\(\int \frac {-12 x^2-6 x \log (x)+(2+7 x+6 x^2) \log ^3(2+3 x)}{(4 x^2+6 x^3+(2 x+3 x^2) \log (x)) \log ^3(2+3 x)} \, dx\) [8367]

Optimal result

Integrand size = 64, antiderivative size = 16 \[ \int \frac {-12 x^2-6 x \log (x)+\left (2+7 x+6 x^2\right ) \log ^3(2+3 x)}{\left (4 x^2+6 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log ^3(2+3 x)} \, dx=\frac {1}{\log ^2(2+3 x)}+\log (2 x+\log (x)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6873, 6874, 6816, 2437, 2339, 30} \[ \int \frac {-12 x^2-6 x \log (x)+\left (2+7 x+6 x^2\right ) \log ^3(2+3 x)}{\left (4 x^2+6 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log ^3(2+3 x)} \, dx=\frac {1}{\log ^2(3 x+2)}+\log (2 x+\log (x)) \]

[In]

Int[(-12*x^2 - 6*x*Log[x] + (2 + 7*x + 6*x^2)*Log[2 + 3*x]^3)/((4*x^2 + 6*x^3 + (2*x + 3*x^2)*Log[x])*Log[2 +
3*x]^3),x]

[Out]

Log[2 + 3*x]^(-2) + Log[2*x + Log[x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 {-12 x^2-6 x \log (x)+\left (2+7 x+6 x^2\right ) \log ^3(2+3 x)}{x (2+3 x) (2 x+\log (x)) \log ^3(2+3 x)} \, dx \\ & = \int \left (\frac {1+2 x}{x (2 x+\log (x))}-\frac {6}{(2+3 x) \log ^3(2+3 x)}\right ) \, dx \\ & = -\left (6 \int \frac {1}{(2+3 x) \log ^3(2+3 x)} \, dx\right )+\int \frac {1+2 x}{x (2 x+\log (x))} \, dx \\ & = \log (2 x+\log (x))-2 \text {Subst}\left (\int \frac {1}{x \log ^3(x)} \, dx,x,2+3 x\right ) \\ & = \log (2 x+\log (x))-2 \text {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (2+3 x)\right ) \\ & = \frac {1}{\log ^2(2+3 x)}+\log (2 x+\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-12 x^2-6 x \log (x)+\left (2+7 x+6 x^2\right ) \log ^3(2+3 x)}{\left (4 x^2+6 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log ^3(2+3 x)} \, dx=\frac {1}{\log ^2(2+3 x)}+\log (2 x+\log (x)) \]

[In]

Integrate[(-12*x^2 - 6*x*Log[x] + (2 + 7*x + 6*x^2)*Log[2 + 3*x]^3)/((4*x^2 + 6*x^3 + (2*x + 3*x^2)*Log[x])*Lo
g[2 + 3*x]^3),x]

[Out]

Log[2 + 3*x]^(-2) + Log[2*x + Log[x]]

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {1}{\ln \left (2+3 x \right )^{2}}+\ln \left (2 x +\ln \left (x \right )\right )\)	\(17\)
risch	\(\frac {1}{\ln \left (2+3 x \right )^{2}}+\ln \left (2 x +\ln \left (x \right )\right )\)	\(17\)
parallelrisch	\(\frac {12+12 \ln \left (x +\frac {\ln \left (x \right )}{2}\right ) \ln \left (2+3 x \right )^{2}}{12 \ln \left (2+3 x \right )^{2}}\)	\(30\)

[In]

int(((6*x^2+7*x+2)*ln(2+3*x)^3-6*x*ln(x)-12*x^2)/((3*x^2+2*x)*ln(x)+6*x^3+4*x^2)/ln(2+3*x)^3,x,method=_RETURNV
ERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69 \[ \int \frac {-12 x^2-6 x \log (x)+\left (2+7 x+6 x^2\right ) \log ^3(2+3 x)}{\left (4 x^2+6 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log ^3(2+3 x)} \, dx=\frac {\log \left (3 \, x + 2\right )^{2} \log \left (2 \, x + \log \left (x\right )\right ) + 1}{\log \left (3 \, x + 2\right )^{2}} \]

[In]

integrate(((6*x^2+7*x+2)*log(2+3*x)^3-6*x*log(x)-12*x^2)/((3*x^2+2*x)*log(x)+6*x^3+4*x^2)/log(2+3*x)^3,x, algo
rithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-12 x^2-6 x \log (x)+\left (2+7 x+6 x^2\right ) \log ^3(2+3 x)}{\left (4 x^2+6 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log ^3(2+3 x)} \, dx=\log {\left (2 x + \log {\left (x \right )} \right )} + \frac {1}{\log {\left (3 x + 2 \right )}^{2}} \]

[In]

integrate(((6*x**2+7*x+2)*ln(2+3*x)**3-6*x*ln(x)-12*x**2)/((3*x**2+2*x)*ln(x)+6*x**3+4*x**2)/ln(2+3*x)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-12 x^2-6 x \log (x)+\left (2+7 x+6 x^2\right ) \log ^3(2+3 x)}{\left (4 x^2+6 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log ^3(2+3 x)} \, dx=\frac {1}{\log \left (3 \, x + 2\right )^{2}} + \log \left (2 \, x + \log \left (x\right )\right ) \]

[In]

integrate(((6*x^2+7*x+2)*log(2+3*x)^3-6*x*log(x)-12*x^2)/((3*x^2+2*x)*log(x)+6*x^3+4*x^2)/log(2+3*x)^3,x, algo
rithm="maxima")

[Out]

1/log(3*x + 2)^2 + log(2*x + log(x))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-12 x^2-6 x \log (x)+\left (2+7 x+6 x^2\right ) \log ^3(2+3 x)}{\left (4 x^2+6 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log ^3(2+3 x)} \, dx=\frac {1}{\log \left (3 \, x + 2\right )^{2}} + \log \left (2 \, x + \log \left (x\right )\right ) \]

[In]

integrate(((6*x^2+7*x+2)*log(2+3*x)^3-6*x*log(x)-12*x^2)/((3*x^2+2*x)*log(x)+6*x^3+4*x^2)/log(2+3*x)^3,x, algo
rithm="giac")

[Out]

1/log(3*x + 2)^2 + log(2*x + log(x))

Mupad [B] (verification not implemented)

Time = 14.75 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-12 x^2-6 x \log (x)+\left (2+7 x+6 x^2\right ) \log ^3(2+3 x)}{\left (4 x^2+6 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log ^3(2+3 x)} \, dx=\ln \left (2\,x+\ln \left (x\right )\right )+\frac {1}{{\ln \left (3\,x+2\right )}^2} \]

[In]

int(-(6*x*log(x) - log(3*x + 2)^3*(7*x + 6*x^2 + 2) + 12*x^2)/(log(3*x + 2)^3*(log(x)*(2*x + 3*x^2) + 4*x^2 +
6*x^3)),x)

[Out]

log(2*x + log(x)) + 1/log(3*x + 2)^2