3.29.11 \(\int \frac {1}{3-x+(-3+x) \log (-3+x)} \, dx\)

Optimal. Leaf size=7 \[ \log (-1+\log (-3+x)) \]

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Rubi [A]  time = 0.06, antiderivative size = 9, normalized size of antiderivative = 1.29, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6741, 2390, 2302, 29} \begin {gather*} \log (1-\log (x-3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - x + (-3 + x)*Log[-3 + x])^(-1),x]

[Out]

Log[1 - Log[-3 + 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 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{(3-x) (1-\log (-3+x))} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {1}{x (1-\log (x))} \, dx,x,-3+x\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,1-\log (-3+x)\right )\\ &=\log (1-\log (-3+x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 7, normalized size = 1.00 \begin {gather*} \log (-1+\log (-3+x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - x + (-3 + x)*Log[-3 + x])^(-1),x]

[Out]

Log[-1 + Log[-3 + x]]

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fricas [A]  time = 1.02, size = 7, normalized size = 1.00 \begin {gather*} \log \left (\log \left (x - 3\right ) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((x-3)*log(x-3)+3-x),x, algorithm="fricas")

[Out]

log(log(x - 3) - 1)

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giac [A]  time = 0.23, size = 7, normalized size = 1.00 \begin {gather*} \log \left (\log \left (x - 3\right ) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((x-3)*log(x-3)+3-x),x, algorithm="giac")

[Out]

log(log(x - 3) - 1)

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maple [A]  time = 0.05, size = 8, normalized size = 1.14

method	result	size
derivativedivides	\(\ln \left (\ln \left (x -3\right )-1\right )\)	\(8\)
default	\(\ln \left (\ln \left (x -3\right )-1\right )\)	\(8\)
norman	\(\ln \left (\ln \left (x -3\right )-1\right )\)	\(8\)
risch	\(\ln \left (\ln \left (x -3\right )-1\right )\)	\(8\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x-3)*ln(x-3)+3-x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x-3)-1)

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maxima [A]  time = 0.54, size = 7, normalized size = 1.00 \begin {gather*} \log \left (\log \left (x - 3\right ) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((x-3)*log(x-3)+3-x),x, algorithm="maxima")

[Out]

log(log(x - 3) - 1)

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mupad [B]  time = 1.71, size = 7, normalized size = 1.00 \begin {gather*} \ln \left (\ln \left (x-3\right )-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(log(x - 3)*(x - 3) - x + 3),x)

[Out]

log(log(x - 3) - 1)

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sympy [A]  time = 0.10, size = 7, normalized size = 1.00 \begin {gather*} \log {\left (\log {\left (x - 3 \right )} - 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((x-3)*ln(x-3)+3-x),x)

[Out]

log(log(x - 3) - 1)

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