3.43.17 \(\int \frac {2}{-4 x-5 x \log (3)+x \log (2 x)} \, dx\)

Optimal. Leaf size=15 \[ \log \left ((4+5 \log (3)-\log (2 x))^2\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {6, 12, 31} \begin {gather*} 2 \log (-\log (2 x)+4+\log (243)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2/(-4*x - 5*x*Log[3] + x*Log[2*x]),x]

[Out]

2*Log[4 + Log[243] - Log[2*x]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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} &=\int \frac {2}{x (-4-5 \log (3))+x \log (2 x)} \, dx\\ &=2 \int \frac {1}{x (-4-5 \log (3))+x \log (2 x)} \, dx\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{4-x+5 \log (3)} \, dx,x,\log (2 x)\right )\right )\\ &=2 \log (4+\log (243)-\log (2 x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 11, normalized size = 0.73 \begin {gather*} 2 \log \left (-4+\log \left (\frac {2 x}{243}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2/(-4*x - 5*x*Log[3] + x*Log[2*x]),x]

[Out]

2*Log[-4 + Log[(2*x)/243]]

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fricas [A]  time = 1.50, size = 13, normalized size = 0.87 \begin {gather*} 2 \, \log \left (-5 \, \log \relax (3) + \log \left (2 \, x\right ) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(x*log(2*x)-5*x*log(3)-4*x),x, algorithm="fricas")

[Out]

2*log(-5*log(3) + log(2*x) - 4)

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giac [A]  time = 0.32, size = 15, normalized size = 1.00 \begin {gather*} 2 \, \log \left (5 \, \log \relax (3) - \log \left (2 \, x\right ) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(x*log(2*x)-5*x*log(3)-4*x),x, algorithm="giac")

[Out]

2*log(5*log(3) - log(2*x) + 4)

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maple [A]  time = 0.04, size = 14, normalized size = 0.93

method	result	size
risch	\(2 \ln \left (-4+\ln \left (2 x \right )-5 \ln \relax (3)\right )\)	\(14\)
derivativedivides	\(2 \ln \left (4-\ln \left (2 x \right )+5 \ln \relax (3)\right )\)	\(16\)
default	\(2 \ln \left (4-\ln \left (2 x \right )+5 \ln \relax (3)\right )\)	\(16\)
norman	\(2 \ln \left (4-\ln \left (2 x \right )+5 \ln \relax (3)\right )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/(x*ln(2*x)-5*x*ln(3)-4*x),x,method=_RETURNVERBOSE)

[Out]

2*ln(-4+ln(2*x)-5*ln(3))

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maxima [A]  time = 0.43, size = 13, normalized size = 0.87 \begin {gather*} 2 \, \log \left (-5 \, \log \relax (3) + \log \relax (2) + \log \relax (x) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(x*log(2*x)-5*x*log(3)-4*x),x, algorithm="maxima")

[Out]

2*log(-5*log(3) + log(2) + log(x) - 4)

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mupad [B]  time = 3.29, size = 9, normalized size = 0.60 \begin {gather*} 2\,\ln \left (\ln \left (\frac {2\,x}{243}\right )-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2/(4*x - x*log(2*x) + 5*x*log(3)),x)

[Out]

2*log(log((2*x)/243) - 4)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.93 \begin {gather*} 2 \log {\left (\log {\left (2 x \right )} - 5 \log {\relax (3 )} - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(x*ln(2*x)-5*x*ln(3)-4*x),x)

[Out]

2*log(log(2*x) - 5*log(3) - 4)

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