\(\int \frac {2}{(-4+x) \log (-8+2 x)} \, dx\) [2890]

Optimal result

Integrand size = 15, antiderivative size = 14 \[ \int \frac {2}{(-4+x) \log (-8+2 x)} \, dx=3+\log \left (\log ^2\left (-8+\log \left (e^{2 x}\right )\right )\right ) \]

[Out]

ln(ln(-8+ln(exp(x)^2))^2)+3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {12, 2437, 2339, 29} \[ \int \frac {2}{(-4+x) \log (-8+2 x)} \, dx=2 \log (\log (-2 (4-x))) \]

[In]

Int[2/((-4 + x)*Log[-8 + 2*x]),x]

[Out]

2*Log[Log[-2*(4 - x)]]

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {1}{(-4+x) \log (-8+2 x)} \, dx \\ & = \text {Subst}\left (\int \frac {2}{x \log (x)} \, dx,x,-8+2 x\right ) \\ & = 2 \text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-8+2 x\right ) \\ & = 2 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (2 (-4+x))\right ) \\ & = 2 \log (\log (-2 (4-x))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {2}{(-4+x) \log (-8+2 x)} \, dx=2 \log (\log (2 (-4+x))) \]

[In]

Integrate[2/((-4 + x)*Log[-8 + 2*x]),x]

[Out]

2*Log[Log[2*(-4 + x)]]

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(2 \ln \left (\ln \left (2 x -8\right )\right )\)	\(10\)
default	\(2 \ln \left (\ln \left (2 x -8\right )\right )\)	\(10\)
norman	\(2 \ln \left (\ln \left (2 x -8\right )\right )\)	\(10\)
risch	\(2 \ln \left (\ln \left (2 x -8\right )\right )\)	\(10\)
parallelrisch	\(2 \ln \left (\ln \left (2 x -8\right )\right )\)	\(10\)

[In]

int(2/(x-4)/ln(2*x-8),x,method=_RETURNVERBOSE)

[Out]

2*ln(ln(2*x-8))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {2}{(-4+x) \log (-8+2 x)} \, dx=2 \, \log \left (\log \left (2 \, x - 8\right )\right ) \]

[In]

integrate(2/(x-4)/log(2*x-8),x, algorithm="fricas")

[Out]

2*log(log(2*x - 8))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {2}{(-4+x) \log (-8+2 x)} \, dx=2 \log {\left (\log {\left (2 x - 8 \right )} \right )} \]

[In]

integrate(2/(x-4)/ln(2*x-8),x)

[Out]

2*log(log(2*x - 8))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {2}{(-4+x) \log (-8+2 x)} \, dx=2 \, \log \left (\log \left (2 \, x - 8\right )\right ) \]

[In]

integrate(2/(x-4)/log(2*x-8),x, algorithm="maxima")

[Out]

2*log(log(2*x - 8))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {2}{(-4+x) \log (-8+2 x)} \, dx=2 \, \log \left (\log \left (2 \, x - 8\right )\right ) \]

[In]

integrate(2/(x-4)/log(2*x-8),x, algorithm="giac")

[Out]

2*log(log(2*x - 8))

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {2}{(-4+x) \log (-8+2 x)} \, dx=2\,\ln \left (\ln \left (2\,x-8\right )\right ) \]

[In]

int(2/(log(2*x - 8)*(x - 4)),x)

[Out]

2*log(log(2*x - 8))