\(\int \frac {81+54 x-24 x^3-16 x^4}{729-64 x^6} \, dx\) [558]

Optimal result

Integrand size = 25, antiderivative size = 24 \[ \int \frac {81+54 x-24 x^3-16 x^4}{729-64 x^6} \, dx=-\frac {\arctan \left (\frac {3-4 x}{3 \sqrt {3}}\right )}{3 \sqrt {3}} \]

[Out]

-1/9*arctan(1/9*(3-4*x)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1600, 632, 210} \[ \int \frac {81+54 x-24 x^3-16 x^4}{729-64 x^6} \, dx=-\frac {\arctan \left (\frac {3-4 x}{3 \sqrt {3}}\right )}{3 \sqrt {3}} \]

[In]

Int[(81 + 54*x - 24*x^3 - 16*x^4)/(729 - 64*x^6),x]

[Out]

-1/3*ArcTan[(3 - 4*x)/(3*Sqrt[3])]/Sqrt[3]

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 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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{9-6 x+4 x^2} \, dx \\ & = -\left (2 \text {Subst}\left (\int \frac {1}{-108-x^2} \, dx,x,-6+8 x\right )\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {3-4 x}{3 \sqrt {3}}\right )}{3 \sqrt {3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {81+54 x-24 x^3-16 x^4}{729-64 x^6} \, dx=\frac {\arctan \left (\frac {-3+4 x}{3 \sqrt {3}}\right )}{3 \sqrt {3}} \]

[In]

Integrate[(81 + 54*x - 24*x^3 - 16*x^4)/(729 - 64*x^6),x]

[Out]

ArcTan[(-3 + 4*x)/(3*Sqrt[3])]/(3*Sqrt[3])

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {\sqrt {3}\, \arctan \left (\frac {\left (8 x -6\right ) \sqrt {3}}{18}\right )}{9}\)	\(17\)
risch	\(\frac {\sqrt {3}\, \arctan \left (\frac {\left (4 x -3\right ) \sqrt {3}}{9}\right )}{9}\)	\(17\)
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 )}{36 \left (x^{6}\right )^{\frac {1}{6}}}+\frac {x^{5} \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 )}{36 \left (x^{6}\right )^{\frac {5}{6}}}+\frac {x^{4} \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 )}{36 \left (x^{6}\right )^{\frac {2}{3}}}-\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 )}{36 \left (x^{6}\right )^{\frac {1}{3}}}\)	\(381\)

[In]

int((-16*x^4-24*x^3+54*x+81)/(-64*x^6+729),x,method=_RETURNVERBOSE)

[Out]

1/9*3^(1/2)*arctan(1/18*(8*x-6)*3^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {81+54 x-24 x^3-16 x^4}{729-64 x^6} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x - 3\right )}\right ) \]

[In]

integrate((-16*x^4-24*x^3+54*x+81)/(-64*x^6+729),x, algorithm="fricas")

[Out]

1/9*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {81+54 x-24 x^3-16 x^4}{729-64 x^6} \, dx=\frac {\sqrt {3} \operatorname {atan}{\left (\frac {4 \sqrt {3} x}{9} - \frac {\sqrt {3}}{3} \right )}}{9} \]

[In]

integrate((-16*x**4-24*x**3+54*x+81)/(-64*x**6+729),x)

[Out]

sqrt(3)*atan(4*sqrt(3)*x/9 - sqrt(3)/3)/9

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {81+54 x-24 x^3-16 x^4}{729-64 x^6} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x - 3\right )}\right ) \]

[In]

integrate((-16*x^4-24*x^3+54*x+81)/(-64*x^6+729),x, algorithm="maxima")

[Out]

1/9*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {81+54 x-24 x^3-16 x^4}{729-64 x^6} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x - 3\right )}\right ) \]

[In]

integrate((-16*x^4-24*x^3+54*x+81)/(-64*x^6+729),x, algorithm="giac")

[Out]

1/9*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {81+54 x-24 x^3-16 x^4}{729-64 x^6} \, dx=\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,\left (4\,x-3\right )}{9}\right )}{9} \]

[In]

int(-(54*x - 24*x^3 - 16*x^4 + 81)/(64*x^6 - 729),x)

[Out]

(3^(1/2)*atan((3^(1/2)*(4*x - 3))/9))/9