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

Optimal. Leaf size=26 \[ \frac {1}{3} \left (x^2+\log \left (\frac {x^3}{2+x+\log ^2\left (3 x^2\right )}\right )\right ) \]

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Rubi [A]  time = 0.40, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6741, 12, 6742, 14, 6684} \begin {gather*} \frac {x^2}{3}-\frac {1}{3} \log \left (\log ^2\left (3 x^2\right )+x+2\right )+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^2/3 + Log[x] - Log[2 + x + Log[3*x^2]^2]/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 {6+2 x+4 x^2+2 x^3-4 \log \left (3 x^2\right )+\left (3+2 x^2\right ) \log ^2\left (3 x^2\right )}{3 x \left (2+x+\log ^2\left (3 x^2\right )\right )} \, dx\\ &=\frac {1}{3} \int \frac {6+2 x+4 x^2+2 x^3-4 \log \left (3 x^2\right )+\left (3+2 x^2\right ) \log ^2\left (3 x^2\right )}{x \left (2+x+\log ^2\left (3 x^2\right )\right )} \, dx\\ &=\frac {1}{3} \int \left (\frac {3+2 x^2}{x}+\frac {-x-4 \log \left (3 x^2\right )}{x \left (2+x+\log ^2\left (3 x^2\right )\right )}\right ) \, dx\\ &=\frac {1}{3} \int \frac {3+2 x^2}{x} \, dx+\frac {1}{3} \int \frac {-x-4 \log \left (3 x^2\right )}{x \left (2+x+\log ^2\left (3 x^2\right )\right )} \, dx\\ &=-\frac {1}{3} \log \left (2+x+\log ^2\left (3 x^2\right )\right )+\frac {1}{3} \int \left (\frac {3}{x}+2 x\right ) \, dx\\ &=\frac {x^2}{3}+\log (x)-\frac {1}{3} \log \left (2+x+\log ^2\left (3 x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 26, normalized size = 1.00 \begin {gather*} \frac {1}{3} \left (x^2+3 \log (x)-\log \left (2+x+\log ^2\left (3 x^2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.02, size = 28, normalized size = 1.08 \begin {gather*} \frac {1}{3} \, x^{2} + \frac {1}{2} \, \log \left (3 \, x^{2}\right ) - \frac {1}{3} \, \log \left (\log \left (3 \, x^{2}\right )^{2} + x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+3)*log(3*x^2)^2-4*log(3*x^2)+2*x^3+4*x^2+2*x+6)/(3*x*log(3*x^2)^2+3*x^2+6*x),x, algorithm="f
ricas")

[Out]

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

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giac [A]  time = 0.15, size = 26, normalized size = 1.00 \begin {gather*} \frac {1}{3} \, x^{2} - \frac {1}{3} \, \log \left (-\log \left (3 \, x^{2}\right )^{2} - x - 2\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+3)*log(3*x^2)^2-4*log(3*x^2)+2*x^3+4*x^2+2*x+6)/(3*x*log(3*x^2)^2+3*x^2+6*x),x, algorithm="g
iac")

[Out]

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

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maple [A]  time = 0.03, size = 23, normalized size = 0.88

method	result	size
norman	\(\frac {x^{2}}{3}+\ln \relax (x )-\frac {\ln \left (2+x +\ln \left (3 x^{2}\right )^{2}\right )}{3}\)	\(23\)
risch	\(\frac {x^{2}}{3}+\ln \relax (x )-\frac {\ln \left (2+x +\ln \left (3 x^{2}\right )^{2}\right )}{3}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+3)*ln(3*x^2)^2-4*ln(3*x^2)+2*x^3+4*x^2+2*x+6)/(3*x*ln(3*x^2)^2+3*x^2+6*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 31, normalized size = 1.19 \begin {gather*} \frac {1}{3} \, x^{2} - \frac {1}{3} \, \log \left (\frac {1}{4} \, \log \relax (3)^{2} + \log \relax (3) \log \relax (x) + \log \relax (x)^{2} + \frac {1}{4} \, x + \frac {1}{2}\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+3)*log(3*x^2)^2-4*log(3*x^2)+2*x^3+4*x^2+2*x+6)/(3*x*log(3*x^2)^2+3*x^2+6*x),x, algorithm="m
axima")

[Out]

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

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mupad [B]  time = 3.15, size = 26, normalized size = 1.00 \begin {gather*} \frac {\ln \left (x^2\right )}{2}-\frac {\ln \left ({\ln \left (3\,x^2\right )}^2+x+2\right )}{3}+\frac {x^2}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.19, size = 22, normalized size = 0.85 \begin {gather*} \frac {x^{2}}{3} + \log {\relax (x )} - \frac {\log {\left (x + \log {\left (3 x^{2} \right )}^{2} + 2 \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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