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

Optimal. Leaf size=88 \[ \frac {\log \left (x^2-\sqrt {3} x+1\right )}{4 \sqrt {3}}-\frac {\log \left (x^2+\sqrt {3} x+1\right )}{4 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

[Out]

-1/6*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/6*arctan(1/3*(1+2*x)*3^(1/2))*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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Rubi [A]  time = 0.06, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1372, 1164, 628, 1161, 618, 204} \[ \frac {\log \left (x^2-\sqrt {3} x+1\right )}{4 \sqrt {3}}-\frac {\log \left (x^2+\sqrt {3} x+1\right )}{4 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(1 + x^4 + x^8),x]

[Out]

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

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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {x^6}{1+x^4+x^8} \, dx &=-\left (\frac {1}{2} \int \frac {1-x^2}{1-x^2+x^4} \, dx\right )+\frac {1}{2} \int \frac {1+x^2}{1+x^2+x^4} \, dx\\ &=\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 {\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 {\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*}

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Mathematica [A]  time = 0.02, size = 68, normalized size = 0.77 \[ \frac {\log \left (-x^2+\sqrt {3} x-1\right )-\log \left (x^2+\sqrt {3} x+1\right )+2 \tan ^{-1}\left (\frac {2 x-1}{\sqrt {3}}\right )+2 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{4 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(1 + x^4 + x^8),x]

[Out]

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

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fricas [A]  time = 0.82, size = 70, normalized size = 0.80 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{3} + 2 \, x\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} x\right ) + \frac {1}{12} \, \sqrt {3} \log \left (\frac {x^{4} + 5 \, x^{2} - 2 \, \sqrt {3} {\left (x^{3} + x\right )} + 1}{x^{4} - x^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 66, normalized size = 0.75 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.01, size = 67, normalized size = 0.76 \[ \frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{12}-\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(x^8+x^4+1),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{2} \, \int \frac {x^{2} - 1}{x^{4} - x^{2} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)) + 1/2*integrate((x^2 - 1
)/(x^4 - x^2 + 1), x)

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mupad [B]  time = 1.31, size = 38, normalized size = 0.43 \[ -\frac {\sqrt {3}\,\left (\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}-\frac {2}{3}\right )}\right )+\mathrm {atanh}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}+\frac {2}{3}\right )}\right )\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(x^4 + x^8 + 1),x)

[Out]

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

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sympy [A]  time = 0.18, size = 82, normalized size = 0.93 \[ \frac {\sqrt {3} \left (2 \operatorname {atan}{\left (\frac {\sqrt {3} x}{3} \right )} + 2 \operatorname {atan}{\left (\frac {\sqrt {3} x^{3}}{3} + \frac {2 \sqrt {3} x}{3} \right )}\right )}{12} + \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} - \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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

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