3.4.33 \(\int \frac {x}{1+x^4+x^8} \, dx\) [333]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1373, 1108, 648, 632, 210, 642} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {1-2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\text {ArcTan}\left (\frac {2 x^2+1}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{8} \log \left (x^4-x^2+1\right )+\frac {1}{8} \log \left (x^4+x^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(1 + x^4 + x^8),x]

[Out]

-1/4*ArcTan[(1 - 2*x^2)/Sqrt[3]]/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 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 642

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 648

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 1108

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 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*} \int \frac {x}{1+x^4+x^8} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2+x^4} \, dx,x,x^2\right )\\ &=\frac {1}{4} \text {Subst}\left (\int \frac {1-x}{1-x+x^2} \, dx,x,x^2\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1+x}{1+x+x^2} \, dx,x,x^2\right )\\ &=\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,x^2\right )-\frac {1}{8} \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^2\right )+\frac {1}{8} \text {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} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x^2\right )-\frac {1}{4} \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.03, size = 79, normalized size = 1.05 \begin {gather*} \frac {i \left (\sqrt {1-i \sqrt {3}} \tan ^{-1}\left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x^2\right )-\sqrt {1+i \sqrt {3}} \tan ^{-1}\left (\frac {1}{2} \left (i+\sqrt {3}\right ) x^2\right )\right )}{2 \sqrt {6}} \end {gather*}

Antiderivative was successfully verified.

[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]

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Maple [A]
time = 0.02, size = 62, normalized size = 0.83

method	result	size
default	\(\frac {\ln \left (x^{4}+x^{2}+1\right )}{8}+\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\ln \left (x^{4}-x^{2}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}\)	\(62\)
risch	\(-\frac {\ln \left (4 x^{4}-4 x^{2}+4\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (4 x^{4}+4 x^{2}+4\right )}{8}+\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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Maxima [A]
time = 0.48, size = 61, normalized size = 0.81 \begin {gather*} \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 {gather*}

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)

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Fricas [A]
time = 0.35, size = 61, normalized size = 0.81 \begin {gather*} \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 {gather*}

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)

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Sympy [A]
time = 0.09, size = 76, normalized size = 1.01 \begin {gather*} - \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 {gather*}

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

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Giac [A]
time = 3.32, size = 61, normalized size = 0.81 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 1.28, size = 51, normalized size = 0.68 \begin {gather*} \mathrm {atan}\left (\frac {\sqrt {3}\,x^2}{2}-\frac {x^2\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {3}}{12}+\frac {1}{4}{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {\sqrt {3}\,x^2}{2}+\frac {x^2\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {3}}{12}-\frac {1}{4}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^4 + x^8 + 1),x)

[Out]

atan((3^(1/2)*x^2)/2 - (x^2*1i)/2)*(3^(1/2)/12 + 1i/4) + atan((3^(1/2)*x^2)/2 + (x^2*1i)/2)*(3^(1/2)/12 - 1i/4
)

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