\(\int \frac {-3 x \log (x)+(-2+3 x) \log (-2+3 x)}{(-2 x+3 x^2) \log (x) \log (-2+3 x)} \, dx\) [12]

Optimal result

Integrand size = 42, antiderivative size = 23 \[ \int \frac {-3 x \log (x)+(-2+3 x) \log (-2+3 x)}{\left (-2 x+3 x^2\right ) \log (x) \log (-2+3 x)} \, dx=\log \left (\frac {4 \left (x+e^2 x\right ) \log (x)}{x \log (-2+3 x)}\right ) \]

[Out]

ln(4*ln(x)*(x+exp(2)*x)/ln(-2+3*x)/x)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {1607, 6820, 2339, 29, 2437} \[ \int \frac {-3 x \log (x)+(-2+3 x) \log (-2+3 x)}{\left (-2 x+3 x^2\right ) \log (x) \log (-2+3 x)} \, dx=\log (\log (x))-\log (\log (3 x-2)) \]

[In]

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

[Out]

Log[Log[x]] - Log[Log[-2 + 3*x]]

Rule 29

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

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 2437

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 6820

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {-3 x \log (x)+(-2+3 x) \log (-2+3 x)}{\left (-2 x+3 x^2\right ) \log (x) \log (-2+3 x)} \, dx=\log (\log (x))-\log (\log (-2+3 x)) \]

[In]

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

[Out]

Log[Log[x]] - Log[Log[-2 + 3*x]]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61

method	result	size
default	\(\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (-2+3 x \right )\right )\)	\(14\)
norman	\(\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (-2+3 x \right )\right )\)	\(14\)
risch	\(\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (-2+3 x \right )\right )\)	\(14\)
parallelrisch	\(\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (-2+3 x \right )\right )\)	\(14\)
parts	\(\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (-2+3 x \right )\right )\)	\(14\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {-3 x \log (x)+(-2+3 x) \log (-2+3 x)}{\left (-2 x+3 x^2\right ) \log (x) \log (-2+3 x)} \, dx=-\log \left (\log \left (3 \, x - 2\right )\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

-log(log(3*x - 2)) + log(log(x))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.52 \[ \int \frac {-3 x \log (x)+(-2+3 x) \log (-2+3 x)}{\left (-2 x+3 x^2\right ) \log (x) \log (-2+3 x)} \, dx=\log {\left (\log {\left (x \right )} \right )} - \log {\left (\log {\left (3 x - 2 \right )} \right )} \]

[In]

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

[Out]

log(log(x)) - log(log(3*x - 2))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {-3 x \log (x)+(-2+3 x) \log (-2+3 x)}{\left (-2 x+3 x^2\right ) \log (x) \log (-2+3 x)} \, dx=-\log \left (\log \left (3 \, x - 2\right )\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

-log(log(3*x - 2)) + log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {-3 x \log (x)+(-2+3 x) \log (-2+3 x)}{\left (-2 x+3 x^2\right ) \log (x) \log (-2+3 x)} \, dx=-\log \left (\log \left (3 \, x - 2\right )\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

-log(log(3*x - 2)) + log(log(x))

Mupad [B] (verification not implemented)

Time = 7.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {-3 x \log (x)+(-2+3 x) \log (-2+3 x)}{\left (-2 x+3 x^2\right ) \log (x) \log (-2+3 x)} \, dx=\ln \left (\ln \left (x\right )\right )-\ln \left (\ln \left (3\,x-2\right )\right ) \]

[In]

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

[Out]

log(log(x)) - log(log(3*x - 2))