\(\int \frac {3-2 x}{729-64 x^6} \, dx\) [559]

Optimal result

Integrand size = 15, antiderivative size = 50 \[ \int \frac {3-2 x}{729-64 x^6} \, dx=\frac {\arctan \left (\frac {3+4 x}{3 \sqrt {3}}\right )}{162 \sqrt {3}}+\frac {1}{486} \log (3+2 x)-\frac {1}{972} \log \left (9-6 x+4 x^2\right ) \]

[Out]

1/486*ln(3+2*x)-1/972*ln(4*x^2-6*x+9)+1/486*arctan(1/9*(3+4*x)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1600, 2083, 642, 632, 210} \[ \int \frac {3-2 x}{729-64 x^6} \, dx=\frac {\arctan \left (\frac {4 x+3}{3 \sqrt {3}}\right )}{162 \sqrt {3}}-\frac {1}{972} \log \left (4 x^2-6 x+9\right )+\frac {1}{486} \log (2 x+3) \]

[In]

Int[(3 - 2*x)/(729 - 64*x^6),x]

[Out]

ArcTan[(3 + 4*x)/(3*Sqrt[3])]/(162*Sqrt[3]) + Log[3 + 2*x]/486 - Log[9 - 6*x + 4*x^2]/972

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 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}{243+162 x+108 x^2+72 x^3+48 x^4+32 x^5} \, dx \\ & = \int \left (\frac {1}{243 (3+2 x)}+\frac {3-4 x}{486 \left (9-6 x+4 x^2\right )}+\frac {1}{54 \left (9+6 x+4 x^2\right )}\right ) \, dx \\ & = \frac {1}{486} \log (3+2 x)+\frac {1}{486} \int \frac {3-4 x}{9-6 x+4 x^2} \, dx+\frac {1}{54} \int \frac {1}{9+6 x+4 x^2} \, dx \\ & = \frac {1}{486} \log (3+2 x)-\frac {1}{972} \log \left (9-6 x+4 x^2\right )-\frac {1}{27} \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 )}{162 \sqrt {3}}+\frac {1}{486} \log (3+2 x)-\frac {1}{972} \log \left (9-6 x+4 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {3-2 x}{729-64 x^6} \, dx=\frac {\arctan \left (\frac {3+4 x}{3 \sqrt {3}}\right )}{162 \sqrt {3}}+\frac {1}{486} \log (3+2 x)-\frac {1}{972} \log \left (9-6 x+4 x^2\right ) \]

[In]

Integrate[(3 - 2*x)/(729 - 64*x^6),x]

[Out]

ArcTan[(3 + 4*x)/(3*Sqrt[3])]/(162*Sqrt[3]) + Log[3 + 2*x]/486 - Log[9 - 6*x + 4*x^2]/972

Maple [A] (verified)

Time = 1.55 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {\ln \left (4 x^{2}-6 x +9\right )}{972}+\frac {\ln \left (2 x +3\right )}{486}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (8 x +6\right ) \sqrt {3}}{18}\right )}{486}\)	\(39\)
risch	\(\frac {\ln \left (2 x +3\right )}{486}-\frac {\ln \left (4 x^{2}-6 x +9\right )}{972}+\frac {\arctan \left (\frac {\left (4 x +3\right ) \sqrt {3}}{9}\right ) \sqrt {3}}{486}\)	\(39\)
meijerg	\(-\frac {x \left (\ln \left (1-\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}\right )-\ln \left (1+\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}\right )+\frac {\ln \left (1-\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}+\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{3-\left (x^{6}\right )^{\frac {1}{6}}}\right )-\frac {\ln \left (1+\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}+\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{3+\left (x^{6}\right )^{\frac {1}{6}}}\right )\right )}{972 \left (x^{6}\right )^{\frac {1}{6}}}+\frac {x^{2} \left (\ln \left (1-\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )-\frac {\ln \left (1+\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}+\frac {16 \left (x^{6}\right )^{\frac {2}{3}}}{81}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (x^{6}\right )^{\frac {1}{3}}}{9 \left (1+\frac {2 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )}\right )\right )}{972 \left (x^{6}\right )^{\frac {1}{3}}}\)	\(192\)

[In]

int((3-2*x)/(-64*x^6+729),x,method=_RETURNVERBOSE)

[Out]

-1/972*ln(4*x^2-6*x+9)+1/486*ln(2*x+3)+1/486*3^(1/2)*arctan(1/18*(8*x+6)*3^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int \frac {3-2 x}{729-64 x^6} \, dx=\frac {1}{486} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x + 3\right )}\right ) - \frac {1}{972} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) + \frac {1}{486} \, \log \left (2 \, x + 3\right ) \]

[In]

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

[Out]

1/486*sqrt(3)*arctan(1/9*sqrt(3)*(4*x + 3)) - 1/972*log(4*x^2 - 6*x + 9) + 1/486*log(2*x + 3)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int \frac {3-2 x}{729-64 x^6} \, dx=\frac {\log {\left (x + \frac {3}{2} \right )}}{486} - \frac {\log {\left (4 x^{2} - 6 x + 9 \right )}}{972} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {4 \sqrt {3} x}{9} + \frac {\sqrt {3}}{3} \right )}}{486} \]

[In]

integrate((3-2*x)/(-64*x**6+729),x)

[Out]

log(x + 3/2)/486 - log(4*x**2 - 6*x + 9)/972 + sqrt(3)*atan(4*sqrt(3)*x/9 + sqrt(3)/3)/486

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int \frac {3-2 x}{729-64 x^6} \, dx=\frac {1}{486} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x + 3\right )}\right ) - \frac {1}{972} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) + \frac {1}{486} \, \log \left (2 \, x + 3\right ) \]

[In]

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

[Out]

1/486*sqrt(3)*arctan(1/9*sqrt(3)*(4*x + 3)) - 1/972*log(4*x^2 - 6*x + 9) + 1/486*log(2*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.78 \[ \int \frac {3-2 x}{729-64 x^6} \, dx=\frac {1}{486} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x + 3\right )}\right ) - \frac {1}{972} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) + \frac {1}{486} \, \log \left ({\left | 2 \, x + 3 \right |}\right ) \]

[In]

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

[Out]

1/486*sqrt(3)*arctan(1/9*sqrt(3)*(4*x + 3)) - 1/972*log(4*x^2 - 6*x + 9) + 1/486*log(abs(2*x + 3))

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98 \[ \int \frac {3-2 x}{729-64 x^6} \, dx=\frac {\ln \left (x+\frac {3}{2}\right )}{486}-\frac {\ln \left (x^2-\frac {3\,x}{2}+\frac {9}{4}\right )}{972}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}}{1327104\,\left (\frac {x}{884736}+\frac {1}{884736}\right )}-\frac {\sqrt {3}\,x}{7962624\,\left (\frac {x}{884736}+\frac {1}{884736}\right )}\right )}{486} \]

[In]

int((2*x - 3)/(64*x^6 - 729),x)

[Out]

log(x + 3/2)/486 - log(x^2 - (3*x)/2 + 9/4)/972 - (3^(1/2)*atan(3^(1/2)/(1327104*(x/884736 + 1/884736)) - (3^(
1/2)*x)/(7962624*(x/884736 + 1/884736))))/486