3.22.90 \(\int \frac {-1+2 \log (4)}{2 x^2 \log (4)} \, dx\) [2190]

Optimal. Leaf size=21 \[ \frac {-2+x-\frac {-1+2 x}{\log (4)}}{2 x} \]

[Out]

1/2*(x-2-1/2*(-1+2*x)/ln(2))/x

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Rubi [A]
time = 0.00, antiderivative size = 17, normalized size of antiderivative = 0.81, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 30} \begin {gather*} \frac {1-\log (16)}{2 x \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 2*Log[4])/(2*x^2*Log[4]),x]

[Out]

(1 - Log[16])/(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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {(-1+2 \log (4)) \int \frac {1}{x^2} \, dx}{2 \log (4)}\\ &=\frac {1-\log (16)}{2 x \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 13, normalized size = 0.62 \begin {gather*} -\frac {-1+\log (16)}{x \log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 2*Log[4])/(2*x^2*Log[4]),x]

[Out]

-((-1 + Log[16])/(x*Log[16]))

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Maple [A]
time = 0.07, size = 16, normalized size = 0.76

method	result	size
gosper	\(-\frac {4 \ln \left (2\right )-1}{4 x \ln \left (2\right )}\)	\(16\)
default	\(-\frac {4 \ln \left (2\right )-1}{4 x \ln \left (2\right )}\)	\(16\)
norman	\(-\frac {4 \ln \left (2\right )-1}{4 x \ln \left (2\right )}\)	\(16\)
risch	\(-\frac {1}{x}+\frac {1}{4 x \ln \left (2\right )}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(4*ln(2)-1)/x^2/ln(2),x,method=_RETURNVERBOSE)

[Out]

-1/4/x*(4*ln(2)-1)/ln(2)

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Maxima [A]
time = 0.27, size = 15, normalized size = 0.71 \begin {gather*} -\frac {4 \, \log \left (2\right ) - 1}{4 \, x \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*log(2)-1)/x^2/log(2),x, algorithm="maxima")

[Out]

-1/4*(4*log(2) - 1)/(x*log(2))

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Fricas [A]
time = 0.40, size = 15, normalized size = 0.71 \begin {gather*} -\frac {4 \, \log \left (2\right ) - 1}{4 \, x \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*log(2)-1)/x^2/log(2),x, algorithm="fricas")

[Out]

-1/4*(4*log(2) - 1)/(x*log(2))

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Sympy [A]
time = 0.01, size = 14, normalized size = 0.67 \begin {gather*} - \frac {-1 + 4 \log {\left (2 \right )}}{4 x \log {\left (2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*ln(2)-1)/x**2/ln(2),x)

[Out]

-(-1 + 4*log(2))/(4*x*log(2))

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Giac [A]
time = 0.42, size = 15, normalized size = 0.71 \begin {gather*} -\frac {4 \, \log \left (2\right ) - 1}{4 \, x \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*log(2)-1)/x^2/log(2),x, algorithm="giac")

[Out]

-1/4*(4*log(2) - 1)/(x*log(2))

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Mupad [B]
time = 0.02, size = 13, normalized size = 0.62 \begin {gather*} -\frac {\ln \left (16\right )-1}{4\,x\,\ln \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2) - 1/4)/(x^2*log(2)),x)

[Out]

-(log(16) - 1)/(4*x*log(2))

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