\(\int \frac {1}{x^3 (1+x^4+x^8)} \, dx\) [335]

Optimal result

Integrand size = 14, antiderivative size = 54 \[ \int \frac {1}{x^3 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}+\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\arctan \left (\frac {1+2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

[Out]

-1/2/x^2+1/6*arctan(1/3*(-2*x^2+1)*3^(1/2))*3^(1/2)-1/6*arctan(1/3*(2*x^2+1)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1373, 1137, 1175, 632, 210} \[ \int \frac {1}{x^3 \left (1+x^4+x^8\right )} \, dx=\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{2 x^2} \]

[In]

Int[1/(x^3*(1 + x^4 + x^8)),x]

[Out]

-1/2*1/x^2 + ArcTan[(1 - 2*x^2)/Sqrt[3]]/(2*Sqrt[3]) - ArcTan[(1 + 2*x^2)/Sqrt[3]]/(2*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 1137

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^2 +
c*x^4)^(p + 1)/(a*d*(m + 1))), x] - Dist[1/(a*d^2*(m + 1)), Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p +
5)*x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && In
tegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1175

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e) - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !Lt
Q[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1373

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[
1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k) + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b,
c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.85 \[ \int \frac {1}{x^3 \left (1+x^4+x^8\right )} \, dx=\frac {1}{12} \left (-\frac {6}{x^2}-2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-i \sqrt {3} \log \left (i+\sqrt {3}-2 i x^2\right )+i \sqrt {3} \log \left (-i+\sqrt {3}+2 i x^2\right )\right ) \]

[In]

Integrate[1/(x^3*(1 + x^4 + x^8)),x]

[Out]

(-6/x^2 - 2*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] + 2*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - I*Sqrt[3]*Log[I + Sqrt[
3] - (2*I)*x^2] + I*Sqrt[3]*Log[-I + Sqrt[3] + (2*I)*x^2])/12

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81

method	result	size
risch	\(-\frac {1}{2 x^{2}}-\frac {\sqrt {3}\, \arctan \left (\frac {x^{6} \sqrt {3}}{3}+\frac {2 x^{2} \sqrt {3}}{3}\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {x^{2} \sqrt {3}}{3}\right )}{6}\)	\(44\)
default	\(-\frac {1}{2 x^{2}}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{6}\)	\(57\)

[In]

int(1/x^3/(x^8+x^4+1),x,method=_RETURNVERBOSE)

[Out]

-1/2/x^2-1/6*3^(1/2)*arctan(1/3*x^6*3^(1/2)+2/3*x^2*3^(1/2))-1/6*3^(1/2)*arctan(1/3*x^2*3^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \left (1+x^4+x^8\right )} \, dx=-\frac {\sqrt {3} x^{2} \arctan \left (\frac {1}{3} \, \sqrt {3} x^{2}\right ) + \sqrt {3} x^{2} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{6} + 2 \, x^{2}\right )}\right ) + 3}{6 \, x^{2}} \]

[In]

integrate(1/x^3/(x^8+x^4+1),x, algorithm="fricas")

[Out]

-1/6*(sqrt(3)*x^2*arctan(1/3*sqrt(3)*x^2) + sqrt(3)*x^2*arctan(1/3*sqrt(3)*(x^6 + 2*x^2)) + 3)/x^2

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^3 \left (1+x^4+x^8\right )} \, dx=\frac {\sqrt {3} \left (- 2 \operatorname {atan}{\left (\frac {\sqrt {3} x^{2}}{3} \right )} - 2 \operatorname {atan}{\left (\frac {\sqrt {3} x^{6}}{3} + \frac {2 \sqrt {3} x^{2}}{3} \right )}\right )}{12} - \frac {1}{2 x^{2}} \]

[In]

integrate(1/x**3/(x**8+x**4+1),x)

[Out]

sqrt(3)*(-2*atan(sqrt(3)*x**2/3) - 2*atan(sqrt(3)*x**6/3 + 2*sqrt(3)*x**2/3))/12 - 1/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{2 \, x^{2}} \]

[In]

integrate(1/x^3/(x^8+x^4+1),x, algorithm="maxima")

[Out]

-1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/2/x^2

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{2 \, x^{2}} \]

[In]

integrate(1/x^3/(x^8+x^4+1),x, algorithm="giac")

[Out]

-1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/2/x^2

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^3 \left (1+x^4+x^8\right )} \, dx=-\frac {\sqrt {3}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {3}\,x^6}{3}+\frac {2\,\sqrt {3}\,x^2}{3}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {3}\,x^2}{3}\right )\right )}{12}-\frac {1}{2\,x^2} \]

[In]

int(1/(x^3*(x^4 + x^8 + 1)),x)

[Out]

- (3^(1/2)*(2*atan((2*3^(1/2)*x^2)/3 + (3^(1/2)*x^6)/3) + 2*atan((3^(1/2)*x^2)/3)))/12 - 1/(2*x^2)