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

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

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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} $$\begin{array}{c}\frac{x^2}{3}-\frac{1}{3}\log ({\log }^2\left(3x^2\right)+x+2)+\log (x)\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{6+2x+4x^2+2x^3-4\log \left(3x^2\right)+(3+2x^2){\log }^2\left(3x^2\right)}{3x(2+x+{\log }^2\left(3x^2\right))}dx\\ & =\frac{1}{3}\int \frac{6+2x+4x^2+2x^3-4\log \left(3x^2\right)+(3+2x^2){\log }^2\left(3x^2\right)}{x(2+x+{\log }^2\left(3x^2\right))}dx\\ & =\frac{1}{3}\int (\frac{3+2x^2}{x}+\frac{-x-4\log \left(3x^2\right)}{x(2+x+{\log }^2\left(3x^2\right))})dx\\ & =\frac{1}{3}\int \frac{3+2x^2}{x}dx+\frac{1}{3}\int \frac{-x-4\log \left(3x^2\right)}{x(2+x+{\log }^2\left(3x^2\right))}dx\\ & =-\frac{1}{3}\log (2+x+{\log }^2\left(3x^2\right))+\frac{1}{3}\int (\frac{3}{x}+2x)dx\\ & =\frac{x^2}{3}+\log (x)-\frac{1}{3}\log (2+x+{\log }^2\left(3x^2\right))\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{1}{3}{x}^2+\frac{1}{2}\log \left(3x^2\right)-\frac{1}{3}\log (\log {\left(3x^2\right)}^2+x+2)\end{array}$$

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 $$\begin{array}{c}\frac{1}{3}{x}^2-\frac{1}{3}\log (-\log {\left(3x^2\right)}^2-x-2)+\log (x)\end{array}$$

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	$\frac{x^2}{3}+\ln (x)-\frac{\ln (2+x+\ln {\left(3x^2\right)}^2)}{3}$	$23$
risch	$\frac{x^2}{3}+\ln (x)-\frac{\ln (2+x+\ln {\left(3x^2\right)}^2)}{3}$	$23$

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 $$\begin{array}{c}\frac{1}{3}{x}^2-\frac{1}{3}\log (\frac{1}{4}\log (3{)}^2+\log (3)\log (x)+\log (x{)}^2+\frac{1}{4}x+\frac{1}{2})+\log (x)\end{array}$$

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 $$\begin{array}{c}\frac{\ln \left(x^2\right)}{2}-\frac{\ln ({\ln \left(3x^2\right)}^2+x+2)}{3}+\frac{x^2}{3}\end{array}$$

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 $$\begin{array}{c}\frac{x^2}{3}+\log (x)-\frac{\log (x+\log {\left(3x^2\right)}^2+2)}{3}\end{array}$$

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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