3.564 \(\int \frac{27+36 x+24 x^2+8 x^3}{729-64 x^6} \, dx\)

Optimal. Leaf size=50 \[ \frac{1}{36} \log \left (4 x^2-6 x+9\right )-\frac{1}{18} \log (3-2 x)-\frac{\tan ^{-1}\left (\frac{3-4 x}{3 \sqrt{3}}\right )}{18 \sqrt{3}} \]

[Out]

-ArcTan[(3 - 4*x)/(3*Sqrt[3])]/(18*Sqrt[3]) - Log[3 - 2*x]/18 + Log[9 - 6*x + 4*x^2]/36

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Rubi [A]  time = 0.0428354, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1586, 2058, 634, 618, 204, 628} \[ \frac{1}{36} \log \left (4 x^2-6 x+9\right )-\frac{1}{18} \log (3-2 x)-\frac{\tan ^{-1}\left (\frac{3-4 x}{3 \sqrt{3}}\right )}{18 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(27 + 36*x + 24*x^2 + 8*x^3)/(729 - 64*x^6),x]

[Out]

-ArcTan[(3 - 4*x)/(3*Sqrt[3])]/(18*Sqrt[3]) - Log[3 - 2*x]/18 + Log[9 - 6*x + 4*x^2]/36

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 2058

Int[(P_)^(p_), x_Symbol] :> With[{u = Factor[P]}, Int[ExpandIntegrand[u^p, x], x] /;  !SumQ[NonfreeFactors[u,
x]]] /; PolyQ[P, x] && ILtQ[p, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{27+36 x+24 x^2+8 x^3}{729-64 x^6} \, dx &=\int \frac{1}{27-36 x+24 x^2-8 x^3} \, dx\\ &=\int \left (-\frac{1}{9 (-3+2 x)}+\frac{2 x}{9 \left (9-6 x+4 x^2\right )}\right ) \, dx\\ &=-\frac{1}{18} \log (3-2 x)+\frac{2}{9} \int \frac{x}{9-6 x+4 x^2} \, dx\\ &=-\frac{1}{18} \log (3-2 x)+\frac{1}{36} \int \frac{-6+8 x}{9-6 x+4 x^2} \, dx+\frac{1}{6} \int \frac{1}{9-6 x+4 x^2} \, dx\\ &=-\frac{1}{18} \log (3-2 x)+\frac{1}{36} \log \left (9-6 x+4 x^2\right )-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-108-x^2} \, dx,x,-6+8 x\right )\\ &=-\frac{\tan ^{-1}\left (\frac{3-4 x}{3 \sqrt{3}}\right )}{18 \sqrt{3}}-\frac{1}{18} \log (3-2 x)+\frac{1}{36} \log \left (9-6 x+4 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0097146, size = 50, normalized size = 1. \[ \frac{1}{36} \log \left (4 x^2-6 x+9\right )-\frac{1}{18} \log (3-2 x)+\frac{\tan ^{-1}\left (\frac{4 x-3}{3 \sqrt{3}}\right )}{18 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(27 + 36*x + 24*x^2 + 8*x^3)/(729 - 64*x^6),x]

[Out]

ArcTan[(-3 + 4*x)/(3*Sqrt[3])]/(18*Sqrt[3]) - Log[3 - 2*x]/18 + Log[9 - 6*x + 4*x^2]/36

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Maple [A]  time = 0.004, size = 39, normalized size = 0.8 \begin{align*} -{\frac{\ln \left ( -3+2\,x \right ) }{18}}+{\frac{\ln \left ( 4\,{x}^{2}-6\,x+9 \right ) }{36}}+{\frac{\sqrt{3}}{54}\arctan \left ({\frac{ \left ( 8\,x-6 \right ) \sqrt{3}}{18}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^3+24*x^2+36*x+27)/(-64*x^6+729),x)

[Out]

-1/18*ln(-3+2*x)+1/36*ln(4*x^2-6*x+9)+1/54*3^(1/2)*arctan(1/18*(8*x-6)*3^(1/2))

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Maxima [A]  time = 1.36732, size = 51, normalized size = 1.02 \begin{align*} \frac{1}{54} \, \sqrt{3} \arctan \left (\frac{1}{9} \, \sqrt{3}{\left (4 \, x - 3\right )}\right ) + \frac{1}{36} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) - \frac{1}{18} \, \log \left (2 \, x - 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3+24*x^2+36*x+27)/(-64*x^6+729),x, algorithm="maxima")

[Out]

1/54*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3)) + 1/36*log(4*x^2 - 6*x + 9) - 1/18*log(2*x - 3)

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Fricas [A]  time = 1.41497, size = 124, normalized size = 2.48 \begin{align*} \frac{1}{54} \, \sqrt{3} \arctan \left (\frac{1}{9} \, \sqrt{3}{\left (4 \, x - 3\right )}\right ) + \frac{1}{36} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) - \frac{1}{18} \, \log \left (2 \, x - 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3+24*x^2+36*x+27)/(-64*x^6+729),x, algorithm="fricas")

[Out]

1/54*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3)) + 1/36*log(4*x^2 - 6*x + 9) - 1/18*log(2*x - 3)

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Sympy [A]  time = 0.145883, size = 48, normalized size = 0.96 \begin{align*} - \frac{\log{\left (x - \frac{3}{2} \right )}}{18} + \frac{\log{\left (x^{2} - \frac{3 x}{2} + \frac{9}{4} \right )}}{36} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{4 \sqrt{3} x}{9} - \frac{\sqrt{3}}{3} \right )}}{54} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**3+24*x**2+36*x+27)/(-64*x**6+729),x)

[Out]

-log(x - 3/2)/18 + log(x**2 - 3*x/2 + 9/4)/36 + sqrt(3)*atan(4*sqrt(3)*x/9 - sqrt(3)/3)/54

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Giac [A]  time = 1.06169, size = 53, normalized size = 1.06 \begin{align*} \frac{1}{54} \, \sqrt{3} \arctan \left (\frac{1}{9} \, \sqrt{3}{\left (4 \, x - 3\right )}\right ) + \frac{1}{36} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) - \frac{1}{18} \, \log \left ({\left | 2 \, x - 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3+24*x^2+36*x+27)/(-64*x^6+729),x, algorithm="giac")

[Out]

1/54*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3)) + 1/36*log(4*x^2 - 6*x + 9) - 1/18*log(abs(2*x - 3))