\(\int \frac {2}{-8+2 x+\log (4)} \, dx\) [9887]

Optimal result

Integrand size = 11, antiderivative size = 8 \[ \int \frac {2}{-8+2 x+\log (4)} \, dx=\log (-8+2 x+\log (4)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.25, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 31} \[ \int \frac {2}{-8+2 x+\log (4)} \, dx=\log (-2 x+8-\log (4)) \]

[In]

Int[2/(-8 + 2*x + Log[4]),x]

[Out]

Log[8 - 2*x - Log[4]]

Rule 12

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

Rule 31

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {2}{-8+2 x+\log (4)} \, dx=\log (-8+2 x+\log (4)) \]

[In]

Integrate[2/(-8 + 2*x + Log[4]),x]

[Out]

Log[-8 + 2*x + Log[4]]

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
default	\(\ln \left (\ln \left (2\right )+x -4\right )\)	\(7\)
norman	\(\ln \left (\ln \left (2\right )+x -4\right )\)	\(7\)
risch	\(\ln \left (\ln \left (2\right )+x -4\right )\)	\(7\)
parallelrisch	\(\ln \left (\ln \left (2\right )+x -4\right )\)	\(7\)
meijerg	\(\frac {2 \left (\ln \left (2\right )-4\right ) \ln \left (1+\frac {2 x}{2 \ln \left (2\right )-8}\right )}{2 \ln \left (2\right )-8}\)	\(29\)

[In]

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

[Out]

ln(ln(2)+x-4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {2}{-8+2 x+\log (4)} \, dx=\log \left (x + \log \left (2\right ) - 4\right ) \]

[In]

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

[Out]

log(x + log(2) - 4)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88 \[ \int \frac {2}{-8+2 x+\log (4)} \, dx=\log {\left (x - 4 + \log {\left (2 \right )} \right )} \]

[In]

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

[Out]

log(x - 4 + log(2))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {2}{-8+2 x+\log (4)} \, dx=\log \left (x + \log \left (2\right ) - 4\right ) \]

[In]

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

[Out]

log(x + log(2) - 4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88 \[ \int \frac {2}{-8+2 x+\log (4)} \, dx=\log \left ({\left | x + \log \left (2\right ) - 4 \right |}\right ) \]

[In]

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

[Out]

log(abs(x + log(2) - 4))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {2}{-8+2 x+\log (4)} \, dx=\ln \left (x+\ln \left (2\right )-4\right ) \]

[In]

int(2/(2*x + 2*log(2) - 8),x)

[Out]

log(x + log(2) - 4)