\(\int \frac {9+6 x+4 x^2}{729-64 x^6} \, dx\) [562]

Optimal result

Integrand size = 20, antiderivative size = 60 \[ \int \frac {9+6 x+4 x^2}{729-64 x^6} \, dx=-\frac {\arctan \left (\frac {3-4 x}{3 \sqrt {3}}\right )}{54 \sqrt {3}}-\frac {1}{108} \log (3-2 x)+\frac {1}{324} \log (3+2 x)+\frac {1}{324} \log \left (9-6 x+4 x^2\right ) \]

[Out]

-1/108*ln(3-2*x)+1/324*ln(3+2*x)+1/324*ln(4*x^2-6*x+9)-1/162*arctan(1/9*(3-4*x)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1600, 2083, 648, 632, 210, 642} \[ \int \frac {9+6 x+4 x^2}{729-64 x^6} \, dx=-\frac {\arctan \left (\frac {3-4 x}{3 \sqrt {3}}\right )}{54 \sqrt {3}}+\frac {1}{324} \log \left (4 x^2-6 x+9\right )-\frac {1}{108} \log (3-2 x)+\frac {1}{324} \log (2 x+3) \]

[In]

Int[(9 + 6*x + 4*x^2)/(729 - 64*x^6),x]

[Out]

-1/54*ArcTan[(3 - 4*x)/(3*Sqrt[3])]/Sqrt[3] - Log[3 - 2*x]/108 + Log[3 + 2*x]/324 + Log[9 - 6*x + 4*x^2]/324

Rule 210

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

Rule 632

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 642

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]

Rule 648

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 1600

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 2083

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{81-54 x+24 x^3-16 x^4} \, dx \\ & = \int \left (-\frac {1}{54 (-3+2 x)}+\frac {1}{162 (3+2 x)}+\frac {3+2 x}{81 \left (9-6 x+4 x^2\right )}\right ) \, dx \\ & = -\frac {1}{108} \log (3-2 x)+\frac {1}{324} \log (3+2 x)+\frac {1}{81} \int \frac {3+2 x}{9-6 x+4 x^2} \, dx \\ & = -\frac {1}{108} \log (3-2 x)+\frac {1}{324} \log (3+2 x)+\frac {1}{324} \int \frac {-6+8 x}{9-6 x+4 x^2} \, dx+\frac {1}{18} \int \frac {1}{9-6 x+4 x^2} \, dx \\ & = -\frac {1}{108} \log (3-2 x)+\frac {1}{324} \log (3+2 x)+\frac {1}{324} \log \left (9-6 x+4 x^2\right )-\frac {1}{9} \text {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 )}{54 \sqrt {3}}-\frac {1}{108} \log (3-2 x)+\frac {1}{324} \log (3+2 x)+\frac {1}{324} \log \left (9-6 x+4 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87 \[ \int \frac {9+6 x+4 x^2}{729-64 x^6} \, dx=\frac {1}{324} \left (2 \sqrt {3} \arctan \left (\frac {-3+4 x}{3 \sqrt {3}}\right )-3 \log (3-2 x)+\log (3+2 x)+\log \left (9-6 x+4 x^2\right )\right ) \]

[In]

Integrate[(9 + 6*x + 4*x^2)/(729 - 64*x^6),x]

[Out]

(2*Sqrt[3]*ArcTan[(-3 + 4*x)/(3*Sqrt[3])] - 3*Log[3 - 2*x] + Log[3 + 2*x] + Log[9 - 6*x + 4*x^2])/324

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {\ln \left (-3+2 x \right )}{108}+\frac {\ln \left (4 x^{2}-6 x +9\right )}{324}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (8 x -6\right ) \sqrt {3}}{18}\right )}{162}+\frac {\ln \left (2 x +3\right )}{324}\)	\(47\)
risch	\(\frac {\ln \left (16 x^{2}-24 x +36\right )}{324}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (4 x -3\right ) \sqrt {3}}{9}\right )}{162}-\frac {\ln \left (-3+2 x \right )}{108}+\frac {\ln \left (2 x +3\right )}{324}\)	\(47\)
meijerg	\(-\frac {x \left (\ln \left (1-\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}\right )-\ln \left (1+\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}\right )+\frac {\ln \left (1-\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}+\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{3-\left (x^{6}\right )^{\frac {1}{6}}}\right )-\frac {\ln \left (1+\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}+\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{3+\left (x^{6}\right )^{\frac {1}{6}}}\right )\right )}{324 \left (x^{6}\right )^{\frac {1}{6}}}+\frac {\operatorname {arctanh}\left (\frac {8 x^{3}}{27}\right )}{162}-\frac {x^{2} \left (\ln \left (1-\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )-\frac {\ln \left (1+\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}+\frac {16 \left (x^{6}\right )^{\frac {2}{3}}}{81}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (x^{6}\right )^{\frac {1}{3}}}{9 \left (1+\frac {2 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )}\right )\right )}{324 \left (x^{6}\right )^{\frac {1}{3}}}\)	\(200\)

[In]

int((4*x^2+6*x+9)/(-64*x^6+729),x,method=_RETURNVERBOSE)

[Out]

-1/108*ln(-3+2*x)+1/324*ln(4*x^2-6*x+9)+1/162*3^(1/2)*arctan(1/18*(8*x-6)*3^(1/2))+1/324*ln(2*x+3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77 \[ \int \frac {9+6 x+4 x^2}{729-64 x^6} \, dx=\frac {1}{162} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x - 3\right )}\right ) + \frac {1}{324} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) + \frac {1}{324} \, \log \left (2 \, x + 3\right ) - \frac {1}{108} \, \log \left (2 \, x - 3\right ) \]

[In]

integrate((4*x^2+6*x+9)/(-64*x^6+729),x, algorithm="fricas")

[Out]

1/162*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3)) + 1/324*log(4*x^2 - 6*x + 9) + 1/324*log(2*x + 3) - 1/108*log(2*x
- 3)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \frac {9+6 x+4 x^2}{729-64 x^6} \, dx=- \frac {\log {\left (x - \frac {3}{2} \right )}}{108} + \frac {\log {\left (x + \frac {3}{2} \right )}}{324} + \frac {\log {\left (x^{2} - \frac {3 x}{2} + \frac {9}{4} \right )}}{324} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {4 \sqrt {3} x}{9} - \frac {\sqrt {3}}{3} \right )}}{162} \]

[In]

integrate((4*x**2+6*x+9)/(-64*x**6+729),x)

[Out]

-log(x - 3/2)/108 + log(x + 3/2)/324 + log(x**2 - 3*x/2 + 9/4)/324 + sqrt(3)*atan(4*sqrt(3)*x/9 - sqrt(3)/3)/1
62

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77 \[ \int \frac {9+6 x+4 x^2}{729-64 x^6} \, dx=\frac {1}{162} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x - 3\right )}\right ) + \frac {1}{324} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) + \frac {1}{324} \, \log \left (2 \, x + 3\right ) - \frac {1}{108} \, \log \left (2 \, x - 3\right ) \]

[In]

integrate((4*x^2+6*x+9)/(-64*x^6+729),x, algorithm="maxima")

[Out]

1/162*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3)) + 1/324*log(4*x^2 - 6*x + 9) + 1/324*log(2*x + 3) - 1/108*log(2*x
- 3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.80 \[ \int \frac {9+6 x+4 x^2}{729-64 x^6} \, dx=\frac {1}{162} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x - 3\right )}\right ) + \frac {1}{324} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) + \frac {1}{324} \, \log \left ({\left | 2 \, x + 3 \right |}\right ) - \frac {1}{108} \, \log \left ({\left | 2 \, x - 3 \right |}\right ) \]

[In]

integrate((4*x^2+6*x+9)/(-64*x^6+729),x, algorithm="giac")

[Out]

1/162*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3)) + 1/324*log(4*x^2 - 6*x + 9) + 1/324*log(abs(2*x + 3)) - 1/108*log
(abs(2*x - 3))

Mupad [B] (verification not implemented)

Time = 9.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87 \[ \int \frac {9+6 x+4 x^2}{729-64 x^6} \, dx=\frac {\ln \left (x+\frac {3}{2}\right )}{324}-\frac {\ln \left (x-\frac {3}{2}\right )}{108}-\ln \left (x-\frac {3}{4}-\frac {\sqrt {3}\,3{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{324}+\frac {\sqrt {3}\,1{}\mathrm {i}}{324}\right )+\ln \left (x-\frac {3}{4}+\frac {\sqrt {3}\,3{}\mathrm {i}}{4}\right )\,\left (\frac {1}{324}+\frac {\sqrt {3}\,1{}\mathrm {i}}{324}\right ) \]

[In]

int(-(6*x + 4*x^2 + 9)/(64*x^6 - 729),x)

[Out]

log(x + 3/2)/324 - log(x - 3/2)/108 - log(x - (3^(1/2)*3i)/4 - 3/4)*((3^(1/2)*1i)/324 - 1/324) + log(x + (3^(1
/2)*3i)/4 - 3/4)*((3^(1/2)*1i)/324 + 1/324)