3.92 \(\int \frac {-1-3 x^2}{1+2 x^2+9 x^4} \, dx\)

Optimal. Leaf size=43 \[ \frac {\tan ^{-1}\left (\frac {1-3 x}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (\frac {3 x+1}{\sqrt {2}}\right )}{2 \sqrt {2}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1161, 618, 204} \[ \frac {\tan ^{-1}\left (\frac {1-3 x}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (\frac {3 x+1}{\sqrt {2}}\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[-((1 + 3*x^2)/(1 + 2*x^2 + 9*x^4)),x]

[Out]

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

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 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]))

Rubi steps

\begin {align*} \int -\frac {1+3 x^2}{1+2 x^2+9 x^4} \, dx &=-\left (\frac {1}{6} \int \frac {1}{\frac {1}{3}-\frac {2 x}{3}+x^2} \, dx\right )-\frac {1}{6} \int \frac {1}{\frac {1}{3}+\frac {2 x}{3}+x^2} \, dx\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-\frac {8}{9}-x^2} \, dx,x,-\frac {2}{3}+2 x\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-\frac {8}{9}-x^2} \, dx,x,\frac {2}{3}+2 x\right )\\ &=\frac {\tan ^{-1}\left (\frac {1-3 x}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (\frac {1+3 x}{\sqrt {2}}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 99, normalized size = 2.30 \[ -\frac {\left (\sqrt {2}-i\right ) \tan ^{-1}\left (\frac {3 x}{\sqrt {1-2 i \sqrt {2}}}\right )}{2 \sqrt {2 \left (1-2 i \sqrt {2}\right )}}-\frac {\left (\sqrt {2}+i\right ) \tan ^{-1}\left (\frac {3 x}{\sqrt {1+2 i \sqrt {2}}}\right )}{2 \sqrt {2 \left (1+2 i \sqrt {2}\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 3*x^2)/(1 + 2*x^2 + 9*x^4),x]

[Out]

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

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fricas [A]  time = 0.39, size = 33, normalized size = 0.77 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{4} \, \sqrt {2} {\left (9 \, x^{3} + 5 \, x\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {3}{4} \, \sqrt {2} x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2-1)/(9*x^4+2*x^2+1),x, algorithm="fricas")

[Out]

-1/4*sqrt(2)*arctan(1/4*sqrt(2)*(9*x^3 + 5*x)) - 1/4*sqrt(2)*arctan(3/4*sqrt(2)*x)

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giac [A]  time = 0.16, size = 33, normalized size = 0.77 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 1\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x - 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2-1)/(9*x^4+2*x^2+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 34, normalized size = 0.79 \[ -\frac {\sqrt {2}\, \arctan \left (\frac {\left (6 x -2\right ) \sqrt {2}}{4}\right )}{4}-\frac {\sqrt {2}\, \arctan \left (\frac {\left (6 x +2\right ) \sqrt {2}}{4}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x^2-1)/(9*x^4+2*x^2+1),x)

[Out]

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

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maxima [A]  time = 2.35, size = 33, normalized size = 0.77 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 1\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x - 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2-1)/(9*x^4+2*x^2+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.38, size = 29, normalized size = 0.67 \[ -\frac {\sqrt {2}\,\left (\mathrm {atan}\left (\frac {9\,\sqrt {2}\,x^3}{4}+\frac {5\,\sqrt {2}\,x}{4}\right )+\mathrm {atan}\left (\frac {3\,\sqrt {2}\,x}{4}\right )\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x^2 + 1)/(2*x^2 + 9*x^4 + 1),x)

[Out]

-(2^(1/2)*(atan((5*2^(1/2)*x)/4 + (9*2^(1/2)*x^3)/4) + atan((3*2^(1/2)*x)/4)))/4

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sympy [A]  time = 0.14, size = 46, normalized size = 1.07 \[ - \frac {\sqrt {2} \left (2 \operatorname {atan}{\left (\frac {3 \sqrt {2} x}{4} \right )} + 2 \operatorname {atan}{\left (\frac {9 \sqrt {2} x^{3}}{4} + \frac {5 \sqrt {2} x}{4} \right )}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x**2-1)/(9*x**4+2*x**2+1),x)

[Out]

-sqrt(2)*(2*atan(3*sqrt(2)*x/4) + 2*atan(9*sqrt(2)*x**3/4 + 5*sqrt(2)*x/4))/8

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