (6)} - \frac {\log \left (x - 3\right )^{2}}{\log \relax (3) + \log \relax (2)}\right )} \log \relax (6) - 27 \, e^{2} \log \left (x - 3\right ) + 2 \, \log \left (2 \, x \log \relax (6) - 6 \, \log \relax (6)\right ) \log \left (x - 3\right ) + 2 \, \log \left (x - 3\right ) \log \left (\frac {\log \relax (2)}{2 \, {\left (x \log \relax (6) - 3 \, \log \relax (6)\right )}}\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.62, size = 25, normalized size = 0.86 \begin {gather*} \frac {{\mathrm {e}}^{-2}\,{\ln \left (\frac {\ln \relax (2)}{\ln \relax (6)\,\left (2\,x-6\right )}\right )}^2}{9}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.16, size = 20, normalized size = 0.69 \begin {gather*} - x + \frac {\log {\left (\frac {\log {\relax (2 )}}{\left (2 x - 6\right ) \log {\relax (6 )}} \right )}^{2}}{9 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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