3.2.39 \(\int \frac {x (2+x^2)}{1+x^2+x^4} \, dx\) [139]

Optimal. Leaf size=37 \[ \frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )+\frac {1}{4} \log \left (1+x^2+x^4\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1261, 648, 632, 210, 642} \begin {gather*} \frac {1}{2} \sqrt {3} \text {ArcTan}\left (\frac {2 x^2+1}{\sqrt {3}}\right )+\frac {1}{4} \log \left (x^4+x^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(2 + x^2))/(1 + x^2 + x^4),x]

[Out]

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

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 1261

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 37, normalized size = 1.00 \begin {gather*} \frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )+\frac {1}{4} \log \left (1+x^2+x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(2 + x^2))/(1 + x^2 + x^4),x]

[Out]

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

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Maple [A]
time = 0.02, size = 31, normalized size = 0.84

method	result	size
default	\(\frac {\ln \left (x^{4}+x^{2}+1\right )}{4}+\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{2}\)	\(31\)
risch	\(\frac {\ln \left (4 x^{4}+4 x^{2}+4\right )}{4}+\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{2}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 30, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{4} \, \log \left (x^{4} + x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 30, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{4} \, \log \left (x^{4} + x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.04, size = 37, normalized size = 1.00 \begin {gather*} \frac {\log {\left (x^{4} + x^{2} + 1 \right )}}{4} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} + \frac {\sqrt {3}}{3} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.56, size = 30, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{4} \, \log \left (x^{4} + x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.21, size = 32, normalized size = 0.86 \begin {gather*} \frac {\ln \left (x^4+x^2+1\right )}{4}+\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x^2}{3}+\frac {\sqrt {3}}{3}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(x^2 + 2))/(x^2 + x^4 + 1),x)

[Out]

log(x^2 + x^4 + 1)/4 + (3^(1/2)*atan(3^(1/2)/3 + (2*3^(1/2)*x^2)/3))/2

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