3.345 \(\int \frac{1}{x^6 (1+x^4+x^8)} \, dx\)

Optimal. Leaf size=98 \[ -\frac{1}{5 x^5}+\frac{\log \left (x^2-\sqrt{3} x+1\right )}{4 \sqrt{3}}-\frac{\log \left (x^2+\sqrt{3} x+1\right )}{4 \sqrt{3}}+\frac{1}{x}-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

[Out]

-1/(5*x^5) + x^(-1) - ArcTan[(1 - 2*x)/Sqrt[3]]/(2*Sqrt[3]) + ArcTan[(1 + 2*x)/Sqrt[3]]/(2*Sqrt[3]) + 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 = 0.0838172, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {1368, 1504, 12, 1372, 1164, 628, 1161, 618, 204} \[ -\frac{1}{5 x^5}+\frac{\log \left (x^2-\sqrt{3} x+1\right )}{4 \sqrt{3}}-\frac{\log \left (x^2+\sqrt{3} x+1\right )}{4 \sqrt{3}}+\frac{1}{x}-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(1 + x^4 + x^8)),x]

[Out]

-1/(5*x^5) + x^(-1) - ArcTan[(1 - 2*x)/Sqrt[3]]/(2*Sqrt[3]) + ArcTan[(1 + 2*x)/Sqrt[3]]/(2*Sqrt[3]) + Log[1 -
Sqrt[3]*x + x^2]/(4*Sqrt[3]) - Log[1 + Sqrt[3]*x + x^2]/(4*Sqrt[3])

Rule 1368

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

Rule 1504

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :>
Simp[(d*(f*x)^(m + 1)*(a + b*x^n + c*x^(2*n))^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^n*(m + 1)), Int[(f*x)^
(m + n)*(a + b*x^n + c*x^(2*n))^p*Simp[a*e*(m + 1) - b*d*(m + n*(p + 1) + 1) - c*d*(m + 2*n*(p + 1) + 1)*x^n,
x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -
1] && IntegerQ[p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1372

Int[(x_)^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[
2*q - b/c, 2]}, -Dist[1/(2*c*r), Int[(x^(m - 3*(n/2))*(q - r*x^(n/2)))/(q - r*x^(n/2) + x^n), x], x] + Dist[1/
(2*c*r), Int[(x^(m - 3*(n/2))*(q + r*x^(n/2)))/(q + r*x^(n/2) + x^n), x], x]]] /; FreeQ[{a, b, c}, x] && EqQ[n
2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n/2, 0] && IGtQ[m, 0] && GeQ[m, (3*n)/2] && LtQ[m, 2*n] && NegQ[b^2 - 4
*a*c]

Rule 1164

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*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/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[b^2
- 4*a*c, 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 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^6 \left (1+x^4+x^8\right )} \, dx &=-\frac{1}{5 x^5}+\frac{1}{5} \int \frac{-5-5 x^4}{x^2 \left (1+x^4+x^8\right )} \, dx\\ &=-\frac{1}{5 x^5}+\frac{1}{x}-\frac{1}{5} \int -\frac{5 x^6}{1+x^4+x^8} \, dx\\ &=-\frac{1}{5 x^5}+\frac{1}{x}+\int \frac{x^6}{1+x^4+x^8} \, dx\\ &=-\frac{1}{5 x^5}+\frac{1}{x}-\frac{1}{2} \int \frac{1-x^2}{1-x^2+x^4} \, dx+\frac{1}{2} \int \frac{1+x^2}{1+x^2+x^4} \, dx\\ &=-\frac{1}{5 x^5}+\frac{1}{x}+\frac{1}{4} \int \frac{1}{1-x+x^2} \, dx+\frac{1}{4} \int \frac{1}{1+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}{5 x^5}+\frac{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}{2} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=-\frac{1}{5 x^5}+\frac{1}{x}-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}+\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 = 0.0381827, size = 95, normalized size = 0.97 \[ \frac{1}{60} \left (-\frac{12}{x^5}+5 \sqrt{3} \log \left (-x^2+\sqrt{3} x-1\right )-5 \sqrt{3} \log \left (x^2+\sqrt{3} x+1\right )+\frac{60}{x}+10 \sqrt{3} \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )+10 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(1 + x^4 + x^8)),x]

[Out]

(-12/x^5 + 60/x + 10*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] + 10*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] + 5*Sqrt[3]*Log
[-1 + Sqrt[3]*x - x^2] - 5*Sqrt[3]*Log[1 + Sqrt[3]*x + x^2])/60

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Maple [A]  time = 0.013, size = 75, normalized size = 0.8 \begin{align*}{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }-{\frac{1}{5\,{x}^{5}}}+{x}^{-1}+{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }-{\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^6/(x^8+x^4+1),x)

[Out]

1/6*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)-1/5/x^5+1/x+1/6*3^(1/2)*arctan(1/3*(2*x-1)*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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{5 \, x^{4} - 1}{5 \, x^{5}} + \frac{1}{2} \, \int \frac{x^{2} - 1}{x^{4} - x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(x^8+x^4+1),x, algorithm="maxima")

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/5*(5*x^4 - 1)/x^5 +
1/2*integrate((x^2 - 1)/(x^4 - x^2 + 1), x)

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Fricas [A]  time = 1.46082, size = 255, normalized size = 2.6 \begin{align*} \frac{10 \, \sqrt{3} x^{5} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (x^{3} + 2 \, x\right )}\right ) + 10 \, \sqrt{3} x^{5} \arctan \left (\frac{1}{3} \, \sqrt{3} x\right ) + 5 \, \sqrt{3} x^{5} \log \left (\frac{x^{4} + 5 \, x^{2} - 2 \, \sqrt{3}{\left (x^{3} + x\right )} + 1}{x^{4} - x^{2} + 1}\right ) + 60 \, x^{4} - 12}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(x^8+x^4+1),x, algorithm="fricas")

[Out]

1/60*(10*sqrt(3)*x^5*arctan(1/3*sqrt(3)*(x^3 + 2*x)) + 10*sqrt(3)*x^5*arctan(1/3*sqrt(3)*x) + 5*sqrt(3)*x^5*lo
g((x^4 + 5*x^2 - 2*sqrt(3)*(x^3 + x) + 1)/(x^4 - x^2 + 1)) + 60*x^4 - 12)/x^5

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Sympy [A]  time = 0.214127, size = 94, normalized size = 0.96 \begin{align*} \frac{\sqrt{3} \left (2 \operatorname{atan}{\left (\frac{\sqrt{3} x}{3} \right )} + 2 \operatorname{atan}{\left (\frac{\sqrt{3} x^{3}}{3} + \frac{2 \sqrt{3} x}{3} \right )}\right )}{12} + \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{5 x^{4} - 1}{5 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(x**8+x**4+1),x)

[Out]

sqrt(3)*(2*atan(sqrt(3)*x/3) + 2*atan(sqrt(3)*x**3/3 + 2*sqrt(3)*x/3))/12 + sqrt(3)*log(x**2 - sqrt(3)*x + 1)/
12 - sqrt(3)*log(x**2 + sqrt(3)*x + 1)/12 + (5*x**4 - 1)/(5*x**5)

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Giac [A]  time = 1.10526, size = 113, normalized size = 1.15 \begin{align*} \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{1}{12} \, \sqrt{3} \log \left (\frac{{\left | 2 \, x - 2 \, \sqrt{3} + \frac{2}{x} \right |}}{{\left | 2 \, x + 2 \, \sqrt{3} + \frac{2}{x} \right |}}\right ) + \frac{5 \, x^{4} - 1}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(x^8+x^4+1),x, algorithm="giac")

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/12*sqrt(3)*log(abs(2
*x - 2*sqrt(3) + 2/x)/abs(2*x + 2*sqrt(3) + 2/x)) + 1/5*(5*x^4 - 1)/x^5