3.36.61 \(\int \frac {x^2+12 x^3+(-12 x+2 x^2+24 x^3) \log (x)+(-1-12 x) \log (\frac {1}{3} (1+12 x))}{3 x+36 x^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {1}{3} \log (x) \left (x^2-\log \left (\frac {1}{3}+4 x\right )\right ) \]

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Rubi [A]  time = 0.24, antiderivative size = 30, normalized size of antiderivative = 1.50, number of steps used = 11, number of rules used = 8, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1593, 6688, 12, 2357, 2304, 2317, 2391, 2392} \begin {gather*} \frac {1}{3} x^2 \log (x)+\frac {1}{3} \log (3) \log (x)-\frac {1}{3} \log (x) \log (12 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + 12*x^3 + (-12*x + 2*x^2 + 24*x^3)*Log[x] + (-1 - 12*x)*Log[(1 + 12*x)/3])/(3*x + 36*x^2),x]

[Out]

(x^2*Log[x])/3 + (Log[3]*Log[x])/3 - (Log[x]*Log[1 + 12*x])/3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1593

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2392

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + (e*x)/d]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2+12 x^3+\left (-12 x+2 x^2+24 x^3\right ) \log (x)+(-1-12 x) \log \left (\frac {1}{3} (1+12 x)\right )}{x (3+36 x)} \, dx\\ &=\int \frac {1}{3} \left (x+\frac {2 \left (-6+x+12 x^2\right ) \log (x)}{1+12 x}-\frac {\log \left (\frac {1}{3}+4 x\right )}{x}\right ) \, dx\\ &=\frac {1}{3} \int \left (x+\frac {2 \left (-6+x+12 x^2\right ) \log (x)}{1+12 x}-\frac {\log \left (\frac {1}{3}+4 x\right )}{x}\right ) \, dx\\ &=\frac {x^2}{6}-\frac {1}{3} \int \frac {\log \left (\frac {1}{3}+4 x\right )}{x} \, dx+\frac {2}{3} \int \frac {\left (-6+x+12 x^2\right ) \log (x)}{1+12 x} \, dx\\ &=\frac {x^2}{6}+\frac {1}{3} \log (3) \log (x)-\frac {1}{3} \int \frac {\log (1+12 x)}{x} \, dx+\frac {2}{3} \int \left (x \log (x)-\frac {6 \log (x)}{1+12 x}\right ) \, dx\\ &=\frac {x^2}{6}+\frac {1}{3} \log (3) \log (x)+\frac {\text {Li}_2(-12 x)}{3}+\frac {2}{3} \int x \log (x) \, dx-4 \int \frac {\log (x)}{1+12 x} \, dx\\ &=\frac {1}{3} x^2 \log (x)+\frac {1}{3} \log (3) \log (x)-\frac {1}{3} \log (x) \log (1+12 x)+\frac {\text {Li}_2(-12 x)}{3}+\frac {1}{3} \int \frac {\log (1+12 x)}{x} \, dx\\ &=\frac {1}{3} x^2 \log (x)+\frac {1}{3} \log (3) \log (x)-\frac {1}{3} \log (x) \log (1+12 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.30 \begin {gather*} \frac {1}{3} \left (x^2 \log (x)+\log (3) \log (x)-\log (x) \log (1+12 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + 12*x^3 + (-12*x + 2*x^2 + 24*x^3)*Log[x] + (-1 - 12*x)*Log[(1 + 12*x)/3])/(3*x + 36*x^2),x]

[Out]

(x^2*Log[x] + Log[3]*Log[x] - Log[x]*Log[1 + 12*x])/3

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fricas [A]  time = 0.75, size = 18, normalized size = 0.90 \begin {gather*} \frac {1}{3} \, x^{2} \log \relax (x) - \frac {1}{3} \, \log \left (4 \, x + \frac {1}{3}\right ) \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x-1)*log(4*x+1/3)+(24*x^3+2*x^2-12*x)*log(x)+12*x^3+x^2)/(36*x^2+3*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 24, normalized size = 1.20 \begin {gather*} \frac {1}{3} \, x^{2} \log \relax (x) + \frac {1}{3} \, \log \relax (3) \log \relax (x) - \frac {1}{3} \, \log \left (12 \, x + 1\right ) \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x-1)*log(4*x+1/3)+(24*x^3+2*x^2-12*x)*log(x)+12*x^3+x^2)/(36*x^2+3*x),x, algorithm="giac")

[Out]

1/3*x^2*log(x) + 1/3*log(3)*log(x) - 1/3*log(12*x + 1)*log(x)

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maple [A]  time = 0.08, size = 19, normalized size = 0.95

method	result	size
risch	\(-\frac {\ln \relax (x ) \ln \left (4 x +\frac {1}{3}\right )}{3}+\frac {x^{2} \ln \relax (x )}{3}\)	\(19\)
default	\(\frac {\ln \relax (3) \ln \relax (x )}{3}+\frac {x^{2} \ln \relax (x )}{3}-\frac {\ln \relax (x ) \ln \left (12 x +1\right )}{3}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-12*x-1)*ln(4*x+1/3)+(24*x^3+2*x^2-12*x)*ln(x)+12*x^3+x^2)/(36*x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.91, size = 21, normalized size = 1.05 \begin {gather*} \frac {1}{3} \, {\left (x^{2} + \log \relax (3)\right )} \log \relax (x) - \frac {1}{3} \, \log \left (12 \, x + 1\right ) \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x-1)*log(4*x+1/3)+(24*x^3+2*x^2-12*x)*log(x)+12*x^3+x^2)/(36*x^2+3*x),x, algorithm="maxima")

[Out]

1/3*(x^2 + log(3))*log(x) - 1/3*log(12*x + 1)*log(x)

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mupad [B]  time = 2.36, size = 16, normalized size = 0.80 \begin {gather*} -\frac {\ln \relax (x)\,\left (\ln \left (4\,x+\frac {1}{3}\right )-x^2\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - log(4*x + 1/3)*(12*x + 1) + 12*x^3 + log(x)*(2*x^2 - 12*x + 24*x^3))/(3*x + 36*x^2),x)

[Out]

-(log(x)*(log(4*x + 1/3) - x^2))/3

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sympy [A]  time = 0.34, size = 20, normalized size = 1.00 \begin {gather*} \frac {x^{2} \log {\relax (x )}}{3} - \frac {\log {\relax (x )} \log {\left (4 x + \frac {1}{3} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x-1)*ln(4*x+1/3)+(24*x**3+2*x**2-12*x)*ln(x)+12*x**3+x**2)/(36*x**2+3*x),x)

[Out]

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

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