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

Optimal. Leaf size=16 \[ \frac {1}{\log ^2(2+3 x)}+\log (2 x+\log (x)) \]

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Rubi [A]  time = 1.04, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6741, 6742, 6684, 2390, 2302, 30} \begin {gather*} \frac {1}{\log ^2(3 x+2)}+\log (2 x+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[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 2302

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 2390

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 6684

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

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 \operatorname {Subst}\left (\int \frac {1}{x \log ^3(x)} \, dx,x,2+3 x\right )\\ &=\log (2 x+\log (x))-2 \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 16, normalized size = 1.00 \begin {gather*} \frac {1}{\log ^2(2+3 x)}+\log (2 x+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[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]]

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fricas [A]  time = 0.70, size = 27, normalized size = 1.69 \begin {gather*} \frac {\log \left (3 \, x + 2\right )^{2} \log \left (2 \, x + \log \relax (x)\right ) + 1}{\log \left (3 \, x + 2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 16, normalized size = 1.00 \begin {gather*} \frac {1}{\log \left (3 \, x + 2\right )^{2}} + \log \left (2 \, x + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/log(3*x + 2)^2 + log(2*x + log(x))

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maple [A]  time = 0.04, size = 17, normalized size = 1.06

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 16, normalized size = 1.00 \begin {gather*} \frac {1}{\log \left (3 \, x + 2\right )^{2}} + \log \left (2 \, x + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/log(3*x + 2)^2 + log(2*x + log(x))

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mupad [B]  time = 5.89, size = 16, normalized size = 1.00 \begin {gather*} \ln \left (2\,x+\ln \relax (x)\right )+\frac {1}{{\ln \left (3\,x+2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.32, size = 17, normalized size = 1.06 \begin {gather*} \log {\left (2 x + \log {\relax (x )} \right )} + \frac {1}{\log {\left (3 x + 2 \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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