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

Optimal. Leaf size=54 \[ -\frac{1}{2 x^2}+\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{2 \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2 x^2+1}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0515594, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {1359, 1123, 1161, 618, 204} \[ -\frac{1}{2 x^2}+\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{2 \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2 x^2+1}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 + x^4 + x^8)),x]

[Out]

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

Rule 1359

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]

Rule 1123

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 1161

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 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])

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (1+x^4+x^8\right )} \, dx &=\frac{1}{2} \operatorname{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} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,x^2\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,x^2\right )\\ &=-\frac{1}{2 x^2}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x^2\right )+\frac{1}{2} \operatorname{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]  time = 0.0507518, size = 100, normalized size = 1.85 \[ \frac{1}{12} \left (-\frac{6}{x^2}+i \sqrt{3} \log \left (2 x^2-i \sqrt{3}-1\right )-i \sqrt{3} \log \left (2 x^2+i \sqrt{3}-1\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[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[-1 - I*Sq
rt[3] + 2*x^2] - I*Sqrt[3]*Log[-1 + I*Sqrt[3] + 2*x^2])/12

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 57, normalized size = 1.1 \begin{align*}{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }-{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }-{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 2\,{x}^{2}-1 \right ) \sqrt{3}}{3}} \right ) }-{\frac{1}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.51486, size = 57, normalized size = 1.06 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]  time = 1.4664, size = 135, normalized size = 2.5 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0.161669, size = 53, normalized size = 0.98 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [A]  time = 1.11086, size = 57, normalized size = 1.06 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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