3.68.91 \(\int \frac {-1+(1-2 x) \log (2 x)}{x \log (2 x)} \, dx\)

Optimal. Leaf size=17 \[ \log (x)-\log \left (5 e^{2 x} \log (2 x)\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 13, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6688, 2302, 29} \begin {gather*} -2 x+\log (x)-\log (\log (2 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + (1 - 2*x)*Log[2*x])/(x*Log[2*x]),x]

[Out]

-2*x + Log[x] - Log[Log[2*x]]

Rule 29

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

Rule 2302

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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-2+\frac {1}{x}-\frac {1}{x \log (2 x)}\right ) \, dx\\ &=-2 x+\log (x)-\int \frac {1}{x \log (2 x)} \, dx\\ &=-2 x+\log (x)-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (2 x)\right )\\ &=-2 x+\log (x)-\log (\log (2 x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.76 \begin {gather*} -2 x+\log (x)-\log (\log (2 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*x + Log[x] - Log[Log[2*x]]

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fricas [A]  time = 0.53, size = 15, normalized size = 0.88 \begin {gather*} -2 \, x + \log \left (2 \, x\right ) - \log \left (\log \left (2 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x + log(2*x) - log(log(2*x))

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giac [A]  time = 0.17, size = 13, normalized size = 0.76 \begin {gather*} -2 \, x + \log \relax (x) - \log \left (\log \left (2 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x + log(x) - log(log(2*x))

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maple [A]  time = 0.02, size = 14, normalized size = 0.82

method	result	size
risch	\(\ln \relax (x )-2 x -\ln \left (\ln \left (2 x \right )\right )\)	\(14\)
derivativedivides	\(-2 x +\ln \left (2 x \right )-\ln \left (\ln \left (2 x \right )\right )\)	\(16\)
default	\(-2 x +\ln \left (2 x \right )-\ln \left (\ln \left (2 x \right )\right )\)	\(16\)
norman	\(-2 x +\ln \left (2 x \right )-\ln \left (\ln \left (2 x \right )\right )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 13, normalized size = 0.76 \begin {gather*} -2 \, x + \log \relax (x) - \log \left (\log \left (2 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x + log(x) - log(log(2*x))

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mupad [B]  time = 4.09, size = 13, normalized size = 0.76 \begin {gather*} \ln \relax (x)-\ln \left (\ln \left (2\,x\right )\right )-2\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*x)*(2*x - 1) + 1)/(x*log(2*x)),x)

[Out]

log(x) - log(log(2*x)) - 2*x

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sympy [A]  time = 0.11, size = 12, normalized size = 0.71 \begin {gather*} - 2 x + \log {\relax (x )} - \log {\left (\log {\left (2 x \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x + log(x) - log(log(2*x))

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