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

Optimal. Leaf size=15 \[ \frac{1}{3} \tan ^{-1}(x)+\frac{1}{3} \tan ^{-1}(2 x) \]

[Out]

ArcTan[x]/3 + ArcTan[2*x]/3

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Rubi [A]  time = 0.0093769, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1163, 203} \[ \frac{1}{3} \tan ^{-1}(x)+\frac{1}{3} \tan ^{-1}(2 x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^2)/(1 + 5*x^2 + 4*x^4),x]

[Out]

ArcTan[x]/3 + ArcTan[2*x]/3

Rule 1163

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), 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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1+2 x^2}{1+5 x^2+4 x^4} \, dx &=\frac{2}{3} \int \frac{1}{1+4 x^2} \, dx+\frac{4}{3} \int \frac{1}{4+4 x^2} \, dx\\ &=\frac{1}{3} \tan ^{-1}(x)+\frac{1}{3} \tan ^{-1}(2 x)\\ \end{align*}

Mathematica [A]  time = 0.0075067, size = 17, normalized size = 1.13 \[ -\frac{1}{3} \tan ^{-1}\left (\frac{3 x}{2 x^2-1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^2)/(1 + 5*x^2 + 4*x^4),x]

[Out]

-ArcTan[(3*x)/(-1 + 2*x^2)]/3

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Maple [A]  time = 0.05, size = 12, normalized size = 0.8 \begin{align*}{\frac{\arctan \left ( x \right ) }{3}}+{\frac{\arctan \left ( 2\,x \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*arctan(x)+1/3*arctan(2*x)

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Maxima [A]  time = 1.45741, size = 15, normalized size = 1. \begin{align*} \frac{1}{3} \, \arctan \left (2 \, x\right ) + \frac{1}{3} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*arctan(2*x) + 1/3*arctan(x)

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Fricas [A]  time = 1.27451, size = 66, normalized size = 4.4 \begin{align*} \frac{1}{3} \, \arctan \left (\frac{4}{3} \, x^{3} + \frac{7}{3} \, x\right ) + \frac{1}{3} \, \arctan \left (\frac{2}{3} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*arctan(4/3*x^3 + 7/3*x) + 1/3*arctan(2/3*x)

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Sympy [B]  time = 0.104221, size = 22, normalized size = 1.47 \begin{align*} \frac{\operatorname{atan}{\left (\frac{2 x}{3} \right )}}{3} + \frac{\operatorname{atan}{\left (\frac{4 x^{3}}{3} + \frac{7 x}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(2*x/3)/3 + atan(4*x**3/3 + 7*x/3)/3

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Giac [A]  time = 1.13615, size = 15, normalized size = 1. \begin{align*} \frac{1}{3} \, \arctan \left (2 \, x\right ) + \frac{1}{3} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*arctan(2*x) + 1/3*arctan(x)