\(\int \frac {-4+6 x+(-4+9 x) \log (x)}{(-2 x+3 x^2) \log (x)} \, dx\) [1365]

Optimal result

Integrand size = 29, antiderivative size = 20 \[ \int \frac {-4+6 x+(-4+9 x) \log (x)}{\left (-2 x+3 x^2\right ) \log (x)} \, dx=\log \left (\frac {1}{5} e^3 (2-3 x) x^2 \log ^2(x)\right ) \]

[Out]

ln(1/5*x^2*exp(ln(2-3*x)+3)*ln(x)^2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1607, 6874, 78, 2339, 29} \[ \int \frac {-4+6 x+(-4+9 x) \log (x)}{\left (-2 x+3 x^2\right ) \log (x)} \, dx=\log (2-3 x)+2 \log (x)+2 \log (\log (x)) \]

[In]

Int[(-4 + 6*x + (-4 + 9*x)*Log[x])/((-2*x + 3*x^2)*Log[x]),x]

[Out]

Log[2 - 3*x] + 2*Log[x] + 2*Log[Log[x]]

Rule 29

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2339

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4+6 x+(-4+9 x) \log (x)}{x (-2+3 x) \log (x)} \, dx \\ & = \int \left (\frac {-4+9 x}{x (-2+3 x)}+\frac {2}{x \log (x)}\right ) \, dx \\ & = 2 \int \frac {1}{x \log (x)} \, dx+\int \frac {-4+9 x}{x (-2+3 x)} \, dx \\ & = 2 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )+\int \left (\frac {2}{x}+\frac {3}{-2+3 x}\right ) \, dx \\ & = \log (2-3 x)+2 \log (x)+2 \log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-4+6 x+(-4+9 x) \log (x)}{\left (-2 x+3 x^2\right ) \log (x)} \, dx=\log (2-3 x)+2 \log (x)+2 \log (\log (x)) \]

[In]

Integrate[(-4 + 6*x + (-4 + 9*x)*Log[x])/((-2*x + 3*x^2)*Log[x]),x]

[Out]

Log[2 - 3*x] + 2*Log[x] + 2*Log[Log[x]]

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
parallelrisch	\(2 \ln \left (\ln \left (x \right )\right )+\ln \left (-\frac {2}{3}+x \right )+2 \ln \left (x \right )\)	\(15\)
default	\(2 \ln \left (x \right )+\ln \left (-2+3 x \right )+2 \ln \left (\ln \left (x \right )\right )\)	\(17\)
norman	\(2 \ln \left (x \right )+\ln \left (-2+3 x \right )+2 \ln \left (\ln \left (x \right )\right )\)	\(17\)
risch	\(2 \ln \left (x \right )+\ln \left (-2+3 x \right )+2 \ln \left (\ln \left (x \right )\right )\)	\(17\)
parts	\(2 \ln \left (x \right )+\ln \left (-2+3 x \right )+2 \ln \left (\ln \left (x \right )\right )\)	\(17\)

[In]

int(((9*x-4)*ln(x)+6*x-4)/(3*x^2-2*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

2*ln(ln(x))+ln(-2/3+x)+2*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-4+6 x+(-4+9 x) \log (x)}{\left (-2 x+3 x^2\right ) \log (x)} \, dx=\log \left (3 \, x - 2\right ) + 2 \, \log \left (x\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((9*x-4)*log(x)+6*x-4)/(3*x^2-2*x)/log(x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-4+6 x+(-4+9 x) \log (x)}{\left (-2 x+3 x^2\right ) \log (x)} \, dx=2 \log {\left (x \right )} + \log {\left (x - \frac {2}{3} \right )} + 2 \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate(((9*x-4)*ln(x)+6*x-4)/(3*x**2-2*x)/ln(x),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-4+6 x+(-4+9 x) \log (x)}{\left (-2 x+3 x^2\right ) \log (x)} \, dx=\log \left (3 \, x - 2\right ) + 2 \, \log \left (x\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((9*x-4)*log(x)+6*x-4)/(3*x^2-2*x)/log(x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-4+6 x+(-4+9 x) \log (x)}{\left (-2 x+3 x^2\right ) \log (x)} \, dx=\log \left (3 \, x - 2\right ) + 2 \, \log \left (x\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((9*x-4)*log(x)+6*x-4)/(3*x^2-2*x)/log(x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.49 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-4+6 x+(-4+9 x) \log (x)}{\left (-2 x+3 x^2\right ) \log (x)} \, dx=\ln \left (x-\frac {2}{3}\right )+2\,\ln \left (\ln \left (x\right )\right )+2\,\ln \left (x\right ) \]

[In]

int(-(6*x + log(x)*(9*x - 4) - 4)/(log(x)*(2*x - 3*x^2)),x)

[Out]

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