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

Optimal. Leaf size=14 \[ \frac {\tan ^{-1}\left (\sqrt {2} x\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {28, 21, 203} \begin {gather*} \frac {\tan ^{-1}\left (\sqrt {2} x\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[Sqrt[2]*x]/Sqrt[2]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\sqrt {2} x\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[Sqrt[2]*x]/Sqrt[2]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+2 x^2}{1+4 x^2+4 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 11, normalized size = 0.79 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*arctan(sqrt(2)*x)

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giac [A]  time = 0.16, size = 11, normalized size = 0.79 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*arctan(sqrt(2)*x)

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maple [A]  time = 0.00, size = 12, normalized size = 0.86 \begin {gather*} \frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x \right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(2^(1/2)*x)*2^(1/2)

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maxima [A]  time = 2.30, size = 11, normalized size = 0.79 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*arctan(sqrt(2)*x)

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mupad [B]  time = 0.03, size = 11, normalized size = 0.79 \begin {gather*} \frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*atan(2^(1/2)*x))/2

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sympy [A]  time = 0.12, size = 14, normalized size = 1.00 \begin {gather*} \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*atan(sqrt(2)*x)/2

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