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

Optimal. Leaf size=35 \[ \frac {\tan ^{-1}\left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1162, 617, 204} \begin {gather*} \frac {\tan ^{-1}\left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^2)/(1 + x^4),x]

[Out]

-(ArcTan[1 - Sqrt[2]*x]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*x]/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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[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, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 30, normalized size = 0.86 \begin {gather*} \frac {\tan ^{-1}\left (\sqrt {2} x+1\right )-\tan ^{-1}\left (1-\sqrt {2} x\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^2)/(1 + x^4),x]

[Out]

(-ArcTan[1 - Sqrt[2]*x] + ArcTan[1 + 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+x^2}{1+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.22, size = 29, normalized size = 0.83 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{3} + x\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 39, normalized size = 1.11 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.00, size = 88, normalized size = 2.51 \begin {gather*} \frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x -1\right )}{2}+\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x +1\right )}{2}+\frac {\sqrt {2}\, \ln \left (\frac {x^{2}-\sqrt {2}\, x +1}{x^{2}+\sqrt {2}\, x +1}\right )}{8}+\frac {\sqrt {2}\, \ln \left (\frac {x^{2}+\sqrt {2}\, x +1}{x^{2}-\sqrt {2}\, x +1}\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(-1+2^(1/2)*x)*2^(1/2)+1/8*2^(1/2)*ln((1+x^2+2^(1/2)*x)/(1+x^2-2^(1/2)*x))+1/2*arctan(1+2^(1/2)*x)*2
^(1/2)+1/8*2^(1/2)*ln((1+x^2-2^(1/2)*x)/(1+x^2+2^(1/2)*x))

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maxima [A]  time = 2.42, size = 39, normalized size = 1.11 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.37, size = 29, normalized size = 0.83 \begin {gather*} \frac {\sqrt {2}\,\left (\mathrm {atan}\left (\frac {\sqrt {2}\,x^3}{2}+\frac {\sqrt {2}\,x}{2}\right )+\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 1)/(x^4 + 1),x)

[Out]

(2^(1/2)*(atan((2^(1/2)*x)/2 + (2^(1/2)*x^3)/2) + atan((2^(1/2)*x)/2)))/2

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sympy [A]  time = 0.12, size = 39, normalized size = 1.11 \begin {gather*} \frac {\sqrt {2} \left (2 \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )} + 2 \operatorname {atan}{\left (\frac {\sqrt {2} x^{3}}{2} + \frac {\sqrt {2} x}{2} \right )}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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