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

Optimal. Leaf size=85 \[ -\frac{\log \left (x^2-\sqrt{2} x+1\right )}{4 \sqrt{2}}+\frac{\log \left (x^2+\sqrt{2} x+1\right )}{4 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} x\right )}{2 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} x+1\right )}{2 \sqrt{2}} \]

[Out]

-ArcTan[1 - Sqrt[2]*x]/(2*Sqrt[2]) + ArcTan[1 + Sqrt[2]*x]/(2*Sqrt[2]) - Log[1 - Sqrt[2]*x + x^2]/(4*Sqrt[2])
+ Log[1 + Sqrt[2]*x + x^2]/(4*Sqrt[2])

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Rubi [A]  time = 0.0394903, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used = {211, 1165, 628, 1162, 617, 204} \[ -\frac{\log \left (x^2-\sqrt{2} x+1\right )}{4 \sqrt{2}}+\frac{\log \left (x^2+\sqrt{2} x+1\right )}{4 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} x\right )}{2 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} x+1\right )}{2 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)^(-1),x]

[Out]

-ArcTan[1 - Sqrt[2]*x]/(2*Sqrt[2]) + ArcTan[1 + Sqrt[2]*x]/(2*Sqrt[2]) - Log[1 - Sqrt[2]*x + x^2]/(4*Sqrt[2])
+ Log[1 + Sqrt[2]*x + x^2]/(4*Sqrt[2])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 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]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0170554, size = 64, normalized size = 0.75 \[ \frac{-\log \left (x^2-\sqrt{2} x+1\right )+\log \left (x^2+\sqrt{2} x+1\right )-2 \tan ^{-1}\left (1-\sqrt{2} x\right )+2 \tan ^{-1}\left (\sqrt{2} x+1\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^4)^(-1),x]

[Out]

(-2*ArcTan[1 - Sqrt[2]*x] + 2*ArcTan[1 + Sqrt[2]*x] - Log[1 - Sqrt[2]*x + x^2] + Log[1 + Sqrt[2]*x + x^2])/(4*
Sqrt[2])

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Maple [A]  time = 0.003, size = 58, normalized size = 0.7 \begin{align*}{\frac{\arctan \left ( 1+x\sqrt{2} \right ) \sqrt{2}}{4}}+{\frac{\arctan \left ( -1+x\sqrt{2} \right ) \sqrt{2}}{4}}+{\frac{\sqrt{2}}{8}\ln \left ({\frac{1+{x}^{2}+x\sqrt{2}}{1+{x}^{2}-x\sqrt{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*arctan(1+x*2^(1/2))*2^(1/2)+1/4*arctan(-1+x*2^(1/2))*2^(1/2)+1/8*2^(1/2)*ln((1+x^2+x*2^(1/2))/(1+x^2-x*2^(
1/2)))

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Maxima [A]  time = 1.42245, size = 97, normalized size = 1.14 \begin{align*} \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x + \sqrt{2}\right )}\right ) + \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x - \sqrt{2}\right )}\right ) + \frac{1}{8} \, \sqrt{2} \log \left (x^{2} + \sqrt{2} x + 1\right ) - \frac{1}{8} \, \sqrt{2} \log \left (x^{2} - \sqrt{2} x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) + 1/8*sqrt(2
)*log(x^2 + sqrt(2)*x + 1) - 1/8*sqrt(2)*log(x^2 - sqrt(2)*x + 1)

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Fricas [A]  time = 1.20729, size = 302, normalized size = 3.55 \begin{align*} -\frac{1}{2} \, \sqrt{2} \arctan \left (-\sqrt{2} x + \sqrt{2} \sqrt{x^{2} + \sqrt{2} x + 1} - 1\right ) - \frac{1}{2} \, \sqrt{2} \arctan \left (-\sqrt{2} x + \sqrt{2} \sqrt{x^{2} - \sqrt{2} x + 1} + 1\right ) + \frac{1}{8} \, \sqrt{2} \log \left (x^{2} + \sqrt{2} x + 1\right ) - \frac{1}{8} \, \sqrt{2} \log \left (x^{2} - \sqrt{2} x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.13663, size = 73, normalized size = 0.86 \begin{align*} - \frac{\sqrt{2} \log{\left (x^{2} - \sqrt{2} x + 1 \right )}}{8} + \frac{\sqrt{2} \log{\left (x^{2} + \sqrt{2} x + 1 \right )}}{8} + \frac{\sqrt{2} \operatorname{atan}{\left (\sqrt{2} x - 1 \right )}}{4} + \frac{\sqrt{2} \operatorname{atan}{\left (\sqrt{2} x + 1 \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*log(x**2 - sqrt(2)*x + 1)/8 + sqrt(2)*log(x**2 + sqrt(2)*x + 1)/8 + sqrt(2)*atan(sqrt(2)*x - 1)/4 + s
qrt(2)*atan(sqrt(2)*x + 1)/4

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Giac [A]  time = 1.08464, size = 97, normalized size = 1.14 \begin{align*} \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x + \sqrt{2}\right )}\right ) + \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x - \sqrt{2}\right )}\right ) + \frac{1}{8} \, \sqrt{2} \log \left (x^{2} + \sqrt{2} x + 1\right ) - \frac{1}{8} \, \sqrt{2} \log \left (x^{2} - \sqrt{2} x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) + 1/8*sqrt(2
)*log(x^2 + sqrt(2)*x + 1) - 1/8*sqrt(2)*log(x^2 - sqrt(2)*x + 1)