3.1.22 \(\int \frac {1-x^4}{1+2 x^4+x^8} \, dx\)

Optimal. Leaf size=97 \[ \frac {x}{2 \left (x^4+1\right )}-\frac {\log \left (x^2-\sqrt {2} x+1\right )}{8 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} x+1\right )}{4 \sqrt {2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {28, 385, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {x}{2 \left (x^4+1\right )}-\frac {\log \left (x^2-\sqrt {2} x+1\right )}{8 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} x+1\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x^4)/(1 + 2*x^4 + x^8),x]

[Out]

x/(2*(1 + x^4)) - ArcTan[1 - Sqrt[2]*x]/(4*Sqrt[2]) + ArcTan[1 + Sqrt[2]*x]/(4*Sqrt[2]) - Log[1 - Sqrt[2]*x +
x^2]/(8*Sqrt[2]) + Log[1 + Sqrt[2]*x + x^2]/(8*Sqrt[2])

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 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 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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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 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 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]

Rubi steps

\begin {align*} \int \frac {1-x^4}{1+2 x^4+x^8} \, dx &=\int \frac {1-x^4}{\left (1+x^4\right )^2} \, dx\\ &=\frac {x}{2 \left (1+x^4\right )}+\frac {1}{2} \int \frac {1}{1+x^4} \, dx\\ &=\frac {x}{2 \left (1+x^4\right )}+\frac {1}{4} \int \frac {1-x^2}{1+x^4} \, dx+\frac {1}{4} \int \frac {1+x^2}{1+x^4} \, dx\\ &=\frac {x}{2 \left (1+x^4\right )}+\frac {1}{8} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx+\frac {1}{8} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx-\frac {\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{8 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{8 \sqrt {2}}\\ &=\frac {x}{2 \left (1+x^4\right )}-\frac {\log \left (1-\sqrt {2} x+x^2\right )}{8 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{8 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{4 \sqrt {2}}\\ &=\frac {x}{2 \left (1+x^4\right )}-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\log \left (1-\sqrt {2} x+x^2\right )}{8 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{8 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 90, normalized size = 0.93 \begin {gather*} \frac {1}{16} \left (\frac {8 x}{x^4+1}-\sqrt {2} \log \left (x^2-\sqrt {2} x+1\right )+\sqrt {2} \log \left (x^2+\sqrt {2} x+1\right )-2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} x\right )+2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} x+1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x^4)/(1 + 2*x^4 + x^8),x]

[Out]

((8*x)/(1 + x^4) - 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] + 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*x] - Sqrt[2]*Log[1 - Sqrt[2]
*x + x^2] + Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/16

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-x^4}{1+2 x^4+x^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(1 - x^4)/(1 + 2*x^4 + x^8),x]

[Out]

IntegrateAlgebraic[(1 - x^4)/(1 + 2*x^4 + x^8), x]

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fricas [A]  time = 1.23, size = 126, normalized size = 1.30 \begin {gather*} -\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} + \sqrt {2} x + 1} - 1\right ) + 4 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} - \sqrt {2} x + 1} + 1\right ) - \sqrt {2} {\left (x^{4} + 1\right )} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \sqrt {2} {\left (x^{4} + 1\right )} \log \left (x^{2} - \sqrt {2} x + 1\right ) - 8 \, x}{16 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4+1)/(x^8+2*x^4+1),x, algorithm="fricas")

[Out]

-1/16*(4*sqrt(2)*(x^4 + 1)*arctan(-sqrt(2)*x + sqrt(2)*sqrt(x^2 + sqrt(2)*x + 1) - 1) + 4*sqrt(2)*(x^4 + 1)*ar
ctan(-sqrt(2)*x + sqrt(2)*sqrt(x^2 - sqrt(2)*x + 1) + 1) - sqrt(2)*(x^4 + 1)*log(x^2 + sqrt(2)*x + 1) + sqrt(2
)*(x^4 + 1)*log(x^2 - sqrt(2)*x + 1) - 8*x)/(x^4 + 1)

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giac [A]  time = 0.30, size = 82, normalized size = 0.85 \begin {gather*} \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {x}{2 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4+1)/(x^8+2*x^4+1),x, algorithm="giac")

[Out]

1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) + 1/16*sqrt(
2)*log(x^2 + sqrt(2)*x + 1) - 1/16*sqrt(2)*log(x^2 - sqrt(2)*x + 1) + 1/2*x/(x^4 + 1)

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maple [A]  time = 0.01, size = 68, normalized size = 0.70 \begin {gather*} \frac {x}{2 x^{4}+2}+\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x -1\right )}{8}+\frac {\sqrt {2}\, \arctan \left (\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 )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^4+1)/(x^8+2*x^4+1),x)

[Out]

1/2/(x^4+1)*x+1/16*2^(1/2)*ln((x^2+2^(1/2)*x+1)/(x^2-2^(1/2)*x+1))+1/8*2^(1/2)*arctan(2^(1/2)*x-1)+1/8*2^(1/2)
*arctan(2^(1/2)*x+1)

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maxima [A]  time = 1.56, size = 82, normalized size = 0.85 \begin {gather*} \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {x}{2 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4+1)/(x^8+2*x^4+1),x, algorithm="maxima")

[Out]

1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) + 1/16*sqrt(
2)*log(x^2 + sqrt(2)*x + 1) - 1/16*sqrt(2)*log(x^2 - sqrt(2)*x + 1) + 1/2*x/(x^4 + 1)

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mupad [B]  time = 1.62, size = 44, normalized size = 0.45 \begin {gather*} \frac {x}{2\,\left (x^4+1\right )}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^4 - 1)/(2*x^4 + x^8 + 1),x)

[Out]

2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(1/8 + 1i/8) + 2^(1/2)*atan(2^(1/2)*x*(1/2 + 1i/2))*(1/8 - 1i/8) + x/(2*(
x^4 + 1))

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sympy [A]  time = 0.18, size = 82, normalized size = 0.85 \begin {gather*} \frac {x}{2 x^{4} + 2} - \frac {\sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{16} + \frac {\sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{16} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{8} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**4+1)/(x**8+2*x**4+1),x)

[Out]

x/(2*x**4 + 2) - sqrt(2)*log(x**2 - sqrt(2)*x + 1)/16 + sqrt(2)*log(x**2 + sqrt(2)*x + 1)/16 + sqrt(2)*atan(sq
rt(2)*x - 1)/8 + sqrt(2)*atan(sqrt(2)*x + 1)/8

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