3.28.72 \(\int \frac {4 x \log (3)-4 \log (3) \log (9)}{(-4 x+5 x \log (3)) \log (9)} \, dx\)

Optimal. Leaf size=26 \[ \frac {4 x \left (-\frac {1}{\log (9)}+\frac {\log (x)}{x}\right )}{-5+\frac {4}{\log (3)}} \]

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.23, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {6, 12, 43} \begin {gather*} \frac {4 \log (3) \log (x)}{4-\log (243)}-\frac {x \log (81)}{\log (9) (4-\log (243))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x*Log[3] - 4*Log[3]*Log[9])/((-4*x + 5*x*Log[3])*Log[9]),x]

[Out]

-((x*Log[81])/(Log[9]*(4 - Log[243]))) + (4*Log[3]*Log[x])/(4 - Log[243])

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x \log (3)-4 \log (3) \log (9)}{x (-4+5 \log (3)) \log (9)} \, dx\\ &=-\frac {\int \frac {4 x \log (3)-4 \log (3) \log (9)}{x} \, dx}{\log (9) (4-\log (243))}\\ &=-\frac {\int \left (-\frac {4 \log (3) \log (9)}{x}+\log (81)\right ) \, dx}{\log (9) (4-\log (243))}\\ &=-\frac {x \log (81)}{\log (9) (4-\log (243))}+\frac {4 \log (3) \log (x)}{4-\log (243)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 22, normalized size = 0.85 \begin {gather*} \frac {4 \log (3) (x-\log (9) \log (x))}{\log (9) (-4+\log (243))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x*Log[3] - 4*Log[3]*Log[9])/((-4*x + 5*x*Log[3])*Log[9]),x]

[Out]

(4*Log[3]*(x - Log[9]*Log[x]))/(Log[9]*(-4 + Log[243]))

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fricas [A]  time = 0.46, size = 20, normalized size = 0.77 \begin {gather*} -\frac {2 \, {\left (2 \, \log \relax (3) \log \relax (x) - x\right )}}{5 \, \log \relax (3) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*log(3)^2+4*x*log(3))/(5*x*log(3)-4*x)/log(3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 37, normalized size = 1.42 \begin {gather*} -\frac {2 \, {\left (\frac {2 \, \log \relax (3)^{2} \log \left ({\left | x \right |}\right )}{5 \, \log \relax (3) - 4} - \frac {x \log \relax (3)}{5 \, \log \relax (3) - 4}\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*log(3)^2+4*x*log(3))/(5*x*log(3)-4*x)/log(3),x, algorithm="giac")

[Out]

-2*(2*log(3)^2*log(abs(x))/(5*log(3) - 4) - x*log(3)/(5*log(3) - 4))/log(3)

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maple [A]  time = 0.05, size = 19, normalized size = 0.73

method	result	size
default	\(\frac {2 x -4 \ln \relax (3) \ln \relax (x )}{5 \ln \relax (3)-4}\)	\(19\)
norman	\(\frac {2 x}{5 \ln \relax (3)-4}-\frac {4 \ln \relax (3) \ln \relax (x )}{5 \ln \relax (3)-4}\)	\(27\)
risch	\(\frac {2 x}{5 \ln \relax (3)-4}-\frac {4 \ln \relax (3) \ln \relax (x )}{5 \ln \relax (3)-4}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-8*ln(3)^2+4*x*ln(3))/(5*x*ln(3)-4*x)/ln(3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 36, normalized size = 1.38 \begin {gather*} -\frac {2 \, {\left (\frac {2 \, \log \relax (3)^{2} \log \relax (x)}{5 \, \log \relax (3) - 4} - \frac {x \log \relax (3)}{5 \, \log \relax (3) - 4}\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*log(3)^2+4*x*log(3))/(5*x*log(3)-4*x)/log(3),x, algorithm="maxima")

[Out]

-2*(2*log(3)^2*log(x)/(5*log(3) - 4) - x*log(3)/(5*log(3) - 4))/log(3)

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mupad [B]  time = 0.06, size = 16, normalized size = 0.62 \begin {gather*} \frac {2\,\left (x-2\,\ln \relax (3)\,\ln \relax (x)\right )}{\ln \left (243\right )-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(x - 2*log(3)*log(x)))/(log(243) - 4)

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sympy [A]  time = 0.10, size = 17, normalized size = 0.65 \begin {gather*} \frac {2 x - 4 \log {\relax (3 )} \log {\relax (x )}}{-4 + 5 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*ln(3)**2+4*x*ln(3))/(5*x*ln(3)-4*x)/ln(3),x)

[Out]

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

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