\(\int \frac {24+12 x+(4-12 x) \log (\frac {3}{-1+3 x})}{-9+27 x+(-12+30 x+18 x^2) \log (\frac {3}{-1+3 x})+(-4+8 x+11 x^2+3 x^3) \log ^2(\frac {3}{-1+3 x})} \, dx\) [1187]

Optimal result

Integrand size = 78, antiderivative size = 22 \[ \int \frac {24+12 x+(4-12 x) \log \left (\frac {3}{-1+3 x}\right )}{-9+27 x+\left (-12+30 x+18 x^2\right ) \log \left (\frac {3}{-1+3 x}\right )+\left (-4+8 x+11 x^2+3 x^3\right ) \log ^2\left (\frac {3}{-1+3 x}\right )} \, dx=2+\frac {4}{3+(2+x) \log \left (\frac {3}{-1+3 x}\right )} \]

[Out]

4/(3+(2+x)*ln(3/(-1+3*x)))+2

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {6820, 12, 6840, 32} \[ \int \frac {24+12 x+(4-12 x) \log \left (\frac {3}{-1+3 x}\right )}{-9+27 x+\left (-12+30 x+18 x^2\right ) \log \left (\frac {3}{-1+3 x}\right )+\left (-4+8 x+11 x^2+3 x^3\right ) \log ^2\left (\frac {3}{-1+3 x}\right )} \, dx=\frac {4}{(x+2) \log \left (-\frac {3}{1-3 x}\right )+3} \]

[In]

Int[(24 + 12*x + (4 - 12*x)*Log[3/(-1 + 3*x)])/(-9 + 27*x + (-12 + 30*x + 18*x^2)*Log[3/(-1 + 3*x)] + (-4 + 8*
x + 11*x^2 + 3*x^3)*Log[3/(-1 + 3*x)]^2),x]

[Out]

4/(3 + (2 + x)*Log[-3/(1 - 3*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 6820

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

Rule 6840

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

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {24+12 x+(4-12 x) \log \left (\frac {3}{-1+3 x}\right )}{-9+27 x+\left (-12+30 x+18 x^2\right ) \log \left (\frac {3}{-1+3 x}\right )+\left (-4+8 x+11 x^2+3 x^3\right ) \log ^2\left (\frac {3}{-1+3 x}\right )} \, dx=\frac {4}{3+(2+x) \log \left (\frac {3}{-1+3 x}\right )} \]

[In]

Integrate[(24 + 12*x + (4 - 12*x)*Log[3/(-1 + 3*x)])/(-9 + 27*x + (-12 + 30*x + 18*x^2)*Log[3/(-1 + 3*x)] + (-
4 + 8*x + 11*x^2 + 3*x^3)*Log[3/(-1 + 3*x)]^2),x]

[Out]

4/(3 + (2 + x)*Log[3/(-1 + 3*x)])

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41

method	result	size
norman	\(\frac {4}{\ln \left (\frac {3}{-1+3 x}\right ) x +2 \ln \left (\frac {3}{-1+3 x}\right )+3}\)	\(31\)
risch	\(\frac {4}{\ln \left (\frac {3}{-1+3 x}\right ) x +2 \ln \left (\frac {3}{-1+3 x}\right )+3}\)	\(31\)
parallelrisch	\(\frac {4}{\ln \left (\frac {3}{-1+3 x}\right ) x +2 \ln \left (\frac {3}{-1+3 x}\right )+3}\)	\(31\)
derivativedivides	\(\frac {36}{\left (-1+3 x \right ) \left (\frac {21 \ln \left (\frac {3}{-1+3 x}\right )}{-1+3 x}+\frac {27}{-1+3 x}+3 \ln \left (\frac {3}{-1+3 x}\right )\right )}\)	\(53\)
default	\(\frac {36}{\left (-1+3 x \right ) \left (\frac {21 \ln \left (\frac {3}{-1+3 x}\right )}{-1+3 x}+\frac {27}{-1+3 x}+3 \ln \left (\frac {3}{-1+3 x}\right )\right )}\)	\(53\)

[In]

int(((-12*x+4)*ln(3/(-1+3*x))+12*x+24)/((3*x^3+11*x^2+8*x-4)*ln(3/(-1+3*x))^2+(18*x^2+30*x-12)*ln(3/(-1+3*x))+
27*x-9),x,method=_RETURNVERBOSE)

[Out]

4/(ln(3/(-1+3*x))*x+2*ln(3/(-1+3*x))+3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {24+12 x+(4-12 x) \log \left (\frac {3}{-1+3 x}\right )}{-9+27 x+\left (-12+30 x+18 x^2\right ) \log \left (\frac {3}{-1+3 x}\right )+\left (-4+8 x+11 x^2+3 x^3\right ) \log ^2\left (\frac {3}{-1+3 x}\right )} \, dx=\frac {4}{{\left (x + 2\right )} \log \left (\frac {3}{3 \, x - 1}\right ) + 3} \]

[In]

integrate(((-12*x+4)*log(3/(-1+3*x))+12*x+24)/((3*x^3+11*x^2+8*x-4)*log(3/(-1+3*x))^2+(18*x^2+30*x-12)*log(3/(
-1+3*x))+27*x-9),x, algorithm="fricas")

[Out]

4/((x + 2)*log(3/(3*x - 1)) + 3)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {24+12 x+(4-12 x) \log \left (\frac {3}{-1+3 x}\right )}{-9+27 x+\left (-12+30 x+18 x^2\right ) \log \left (\frac {3}{-1+3 x}\right )+\left (-4+8 x+11 x^2+3 x^3\right ) \log ^2\left (\frac {3}{-1+3 x}\right )} \, dx=\frac {4}{\left (x + 2\right ) \log {\left (\frac {3}{3 x - 1} \right )} + 3} \]

[In]

integrate(((-12*x+4)*ln(3/(-1+3*x))+12*x+24)/((3*x**3+11*x**2+8*x-4)*ln(3/(-1+3*x))**2+(18*x**2+30*x-12)*ln(3/
(-1+3*x))+27*x-9),x)

[Out]

4/((x + 2)*log(3/(3*x - 1)) + 3)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {24+12 x+(4-12 x) \log \left (\frac {3}{-1+3 x}\right )}{-9+27 x+\left (-12+30 x+18 x^2\right ) \log \left (\frac {3}{-1+3 x}\right )+\left (-4+8 x+11 x^2+3 x^3\right ) \log ^2\left (\frac {3}{-1+3 x}\right )} \, dx=\frac {4}{x \log \left (3\right ) - {\left (x + 2\right )} \log \left (3 \, x - 1\right ) + 2 \, \log \left (3\right ) + 3} \]

[In]

integrate(((-12*x+4)*log(3/(-1+3*x))+12*x+24)/((3*x^3+11*x^2+8*x-4)*log(3/(-1+3*x))^2+(18*x^2+30*x-12)*log(3/(
-1+3*x))+27*x-9),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (22) = 44\).

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.27 \[ \int \frac {24+12 x+(4-12 x) \log \left (\frac {3}{-1+3 x}\right )}{-9+27 x+\left (-12+30 x+18 x^2\right ) \log \left (\frac {3}{-1+3 x}\right )+\left (-4+8 x+11 x^2+3 x^3\right ) \log ^2\left (\frac {3}{-1+3 x}\right )} \, dx=\frac {12}{{\left (3 \, x - 1\right )} {\left (\frac {7 \, \log \left (\frac {3}{3 \, x - 1}\right )}{3 \, x - 1} + \frac {9}{3 \, x - 1} + \log \left (\frac {3}{3 \, x - 1}\right )\right )}} \]

[In]

integrate(((-12*x+4)*log(3/(-1+3*x))+12*x+24)/((3*x^3+11*x^2+8*x-4)*log(3/(-1+3*x))^2+(18*x^2+30*x-12)*log(3/(
-1+3*x))+27*x-9),x, algorithm="giac")

[Out]

12/((3*x - 1)*(7*log(3/(3*x - 1))/(3*x - 1) + 9/(3*x - 1) + log(3/(3*x - 1))))

Mupad [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {24+12 x+(4-12 x) \log \left (\frac {3}{-1+3 x}\right )}{-9+27 x+\left (-12+30 x+18 x^2\right ) \log \left (\frac {3}{-1+3 x}\right )+\left (-4+8 x+11 x^2+3 x^3\right ) \log ^2\left (\frac {3}{-1+3 x}\right )} \, dx=\frac {4}{2\,\ln \left (\frac {3}{3\,x-1}\right )+x\,\ln \left (\frac {3}{3\,x-1}\right )+3} \]

[In]

int((12*x - log(3/(3*x - 1))*(12*x - 4) + 24)/(27*x + log(3/(3*x - 1))*(30*x + 18*x^2 - 12) + log(3/(3*x - 1))
^2*(8*x + 11*x^2 + 3*x^3 - 4) - 9),x)

[Out]

4/(2*log(3/(3*x - 1)) + x*log(3/(3*x - 1)) + 3)