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

Optimal. Leaf size=8 \[ \log (-8+2 x+\log (4)) \]

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Rubi [A]  time = 0.00, antiderivative size = 10, normalized size of antiderivative = 1.25, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 31} \begin {gather*} \log (-2 x+8-\log (4)) \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]  time = 0.00, size = 8, normalized size = 1.00 \begin {gather*} \log (-8+2 x+\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2/(-8 + 2*x + Log[4]),x]

[Out]

Log[-8 + 2*x + Log[4]]

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fricas [A]  time = 0.57, size = 6, normalized size = 0.75 \begin {gather*} \log \left (x + \log \relax (2) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + log(2) - 4)

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giac [A]  time = 0.21, size = 7, normalized size = 0.88 \begin {gather*} \log \left ({\left | x + \log \relax (2) - 4 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(x + log(2) - 4))

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maple [A]  time = 0.13, size = 7, normalized size = 0.88

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(2)+x-4)

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maxima [A]  time = 0.35, size = 6, normalized size = 0.75 \begin {gather*} \log \left (x + \log \relax (2) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + log(2) - 4)

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mupad [B]  time = 0.07, size = 6, normalized size = 0.75 \begin {gather*} \ln \left (x+\ln \relax (2)-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/(2*x + 2*log(2) - 8),x)

[Out]

log(x + log(2) - 4)

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sympy [A]  time = 0.08, size = 7, normalized size = 0.88 \begin {gather*} \log {\left (x - 4 + \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 4 + log(2))

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