3.36.69 \(\int \frac {(4+2 \log (4)) \log (\log (-4+2 x))}{(4-2 x) \log (2) \log (-4+2 x)+(-2+x) \log (-4+2 x) \log ^2(\log (-4+2 x))} \, dx\)

Optimal. Leaf size=28 \[ (2+\log (4)) \left (5+\log \left (2 \log (2)-\log ^2(\log (2-2 (3-x)))\right )\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 3, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {12, 207, 260} \begin {gather*} (2+\log (4)) \log \left (\log (4)-\log ^2(\log (-2 (2-x)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2 + Log[4])*Log[Log[4] - Log[Log[-2*(2 - x)]]^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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=(4+2 \log (4)) \int \frac {\log (\log (-4+2 x))}{(4-2 x) \log (2) \log (-4+2 x)+(-2+x) \log (-4+2 x) \log ^2(\log (-4+2 x))} \, dx\\ &=(4+2 \log (4)) \operatorname {Subst}\left (\int \frac {\log (\log (2 x))}{-2 x \log (2) \log (2 x)+x \log (2 x) \log ^2(\log (2 x))} \, dx,x,-2+x\right )\\ &=(4+2 \log (4)) \operatorname {Subst}\left (\int \frac {\log (x)}{x \left (-2 \log (2)+\log ^2(x)\right )} \, dx,x,\log (2 (-2+x))\right )\\ &=(4+2 \log (4)) \operatorname {Subst}\left (\int \frac {x}{x^2-2 \log (2)} \, dx,x,\log (\log (2 (-2+x)))\right )\\ &=(2+\log (4)) \log \left (\log (4)-\log ^2(\log (-2 (2-x)))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 0.71 \begin {gather*} (2+\log (4)) \log \left (\log (4)-\log ^2(\log (2 (-2+x)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2 + Log[4])*Log[Log[4] - Log[Log[2*(-2 + x)]]^2]

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fricas [A]  time = 0.59, size = 21, normalized size = 0.75 \begin {gather*} 2 \, {\left (\log \relax (2) + 1\right )} \log \left (\log \left (\log \left (2 \, x - 4\right )\right )^{2} - 2 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(2)+4)*log(log(2*x-4))/((x-2)*log(2*x-4)*log(log(2*x-4))^2+(4-2*x)*log(2)*log(2*x-4)),x, algor
ithm="fricas")

[Out]

2*(log(2) + 1)*log(log(log(2*x - 4))^2 - 2*log(2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, {\left (\log \relax (2) + 1\right )} \log \left (\log \left (2 \, x - 4\right )\right )}{{\left (x - 2\right )} \log \left (2 \, x - 4\right ) \log \left (\log \left (2 \, x - 4\right )\right )^{2} - 2 \, {\left (x - 2\right )} \log \relax (2) \log \left (2 \, x - 4\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(2)+4)*log(log(2*x-4))/((x-2)*log(2*x-4)*log(log(2*x-4))^2+(4-2*x)*log(2)*log(2*x-4)),x, algor
ithm="giac")

[Out]

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

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maple [A]  time = 0.17, size = 25, normalized size = 0.89

method	result	size
norman	\(\left (2+2 \ln \relax (2)\right ) \ln \left (2 \ln \relax (2)-\ln \left (\ln \left (2 x -4\right )\right )^{2}\right )\)	\(25\)
risch	\(2 \ln \left (\ln \left (\ln \left (2 x -4\right )\right )^{2}-2 \ln \relax (2)\right ) \ln \relax (2)+2 \ln \left (\ln \left (\ln \left (2 x -4\right )\right )^{2}-2 \ln \relax (2)\right )\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln(2)+4)*ln(ln(2*x-4))/((x-2)*ln(2*x-4)*ln(ln(2*x-4))^2+(4-2*x)*ln(2)*ln(2*x-4)),x,method=_RETURNVERBOS
E)

[Out]

(2+2*ln(2))*ln(2*ln(2)-ln(ln(2*x-4))^2)

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maxima [A]  time = 0.68, size = 22, normalized size = 0.79 \begin {gather*} 2 \, {\left (\log \relax (2) + 1\right )} \log \left (\log \left (\log \relax (2) + \log \left (x - 2\right )\right )^{2} - 2 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(2)+4)*log(log(2*x-4))/((x-2)*log(2*x-4)*log(log(2*x-4))^2+(4-2*x)*log(2)*log(2*x-4)),x, algor
ithm="maxima")

[Out]

2*(log(2) + 1)*log(log(log(2) + log(x - 2))^2 - 2*log(2))

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mupad [B]  time = 3.60, size = 20, normalized size = 0.71 \begin {gather*} \ln \left ({\ln \left (\ln \left (2\,x-4\right )\right )}^2-\ln \relax (4)\right )\,\left (\ln \relax (4)+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(log(2*x - 4))^2 - log(4))*(log(4) + 2)

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sympy [A]  time = 0.39, size = 22, normalized size = 0.79 \begin {gather*} 2 \left (\log {\relax (2 )} + 1\right ) \log {\left (\log {\left (\log {\left (2 x - 4 \right )} \right )}^{2} - 2 \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(log(2) + 1)*log(log(log(2*x - 4))**2 - 2*log(2))

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