3.9.58 \(\int \frac {-1296+1296 \log (x)}{11664 x^2+216 x \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=16 \[ \frac {3 \log (x)}{-27 x-\frac {\log (x)}{4}} \]

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Rubi [A]  time = 0.07, antiderivative size = 13, normalized size of antiderivative = 0.81, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6688, 12, 6711, 32} \begin {gather*} -\frac {12}{\frac {108 x}{\log (x)}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1296 + 1296*Log[x])/(11664*x^2 + 216*x*Log[x] + Log[x]^2),x]

[Out]

-12/(1 + (108*x)/Log[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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6688

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

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1296 (-1+\log (x))}{(108 x+\log (x))^2} \, dx\\ &=1296 \int \frac {-1+\log (x)}{(108 x+\log (x))^2} \, dx\\ &=1296 \operatorname {Subst}\left (\int \frac {1}{(1+108 x)^2} \, dx,x,\frac {x}{\log (x)}\right )\\ &=-\frac {12}{1+\frac {108 x}{\log (x)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 11, normalized size = 0.69 \begin {gather*} \frac {1296 x}{108 x+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1296 + 1296*Log[x])/(11664*x^2 + 216*x*Log[x] + Log[x]^2),x]

[Out]

(1296*x)/(108*x + Log[x])

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fricas [A]  time = 0.62, size = 11, normalized size = 0.69 \begin {gather*} \frac {1296 \, x}{108 \, x + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1296*log(x)-1296)/(log(x)^2+216*x*log(x)+11664*x^2),x, algorithm="fricas")

[Out]

1296*x/(108*x + log(x))

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giac [A]  time = 0.30, size = 11, normalized size = 0.69 \begin {gather*} \frac {1296 \, x}{108 \, x + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1296*log(x)-1296)/(log(x)^2+216*x*log(x)+11664*x^2),x, algorithm="giac")

[Out]

1296*x/(108*x + log(x))

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maple [A]  time = 0.02, size = 12, normalized size = 0.75

method	result	size
norman	\(\frac {1296 x}{\ln \relax (x )+108 x}\)	\(12\)
risch	\(\frac {1296 x}{\ln \relax (x )+108 x}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1296*ln(x)-1296)/(ln(x)^2+216*x*ln(x)+11664*x^2),x,method=_RETURNVERBOSE)

[Out]

1296*x/(ln(x)+108*x)

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maxima [A]  time = 0.42, size = 11, normalized size = 0.69 \begin {gather*} \frac {1296 \, x}{108 \, x + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1296*log(x)-1296)/(log(x)^2+216*x*log(x)+11664*x^2),x, algorithm="maxima")

[Out]

1296*x/(108*x + log(x))

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mupad [B]  time = 0.67, size = 11, normalized size = 0.69 \begin {gather*} \frac {1296\,x}{108\,x+\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1296*log(x) - 1296)/(log(x)^2 + 216*x*log(x) + 11664*x^2),x)

[Out]

(1296*x)/(108*x + log(x))

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sympy [A]  time = 0.08, size = 8, normalized size = 0.50 \begin {gather*} \frac {1296 x}{108 x + \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1296*ln(x)-1296)/(ln(x)**2+216*x*ln(x)+11664*x**2),x)

[Out]

1296*x/(108*x + log(x))

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