3.1371 \(\int \frac{1}{x^2 (1+x^6)} \, dx\)

Optimal. Leaf size=85 \[ -\frac{\log \left (x^2-\sqrt{3} x+1\right )}{4 \sqrt{3}}+\frac{\log \left (x^2+\sqrt{3} x+1\right )}{4 \sqrt{3}}-\frac{1}{x}+\frac{1}{6} \tan ^{-1}\left (\sqrt{3}-2 x\right )-\frac{1}{3} \tan ^{-1}(x)-\frac{1}{6} \tan ^{-1}\left (2 x+\sqrt{3}\right ) \]

[Out]

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

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Rubi [A]  time = 0.249722, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {325, 295, 634, 618, 204, 628, 203} \[ -\frac{\log \left (x^2-\sqrt{3} x+1\right )}{4 \sqrt{3}}+\frac{\log \left (x^2+\sqrt{3} x+1\right )}{4 \sqrt{3}}-\frac{1}{x}+\frac{1}{6} \tan ^{-1}\left (\sqrt{3}-2 x\right )-\frac{1}{3} \tan ^{-1}(x)-\frac{1}{6} \tan ^{-1}\left (2 x+\sqrt{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(1 + x^6)),x]

[Out]

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 295

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/
b, n]], k, u}, Simp[u = Int[(r*Cos[((2*k - 1)*m*Pi)/n] - s*Cos[((2*k - 1)*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(
(2*k - 1)*Pi)/n]*x + s^2*x^2), x] + Int[(r*Cos[((2*k - 1)*m*Pi)/n] + s*Cos[((2*k - 1)*(m + 1)*Pi)/n]*x)/(r^2 +
2*r*s*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x]; (2*(-1)^(m/2)*r^(m + 2)*Int[1/(r^2 + s^2*x^2), x])/(a*n*s^m) +
Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] &&
IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]

Rule 634

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 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])

Rule 628

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 203

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

Rubi steps

\begin{align*} \int \frac{1}{x^2 \left (1+x^6\right )} \, dx &=-\frac{1}{x}-\int \frac{x^4}{1+x^6} \, dx\\ &=-\frac{1}{x}-\frac{1}{3} \int \frac{-\frac{1}{2}+\frac{\sqrt{3} x}{2}}{1-\sqrt{3} x+x^2} \, dx-\frac{1}{3} \int \frac{-\frac{1}{2}-\frac{\sqrt{3} x}{2}}{1+\sqrt{3} x+x^2} \, dx-\frac{1}{3} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{1}{x}-\frac{1}{3} \tan ^{-1}(x)-\frac{1}{12} \int \frac{1}{1-\sqrt{3} x+x^2} \, dx-\frac{1}{12} \int \frac{1}{1+\sqrt{3} x+x^2} \, dx-\frac{\int \frac{-\sqrt{3}+2 x}{1-\sqrt{3} x+x^2} \, dx}{4 \sqrt{3}}+\frac{\int \frac{\sqrt{3}+2 x}{1+\sqrt{3} x+x^2} \, dx}{4 \sqrt{3}}\\ &=-\frac{1}{x}-\frac{1}{3} \tan ^{-1}(x)-\frac{\log \left (1-\sqrt{3} x+x^2\right )}{4 \sqrt{3}}+\frac{\log \left (1+\sqrt{3} x+x^2\right )}{4 \sqrt{3}}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,-\sqrt{3}+2 x\right )+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,\sqrt{3}+2 x\right )\\ &=-\frac{1}{x}+\frac{1}{6} \tan ^{-1}\left (\sqrt{3}-2 x\right )-\frac{1}{3} \tan ^{-1}(x)-\frac{1}{6} \tan ^{-1}\left (\sqrt{3}+2 x\right )-\frac{\log \left (1-\sqrt{3} x+x^2\right )}{4 \sqrt{3}}+\frac{\log \left (1+\sqrt{3} x+x^2\right )}{4 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0207184, size = 82, normalized size = 0.96 \[ -\frac{\sqrt{3} x \log \left (x^2-\sqrt{3} x+1\right )-\sqrt{3} x \log \left (x^2+\sqrt{3} x+1\right )-2 x \tan ^{-1}\left (\sqrt{3}-2 x\right )+4 x \tan ^{-1}(x)+2 x \tan ^{-1}\left (2 x+\sqrt{3}\right )+12}{12 x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(1 + x^6)),x]

[Out]

-(12 - 2*x*ArcTan[Sqrt[3] - 2*x] + 4*x*ArcTan[x] + 2*x*ArcTan[Sqrt[3] + 2*x] + Sqrt[3]*x*Log[1 - Sqrt[3]*x + x
^2] - Sqrt[3]*x*Log[1 + Sqrt[3]*x + x^2])/(12*x)

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Maple [A]  time = 0.012, size = 66, normalized size = 0.8 \begin{align*} -{x}^{-1}-{\frac{\arctan \left ( x \right ) }{3}}-{\frac{\arctan \left ( 2\,x-\sqrt{3} \right ) }{6}}-{\frac{\arctan \left ( 2\,x+\sqrt{3} \right ) }{6}}-{\frac{\ln \left ( 1+{x}^{2}-x\sqrt{3} \right ) \sqrt{3}}{12}}+{\frac{\ln \left ( 1+{x}^{2}+x\sqrt{3} \right ) \sqrt{3}}{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(x^6+1),x)

[Out]

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

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Maxima [A]  time = 1.50672, size = 88, normalized size = 1.04 \begin{align*} \frac{1}{12} \, \sqrt{3} \log \left (x^{2} + \sqrt{3} x + 1\right ) - \frac{1}{12} \, \sqrt{3} \log \left (x^{2} - \sqrt{3} x + 1\right ) - \frac{1}{x} - \frac{1}{6} \, \arctan \left (2 \, x + \sqrt{3}\right ) - \frac{1}{6} \, \arctan \left (2 \, x - \sqrt{3}\right ) - \frac{1}{3} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(x^6+1),x, algorithm="maxima")

[Out]

1/12*sqrt(3)*log(x^2 + sqrt(3)*x + 1) - 1/12*sqrt(3)*log(x^2 - sqrt(3)*x + 1) - 1/x - 1/6*arctan(2*x + sqrt(3)
) - 1/6*arctan(2*x - sqrt(3)) - 1/3*arctan(x)

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Fricas [A]  time = 1.57757, size = 298, normalized size = 3.51 \begin{align*} \frac{\sqrt{3} x \log \left (x^{2} + \sqrt{3} x + 1\right ) - \sqrt{3} x \log \left (x^{2} - \sqrt{3} x + 1\right ) - 4 \, x \arctan \left (x\right ) + 4 \, x \arctan \left (-2 \, x + \sqrt{3} + 2 \, \sqrt{x^{2} - \sqrt{3} x + 1}\right ) + 4 \, x \arctan \left (-2 \, x - \sqrt{3} + 2 \, \sqrt{x^{2} + \sqrt{3} x + 1}\right ) - 12}{12 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(x^6+1),x, algorithm="fricas")

[Out]

1/12*(sqrt(3)*x*log(x^2 + sqrt(3)*x + 1) - sqrt(3)*x*log(x^2 - sqrt(3)*x + 1) - 4*x*arctan(x) + 4*x*arctan(-2*
x + sqrt(3) + 2*sqrt(x^2 - sqrt(3)*x + 1)) + 4*x*arctan(-2*x - sqrt(3) + 2*sqrt(x^2 + sqrt(3)*x + 1)) - 12)/x

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Sympy [A]  time = 0.213678, size = 71, normalized size = 0.84 \begin{align*} - \frac{\sqrt{3} \log{\left (x^{2} - \sqrt{3} x + 1 \right )}}{12} + \frac{\sqrt{3} \log{\left (x^{2} + \sqrt{3} x + 1 \right )}}{12} - \frac{\operatorname{atan}{\left (x \right )}}{3} - \frac{\operatorname{atan}{\left (2 x - \sqrt{3} \right )}}{6} - \frac{\operatorname{atan}{\left (2 x + \sqrt{3} \right )}}{6} - \frac{1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(x**6+1),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{6} + 1\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(x^6+1),x, algorithm="giac")

[Out]

integrate(1/((x^6 + 1)*x^2), x)