\(\int \frac {9 x-3 x^3+(-3+12 x-10 x^2+2 x^3+3 x^4) \log (-1+3 x)}{(3 x-9 x^2-x^3+3 x^4) \log (-1+3 x)} \, dx\) [3838]

Optimal result

Integrand size = 66, antiderivative size = 28 \[ \int \frac {9 x-3 x^3+\left (-3+12 x-10 x^2+2 x^3+3 x^4\right ) \log (-1+3 x)}{\left (3 x-9 x^2-x^3+3 x^4\right ) \log (-1+3 x)} \, dx=x+\log \left (\frac {\frac {3}{x}-x}{8 \log (5) \log (-1+3 x)}\right ) \]

[Out]

x+ln(1/8*(3/x-x)/ln(-1+3*x)/ln(5))

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {6873, 6874, 1816, 266, 2437, 2339, 29} \[ \int \frac {9 x-3 x^3+\left (-3+12 x-10 x^2+2 x^3+3 x^4\right ) \log (-1+3 x)}{\left (3 x-9 x^2-x^3+3 x^4\right ) \log (-1+3 x)} \, dx=\log \left (3-x^2\right )+x-\log (x)-\log (\log (3 x-1)) \]

[In]

Int[(9*x - 3*x^3 + (-3 + 12*x - 10*x^2 + 2*x^3 + 3*x^4)*Log[-1 + 3*x])/((3*x - 9*x^2 - x^3 + 3*x^4)*Log[-1 + 3
*x]),x]

[Out]

x - Log[x] + Log[3 - x^2] - Log[Log[-1 + 3*x]]

Rule 29

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

Rule 266

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

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {9 x-3 x^3+\left (-3+12 x-10 x^2+2 x^3+3 x^4\right ) \log (-1+3 x)}{x \left (3-9 x-x^2+3 x^3\right ) \log (-1+3 x)} \, dx \\ & = \int \left (\frac {3-3 x+x^2+x^3}{x \left (-3+x^2\right )}-\frac {3}{(-1+3 x) \log (-1+3 x)}\right ) \, dx \\ & = -\left (3 \int \frac {1}{(-1+3 x) \log (-1+3 x)} \, dx\right )+\int \frac {3-3 x+x^2+x^3}{x \left (-3+x^2\right )} \, dx \\ & = \int \left (1-\frac {1}{x}+\frac {2 x}{-3+x^2}\right ) \, dx-\text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-1+3 x\right ) \\ & = x-\log (x)+2 \int \frac {x}{-3+x^2} \, dx-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (-1+3 x)\right ) \\ & = x-\log (x)+\log \left (3-x^2\right )-\log (\log (-1+3 x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {9 x-3 x^3+\left (-3+12 x-10 x^2+2 x^3+3 x^4\right ) \log (-1+3 x)}{\left (3 x-9 x^2-x^3+3 x^4\right ) \log (-1+3 x)} \, dx=x-\log (x)+\log \left (3-x^2\right )-\log (\log (-1+3 x)) \]

[In]

Integrate[(9*x - 3*x^3 + (-3 + 12*x - 10*x^2 + 2*x^3 + 3*x^4)*Log[-1 + 3*x])/((3*x - 9*x^2 - x^3 + 3*x^4)*Log[
-1 + 3*x]),x]

[Out]

x - Log[x] + Log[3 - x^2] - Log[Log[-1 + 3*x]]

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79

method	result	size
norman	\(x -\ln \left (x \right )-\ln \left (\ln \left (-1+3 x \right )\right )+\ln \left (x^{2}-3\right )\)	\(22\)
risch	\(x -\ln \left (x \right )-\ln \left (\ln \left (-1+3 x \right )\right )+\ln \left (x^{2}-3\right )\)	\(22\)
parts	\(x -\ln \left (x \right )-\ln \left (\ln \left (-1+3 x \right )\right )+\ln \left (x^{2}-3\right )\)	\(22\)
parallelrisch	\(\frac {2}{3}-\ln \left (x \right )-\ln \left (\ln \left (-1+3 x \right )\right )+\ln \left (x^{2}-3\right )+x\)	\(23\)
derivativedivides	\(-\frac {1}{3}+x +\ln \left (\left (-1+3 x \right )^{2}-28+6 x \right )-\ln \left (3 x \right )-\ln \left (\ln \left (-1+3 x \right )\right )\)	\(32\)
default	\(-\frac {1}{3}+x +\ln \left (\left (-1+3 x \right )^{2}-28+6 x \right )-\ln \left (3 x \right )-\ln \left (\ln \left (-1+3 x \right )\right )\)	\(32\)

[In]

int(((3*x^4+2*x^3-10*x^2+12*x-3)*ln(-1+3*x)-3*x^3+9*x)/(3*x^4-x^3-9*x^2+3*x)/ln(-1+3*x),x,method=_RETURNVERBOS
E)

[Out]

x-ln(x)-ln(ln(-1+3*x))+ln(x^2-3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {9 x-3 x^3+\left (-3+12 x-10 x^2+2 x^3+3 x^4\right ) \log (-1+3 x)}{\left (3 x-9 x^2-x^3+3 x^4\right ) \log (-1+3 x)} \, dx=x + \log \left (x^{2} - 3\right ) - \log \left (x\right ) - \log \left (\log \left (3 \, x - 1\right )\right ) \]

[In]

integrate(((3*x^4+2*x^3-10*x^2+12*x-3)*log(-1+3*x)-3*x^3+9*x)/(3*x^4-x^3-9*x^2+3*x)/log(-1+3*x),x, algorithm="
fricas")

[Out]

x + log(x^2 - 3) - log(x) - log(log(3*x - 1))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {9 x-3 x^3+\left (-3+12 x-10 x^2+2 x^3+3 x^4\right ) \log (-1+3 x)}{\left (3 x-9 x^2-x^3+3 x^4\right ) \log (-1+3 x)} \, dx=x - \log {\left (x \right )} + \log {\left (x^{2} - 3 \right )} - \log {\left (\log {\left (3 x - 1 \right )} \right )} \]

[In]

integrate(((3*x**4+2*x**3-10*x**2+12*x-3)*ln(-1+3*x)-3*x**3+9*x)/(3*x**4-x**3-9*x**2+3*x)/ln(-1+3*x),x)

[Out]

x - log(x) + log(x**2 - 3) - log(log(3*x - 1))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {9 x-3 x^3+\left (-3+12 x-10 x^2+2 x^3+3 x^4\right ) \log (-1+3 x)}{\left (3 x-9 x^2-x^3+3 x^4\right ) \log (-1+3 x)} \, dx=x + \log \left (x^{2} - 3\right ) - \log \left (x\right ) - \log \left (\log \left (3 \, x - 1\right )\right ) \]

[In]

integrate(((3*x^4+2*x^3-10*x^2+12*x-3)*log(-1+3*x)-3*x^3+9*x)/(3*x^4-x^3-9*x^2+3*x)/log(-1+3*x),x, algorithm="
maxima")

[Out]

x + log(x^2 - 3) - log(x) - log(log(3*x - 1))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {9 x-3 x^3+\left (-3+12 x-10 x^2+2 x^3+3 x^4\right ) \log (-1+3 x)}{\left (3 x-9 x^2-x^3+3 x^4\right ) \log (-1+3 x)} \, dx=x + \log \left (x^{2} - 3\right ) - \log \left (x\right ) - \log \left (\log \left (3 \, x - 1\right )\right ) \]

[In]

integrate(((3*x^4+2*x^3-10*x^2+12*x-3)*log(-1+3*x)-3*x^3+9*x)/(3*x^4-x^3-9*x^2+3*x)/log(-1+3*x),x, algorithm="
giac")

[Out]

x + log(x^2 - 3) - log(x) - log(log(3*x - 1))

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {9 x-3 x^3+\left (-3+12 x-10 x^2+2 x^3+3 x^4\right ) \log (-1+3 x)}{\left (3 x-9 x^2-x^3+3 x^4\right ) \log (-1+3 x)} \, dx=x-\ln \left (\ln \left (3\,x-1\right )\right )+\ln \left (x^2-3\right )-\ln \left (x\right ) \]

[In]

int((9*x - 3*x^3 + log(3*x - 1)*(12*x - 10*x^2 + 2*x^3 + 3*x^4 - 3))/(log(3*x - 1)*(3*x - 9*x^2 - x^3 + 3*x^4)
),x)

[Out]

x - log(log(3*x - 1)) + log(x^2 - 3) - log(x)