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

Optimal. Leaf size=75 \[ -\frac{1}{8} \log \left (x^4-x^2+1\right )+\frac{1}{8} \log \left (x^4+x^2+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{4 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{2 x^2+1}{\sqrt{3}}\right )}{4 \sqrt{3}} \]

[Out]

-ArcTan[(1 - 2*x^2)/Sqrt[3]]/(4*Sqrt[3]) + ArcTan[(1 + 2*x^2)/Sqrt[3]]/(4*Sqrt[3]) - Log[1 - x^2 + x^4]/8 + Lo
g[1 + x^2 + x^4]/8

________________________________________________________________________________________

Rubi [A]  time = 0.0636766, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {1359, 1094, 634, 618, 204, 628} \[ -\frac{1}{8} \log \left (x^4-x^2+1\right )+\frac{1}{8} \log \left (x^4+x^2+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{4 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{2 x^2+1}{\sqrt{3}}\right )}{4 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[x/(1 + x^4 + x^8),x]

[Out]

-ArcTan[(1 - 2*x^2)/Sqrt[3]]/(4*Sqrt[3]) + ArcTan[(1 + 2*x^2)/Sqrt[3]]/(4*Sqrt[3]) - Log[1 - x^2 + x^4]/8 + Lo
g[1 + x^2 + x^4]/8

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 1094

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

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]

Rubi steps

\begin{align*} \int \frac{x}{1+x^4+x^8} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+x^2+x^4} \, dx,x,x^2\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1-x}{1-x+x^2} \, dx,x,x^2\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1+x}{1+x+x^2} \, dx,x,x^2\right )\\ &=\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,x^2\right )-\frac{1}{8} \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,x^2\right )+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,x^2\right )+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1+2 x}{1+x+x^2} \, dx,x,x^2\right )\\ &=-\frac{1}{8} \log \left (1-x^2+x^4\right )+\frac{1}{8} \log \left (1+x^2+x^4\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x^2\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x^2\right )\\ &=-\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{4 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{1+2 x^2}{\sqrt{3}}\right )}{4 \sqrt{3}}-\frac{1}{8} \log \left (1-x^2+x^4\right )+\frac{1}{8} \log \left (1+x^2+x^4\right )\\ \end{align*}

Mathematica [C]  time = 0.0501971, size = 79, normalized size = 1.05 \[ \frac{i \left (\sqrt{1-i \sqrt{3}} \tan ^{-1}\left (\frac{1}{2} \left (\sqrt{3}-i\right ) x^2\right )-\sqrt{1+i \sqrt{3}} \tan ^{-1}\left (\frac{1}{2} \left (\sqrt{3}+i\right ) x^2\right )\right )}{2 \sqrt{6}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x/(1 + x^4 + x^8),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 62, normalized size = 0.8 \begin{align*}{\frac{\ln \left ({x}^{4}+{x}^{2}+1 \right ) }{8}}+{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 2\,{x}^{2}+1 \right ) \sqrt{3}}{3}} \right ) }-{\frac{\ln \left ({x}^{4}-{x}^{2}+1 \right ) }{8}}+{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 2\,{x}^{2}-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*ln(x^4+x^2+1)+1/12*arctan(1/3*(2*x^2+1)*3^(1/2))*3^(1/2)-1/8*ln(x^4-x^2+1)+1/12*3^(1/2)*arctan(1/3*(2*x^2-
1)*3^(1/2))

________________________________________________________________________________________

Maxima [A]  time = 1.50983, size = 82, normalized size = 1.09 \begin{align*} \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + 1\right )}\right ) + \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) + \frac{1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac{1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) + 1/8*log(x^4 + x^
2 + 1) - 1/8*log(x^4 - x^2 + 1)

________________________________________________________________________________________

Fricas [A]  time = 1.46677, size = 193, normalized size = 2.57 \begin{align*} \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + 1\right )}\right ) + \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) + \frac{1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac{1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) + 1/8*log(x^4 + x^
2 + 1) - 1/8*log(x^4 - x^2 + 1)

________________________________________________________________________________________

Sympy [A]  time = 0.190986, size = 76, normalized size = 1.01 \begin{align*} - \frac{\log{\left (x^{4} - x^{2} + 1 \right )}}{8} + \frac{\log{\left (x^{4} + x^{2} + 1 \right )}}{8} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x^{2}}{3} - \frac{\sqrt{3}}{3} \right )}}{12} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x^{2}}{3} + \frac{\sqrt{3}}{3} \right )}}{12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x**4 - x**2 + 1)/8 + log(x**4 + x**2 + 1)/8 + sqrt(3)*atan(2*sqrt(3)*x**2/3 - sqrt(3)/3)/12 + sqrt(3)*ata
n(2*sqrt(3)*x**2/3 + sqrt(3)/3)/12

________________________________________________________________________________________

Giac [A]  time = 1.10085, size = 82, normalized size = 1.09 \begin{align*} \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + 1\right )}\right ) + \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) + \frac{1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac{1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) + 1/8*log(x^4 + x^
2 + 1) - 1/8*log(x^4 - x^2 + 1)