3.113 \(\int \frac {1}{1+x^2+x^4} \, dx\)

Optimal. Leaf size=67 \[ -\frac {1}{4} \log \left (x^2-x+1\right )+\frac {1}{4} \log \left (x^2+x+1\right )-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1094, 634, 618, 204, 628} \[ -\frac {1}{4} \log \left (x^2-x+1\right )+\frac {1}{4} \log \left (x^2+x+1\right )-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^2 + x^4)^(-1),x]

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.06, size = 73, normalized size = 1.09 \[ \frac {i \left (\sqrt {1-i \sqrt {3}} \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}-i\right ) x\right )-\sqrt {1+i \sqrt {3}} \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}+i\right ) x\right )\right )}{\sqrt {6}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + x^2 + x^4)^(-1),x]

[Out]

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

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IntegrateAlgebraic [C]  time = 0.00, size = 58, normalized size = 0.87 \[ \sqrt {\frac {1}{6} \left (-1-i \sqrt {3}\right )} \tan ^{-1}\left (\sqrt [6]{-1} x\right )-\sqrt {\frac {1}{6} \left (-1+i \sqrt {3}\right )} \tan ^{-1}\left ((-1)^{5/6} x\right ) \]

Warning: Unable to verify antiderivative.

[In]

IntegrateAlgebraic[(1 + x^2 + x^4)^(-1),x]

[Out]

Sqrt[(-1 - I*Sqrt[3])/6]*ArcTan[(-1)^(1/6)*x] - Sqrt[(-1 + I*Sqrt[3])/6]*ArcTan[(-1)^(5/6)*x]

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fricas [A]  time = 0.89, size = 53, normalized size = 0.79 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4+x^2+1),x, algorithm="fricas")

[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)) + 1/4*log(x^2 + x + 1) -
1/4*log(x^2 - x + 1)

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giac [A]  time = 0.99, size = 53, normalized size = 0.79 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.05, size = 54, normalized size = 0.81

method	result	size
default	\(\frac {\ln \left (x^{2}+x +1\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\ln \left (x^{2}-x +1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{6}\)	\(54\)
risch	\(\frac {\ln \left (4 x^{2}+4 x +4\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\ln \left (4 x^{2}-4 x +4\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{6}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4+x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/4*ln(x^2+x+1)+1/6*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)-1/4*ln(x^2-x+1)+1/6*3^(1/2)*arctan(1/3*(-1+2*x)*3^(1/2
))

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maxima [A]  time = 0.99, size = 53, normalized size = 0.79 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 0.09, size = 47, normalized size = 0.70 \[ \mathrm {atanh}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\mathrm {atanh}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2 + x^4 + 1),x)

[Out]

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

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sympy [A]  time = 0.21, size = 70, normalized size = 1.04 \[ - \frac {\log {\left (x^{2} - x + 1 \right )}}{4} + \frac {\log {\left (x^{2} + x + 1 \right )}}{4} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x**2 - x + 1)/4 + log(x**2 + x + 1)/4 + sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/6 + sqrt(3)*atan(2*sqrt(3
)*x/3 + sqrt(3)/3)/6

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