3.13 \(\int \frac{1}{9+5 x^2+x^4} \, dx\)

Optimal. Leaf size=67 \[ -\frac{1}{12} \log \left (x^2-x+3\right )+\frac{1}{12} \log \left (x^2+x+3\right )-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{11}}\right )}{6 \sqrt{11}}+\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{11}}\right )}{6 \sqrt{11}} \]

[Out]

-ArcTan[(1 - 2*x)/Sqrt[11]]/(6*Sqrt[11]) + ArcTan[(1 + 2*x)/Sqrt[11]]/(6*Sqrt[11]) - Log[3 - x + x^2]/12 + Log
[3 + x + x^2]/12

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Rubi [A]  time = 0.049631, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {1094, 634, 618, 204, 628} \[ -\frac{1}{12} \log \left (x^2-x+3\right )+\frac{1}{12} \log \left (x^2+x+3\right )-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{11}}\right )}{6 \sqrt{11}}+\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{11}}\right )}{6 \sqrt{11}} \]

Antiderivative was successfully verified.

[In]

Int[(9 + 5*x^2 + x^4)^(-1),x]

[Out]

-ArcTan[(1 - 2*x)/Sqrt[11]]/(6*Sqrt[11]) + ArcTan[(1 + 2*x)/Sqrt[11]]/(6*Sqrt[11]) - Log[3 - x + x^2]/12 + Log
[3 + x + x^2]/12

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{1}{9+5 x^2+x^4} \, dx &=\frac{1}{6} \int \frac{1-x}{3-x+x^2} \, dx+\frac{1}{6} \int \frac{1+x}{3+x+x^2} \, dx\\ &=\frac{1}{12} \int \frac{1}{3-x+x^2} \, dx-\frac{1}{12} \int \frac{-1+2 x}{3-x+x^2} \, dx+\frac{1}{12} \int \frac{1}{3+x+x^2} \, dx+\frac{1}{12} \int \frac{1+2 x}{3+x+x^2} \, dx\\ &=-\frac{1}{12} \log \left (3-x+x^2\right )+\frac{1}{12} \log \left (3+x+x^2\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{-11-x^2} \, dx,x,-1+2 x\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{-11-x^2} \, dx,x,1+2 x\right )\\ &=-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{11}}\right )}{6 \sqrt{11}}+\frac{\tan ^{-1}\left (\frac{1+2 x}{\sqrt{11}}\right )}{6 \sqrt{11}}-\frac{1}{12} \log \left (3-x+x^2\right )+\frac{1}{12} \log \left (3+x+x^2\right )\\ \end{align*}

Mathematica [C]  time = 0.0714477, size = 91, normalized size = 1.36 \[ \frac{i \tan ^{-1}\left (\frac{x}{\sqrt{\frac{1}{2} \left (5+i \sqrt{11}\right )}}\right )}{\sqrt{\frac{11}{2} \left (5+i \sqrt{11}\right )}}-\frac{i \tan ^{-1}\left (\frac{x}{\sqrt{\frac{1}{2} \left (5-i \sqrt{11}\right )}}\right )}{\sqrt{\frac{11}{2} \left (5-i \sqrt{11}\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(9 + 5*x^2 + x^4)^(-1),x]

[Out]

((-I)*ArcTan[x/Sqrt[(5 - I*Sqrt[11])/2]])/Sqrt[(11*(5 - I*Sqrt[11]))/2] + (I*ArcTan[x/Sqrt[(5 + I*Sqrt[11])/2]
])/Sqrt[(11*(5 + I*Sqrt[11]))/2]

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Maple [A]  time = 0.046, size = 54, normalized size = 0.8 \begin{align*}{\frac{\ln \left ({x}^{2}+x+3 \right ) }{12}}+{\frac{\sqrt{11}}{66}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{11}}{11}} \right ) }-{\frac{\ln \left ({x}^{2}-x+3 \right ) }{12}}+{\frac{\sqrt{11}}{66}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{11}}{11}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4+5*x^2+9),x)

[Out]

1/12*ln(x^2+x+3)+1/66*arctan(1/11*(1+2*x)*11^(1/2))*11^(1/2)-1/12*ln(x^2-x+3)+1/66*11^(1/2)*arctan(1/11*(2*x-1
)*11^(1/2))

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Maxima [A]  time = 1.46939, size = 72, normalized size = 1.07 \begin{align*} \frac{1}{66} \, \sqrt{11} \arctan \left (\frac{1}{11} \, \sqrt{11}{\left (2 \, x + 1\right )}\right ) + \frac{1}{66} \, \sqrt{11} \arctan \left (\frac{1}{11} \, \sqrt{11}{\left (2 \, x - 1\right )}\right ) + \frac{1}{12} \, \log \left (x^{2} + x + 3\right ) - \frac{1}{12} \, \log \left (x^{2} - x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4+5*x^2+9),x, algorithm="maxima")

[Out]

1/66*sqrt(11)*arctan(1/11*sqrt(11)*(2*x + 1)) + 1/66*sqrt(11)*arctan(1/11*sqrt(11)*(2*x - 1)) + 1/12*log(x^2 +
x + 3) - 1/12*log(x^2 - x + 3)

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Fricas [A]  time = 1.2528, size = 193, normalized size = 2.88 \begin{align*} \frac{1}{66} \, \sqrt{11} \arctan \left (\frac{1}{11} \, \sqrt{11}{\left (2 \, x + 1\right )}\right ) + \frac{1}{66} \, \sqrt{11} \arctan \left (\frac{1}{11} \, \sqrt{11}{\left (2 \, x - 1\right )}\right ) + \frac{1}{12} \, \log \left (x^{2} + x + 3\right ) - \frac{1}{12} \, \log \left (x^{2} - x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4+5*x^2+9),x, algorithm="fricas")

[Out]

1/66*sqrt(11)*arctan(1/11*sqrt(11)*(2*x + 1)) + 1/66*sqrt(11)*arctan(1/11*sqrt(11)*(2*x - 1)) + 1/12*log(x^2 +
x + 3) - 1/12*log(x^2 - x + 3)

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Sympy [A]  time = 0.233039, size = 70, normalized size = 1.04 \begin{align*} - \frac{\log{\left (x^{2} - x + 3 \right )}}{12} + \frac{\log{\left (x^{2} + x + 3 \right )}}{12} + \frac{\sqrt{11} \operatorname{atan}{\left (\frac{2 \sqrt{11} x}{11} - \frac{\sqrt{11}}{11} \right )}}{66} + \frac{\sqrt{11} \operatorname{atan}{\left (\frac{2 \sqrt{11} x}{11} + \frac{\sqrt{11}}{11} \right )}}{66} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**4+5*x**2+9),x)

[Out]

-log(x**2 - x + 3)/12 + log(x**2 + x + 3)/12 + sqrt(11)*atan(2*sqrt(11)*x/11 - sqrt(11)/11)/66 + sqrt(11)*atan
(2*sqrt(11)*x/11 + sqrt(11)/11)/66

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Giac [A]  time = 1.13146, size = 72, normalized size = 1.07 \begin{align*} \frac{1}{66} \, \sqrt{11} \arctan \left (\frac{1}{11} \, \sqrt{11}{\left (2 \, x + 1\right )}\right ) + \frac{1}{66} \, \sqrt{11} \arctan \left (\frac{1}{11} \, \sqrt{11}{\left (2 \, x - 1\right )}\right ) + \frac{1}{12} \, \log \left (x^{2} + x + 3\right ) - \frac{1}{12} \, \log \left (x^{2} - x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4+5*x^2+9),x, algorithm="giac")

[Out]

1/66*sqrt(11)*arctan(1/11*sqrt(11)*(2*x + 1)) + 1/66*sqrt(11)*arctan(1/11*sqrt(11)*(2*x - 1)) + 1/12*log(x^2 +
x + 3) - 1/12*log(x^2 - x + 3)