3.1.29 \(\int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx\)

Optimal. Leaf size=45 \[ \frac {\tan ^{-1}\left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{\sqrt {10}}+\frac {\tan ^{-1}\left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{\sqrt {10}} \]

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Rubi [A]  time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {\tan ^{-1}\left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{\sqrt {10}}+\frac {\tan ^{-1}\left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{\sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^2)/(1 + 6*x^2 + 4*x^4),x]

[Out]

ArcTan[(2*x)/Sqrt[3 - Sqrt[5]]]/Sqrt[10] + ArcTan[(2*x)/Sqrt[3 + Sqrt[5]]]/Sqrt[10]

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])

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]

Rubi steps

\begin {align*} \int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx &=\frac {1}{5} \left (5-\sqrt {5}\right ) \int \frac {1}{3-\sqrt {5}+4 x^2} \, dx+\frac {1}{5} \left (5+\sqrt {5}\right ) \int \frac {1}{3+\sqrt {5}+4 x^2} \, dx\\ &=\frac {\tan ^{-1}\left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{\sqrt {10}}+\frac {\tan ^{-1}\left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{\sqrt {10}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 83, normalized size = 1.84 \begin {gather*} \frac {\left (\sqrt {5}-1\right ) \tan ^{-1}\left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{2 \sqrt {5 \left (3-\sqrt {5}\right )}}+\frac {\left (1+\sqrt {5}\right ) \tan ^{-1}\left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{2 \sqrt {5 \left (3+\sqrt {5}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^2)/(1 + 6*x^2 + 4*x^4),x]

[Out]

((-1 + Sqrt[5])*ArcTan[(2*x)/Sqrt[3 - Sqrt[5]]])/(2*Sqrt[5*(3 - Sqrt[5])]) + ((1 + Sqrt[5])*ArcTan[(2*x)/Sqrt[
3 + Sqrt[5]]])/(2*Sqrt[5*(3 + Sqrt[5])])

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(1 + 2*x^2)/(1 + 6*x^2 + 4*x^4),x]

[Out]

IntegrateAlgebraic[(1 + 2*x^2)/(1 + 6*x^2 + 4*x^4), x]

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fricas [A]  time = 0.80, size = 31, normalized size = 0.69 \begin {gather*} \frac {1}{10} \, \sqrt {10} \arctan \left (\frac {2}{5} \, \sqrt {10} {\left (x^{3} + 2 \, x\right )}\right ) + \frac {1}{10} \, \sqrt {10} \arctan \left (\frac {1}{5} \, \sqrt {10} x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+1)/(4*x^4+6*x^2+1),x, algorithm="fricas")

[Out]

1/10*sqrt(10)*arctan(2/5*sqrt(10)*(x^3 + 2*x)) + 1/10*sqrt(10)*arctan(1/5*sqrt(10)*x)

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giac [A]  time = 0.17, size = 39, normalized size = 0.87 \begin {gather*} \frac {1}{10} \, \sqrt {10} \arctan \left (\frac {4 \, x}{\sqrt {10} + \sqrt {2}}\right ) + \frac {1}{10} \, \sqrt {10} \arctan \left (\frac {4 \, x}{\sqrt {10} - \sqrt {2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+1)/(4*x^4+6*x^2+1),x, algorithm="giac")

[Out]

1/10*sqrt(10)*arctan(4*x/(sqrt(10) + sqrt(2))) + 1/10*sqrt(10)*arctan(4*x/(sqrt(10) - sqrt(2)))

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maple [B]  time = 0.05, size = 136, normalized size = 3.02 \begin {gather*} -\frac {2 \sqrt {5}\, \arctan \left (\frac {8 x}{2 \sqrt {10}-2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}-2 \sqrt {2}\right )}+\frac {2 \arctan \left (\frac {8 x}{2 \sqrt {10}-2 \sqrt {2}}\right )}{2 \sqrt {10}-2 \sqrt {2}}+\frac {2 \sqrt {5}\, \arctan \left (\frac {8 x}{2 \sqrt {10}+2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}+2 \sqrt {2}\right )}+\frac {2 \arctan \left (\frac {8 x}{2 \sqrt {10}+2 \sqrt {2}}\right )}{2 \sqrt {10}+2 \sqrt {2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+1)/(4*x^4+6*x^2+1),x)

[Out]

2/5*5^(1/2)/(2*10^(1/2)+2*2^(1/2))*arctan(8*x/(2*10^(1/2)+2*2^(1/2)))+2/(2*10^(1/2)+2*2^(1/2))*arctan(8*x/(2*1
0^(1/2)+2*2^(1/2)))-2/5*5^(1/2)/(2*10^(1/2)-2*2^(1/2))*arctan(8*x/(2*10^(1/2)-2*2^(1/2)))+2/(2*10^(1/2)-2*2^(1
/2))*arctan(8*x/(2*10^(1/2)-2*2^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{2} + 1}{4 \, x^{4} + 6 \, x^{2} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+1)/(4*x^4+6*x^2+1),x, algorithm="maxima")

[Out]

integrate((2*x^2 + 1)/(4*x^4 + 6*x^2 + 1), x)

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mupad [B]  time = 0.09, size = 29, normalized size = 0.64 \begin {gather*} \frac {\sqrt {10}\,\left (\mathrm {atan}\left (\frac {2\,\sqrt {10}\,x^3}{5}+\frac {4\,\sqrt {10}\,x}{5}\right )+\mathrm {atan}\left (\frac {\sqrt {10}\,x}{5}\right )\right )}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2 + 1)/(6*x^2 + 4*x^4 + 1),x)

[Out]

(10^(1/2)*(atan((4*10^(1/2)*x)/5 + (2*10^(1/2)*x^3)/5) + atan((10^(1/2)*x)/5)))/10

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sympy [A]  time = 0.15, size = 42, normalized size = 0.93 \begin {gather*} \frac {\sqrt {10} \left (2 \operatorname {atan}{\left (\frac {\sqrt {10} x}{5} \right )} + 2 \operatorname {atan}{\left (\frac {2 \sqrt {10} x^{3}}{5} + \frac {4 \sqrt {10} x}{5} \right )}\right )}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+1)/(4*x**4+6*x**2+1),x)

[Out]

sqrt(10)*(2*atan(sqrt(10)*x/5) + 2*atan(2*sqrt(10)*x**3/5 + 4*sqrt(10)*x/5))/20

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