3.4.73 \(\int \frac {x}{1+3 x^4+x^8} \, dx\) [373]

Optimal. Leaf size=75 \[ -\frac {\tan ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{\sqrt {10 \left (3+\sqrt {5}\right )}}+\frac {1}{2} \sqrt {\frac {1}{10} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1373, 1107, 209} \begin {gather*} \frac {1}{2} \sqrt {\frac {1}{10} \left (3+\sqrt {5}\right )} \text {ArcTan}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )-\frac {\text {ArcTan}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{\sqrt {10 \left (3+\sqrt {5}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(1 + 3*x^4 + x^8),x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 1107

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

Rule 1373

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[
1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k) + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b,
c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x}{1+3 x^4+x^8} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+3 x^2+x^4} \, dx,x,x^2\right )\\ &=\frac {\text {Subst}\left (\int \frac {1}{\frac {3}{2}-\frac {\sqrt {5}}{2}+x^2} \, dx,x,x^2\right )}{2 \sqrt {5}}-\frac {\text {Subst}\left (\int \frac {1}{\frac {3}{2}+\frac {\sqrt {5}}{2}+x^2} \, dx,x,x^2\right )}{2 \sqrt {5}}\\ &=-\frac {\tan ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{\sqrt {10 \left (3+\sqrt {5}\right )}}+\frac {1}{2} \sqrt {\frac {1}{10} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 74, normalized size = 0.99 \begin {gather*} \frac {\tan ^{-1}\left (\sqrt {\frac {2}{3-\sqrt {5}}} x^2\right )}{\sqrt {10 \left (3-\sqrt {5}\right )}}-\frac {\tan ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{\sqrt {10 \left (3+\sqrt {5}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(1 + 3*x^4 + x^8),x]

[Out]

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

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Maple [A]
time = 0.03, size = 60, normalized size = 0.80

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\RootOf \left (25 \textit {\_Z}^{4}+15 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-15 \textit {\_R}^{3}+x^{2}-7 \textit {\_R} \right )\right )}{4}\)	\(34\)
default	\(-\frac {2 \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{2 \sqrt {5}+2}\right )}{5 \left (2 \sqrt {5}+2\right )}+\frac {2 \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{2 \sqrt {5}-2}\right )}{5 \left (2 \sqrt {5}-2\right )}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^8+3*x^4+1),x,method=_RETURNVERBOSE)

[Out]

-2/5*5^(1/2)/(2*5^(1/2)+2)*arctan(4*x^2/(2*5^(1/2)+2))+2/5*5^(1/2)/(2*5^(1/2)-2)*arctan(4*x^2/(2*5^(1/2)-2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^8+3*x^4+1),x, algorithm="maxima")

[Out]

integrate(x/(x^8 + 3*x^4 + 1), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (43) = 86\).
time = 0.37, size = 128, normalized size = 1.71 \begin {gather*} \frac {1}{10} \, \sqrt {10} \sqrt {-\sqrt {5} + 3} \arctan \left (-\frac {1}{10} \, \sqrt {10} \sqrt {5} x^{2} \sqrt {-\sqrt {5} + 3} + \frac {1}{20} \, \sqrt {10} \sqrt {5} \sqrt {2} \sqrt {2 \, x^{4} + \sqrt {5} + 3} \sqrt {-\sqrt {5} + 3}\right ) - \frac {1}{10} \, \sqrt {10} \sqrt {\sqrt {5} + 3} \arctan \left (-\frac {1}{20} \, {\left (2 \, \sqrt {10} \sqrt {5} x^{2} - \sqrt {10} \sqrt {5} \sqrt {2} \sqrt {2 \, x^{4} - \sqrt {5} + 3}\right )} \sqrt {\sqrt {5} + 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^8+3*x^4+1),x, algorithm="fricas")

[Out]

1/10*sqrt(10)*sqrt(-sqrt(5) + 3)*arctan(-1/10*sqrt(10)*sqrt(5)*x^2*sqrt(-sqrt(5) + 3) + 1/20*sqrt(10)*sqrt(5)*
sqrt(2)*sqrt(2*x^4 + sqrt(5) + 3)*sqrt(-sqrt(5) + 3)) - 1/10*sqrt(10)*sqrt(sqrt(5) + 3)*arctan(-1/20*(2*sqrt(1
0)*sqrt(5)*x^2 - sqrt(10)*sqrt(5)*sqrt(2)*sqrt(2*x^4 - sqrt(5) + 3))*sqrt(sqrt(5) + 3))

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Sympy [A]
time = 0.11, size = 49, normalized size = 0.65 \begin {gather*} 2 \left (\frac {\sqrt {5}}{40} + \frac {1}{8}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{-1 + \sqrt {5}} \right )} - 2 \cdot \left (\frac {1}{8} - \frac {\sqrt {5}}{40}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{1 + \sqrt {5}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x**8+3*x**4+1),x)

[Out]

2*(sqrt(5)/40 + 1/8)*atan(2*x**2/(-1 + sqrt(5))) - 2*(1/8 - sqrt(5)/40)*atan(2*x**2/(1 + sqrt(5)))

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Giac [A]
time = 3.22, size = 41, normalized size = 0.55 \begin {gather*} \frac {1}{20} \, {\left (\sqrt {5} - 5\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} + 1}\right ) + \frac {1}{20} \, {\left (\sqrt {5} + 5\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^8+3*x^4+1),x, algorithm="giac")

[Out]

1/20*(sqrt(5) - 5)*arctan(2*x^2/(sqrt(5) + 1)) + 1/20*(sqrt(5) + 5)*arctan(2*x^2/(sqrt(5) - 1))

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Mupad [B]
time = 0.05, size = 125, normalized size = 1.67 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {160\,x^2\,\sqrt {\frac {\sqrt {5}}{160}-\frac {3}{160}}}{8\,\sqrt {5}-18}-\frac {72\,\sqrt {5}\,x^2\,\sqrt {\frac {\sqrt {5}}{160}-\frac {3}{160}}}{8\,\sqrt {5}-18}\right )\,\sqrt {\frac {\sqrt {5}}{160}-\frac {3}{160}}-2\,\mathrm {atanh}\left (\frac {160\,x^2\,\sqrt {-\frac {\sqrt {5}}{160}-\frac {3}{160}}}{8\,\sqrt {5}+18}+\frac {72\,\sqrt {5}\,x^2\,\sqrt {-\frac {\sqrt {5}}{160}-\frac {3}{160}}}{8\,\sqrt {5}+18}\right )\,\sqrt {-\frac {\sqrt {5}}{160}-\frac {3}{160}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(3*x^4 + x^8 + 1),x)

[Out]

2*atanh((160*x^2*(5^(1/2)/160 - 3/160)^(1/2))/(8*5^(1/2) - 18) - (72*5^(1/2)*x^2*(5^(1/2)/160 - 3/160)^(1/2))/
(8*5^(1/2) - 18))*(5^(1/2)/160 - 3/160)^(1/2) - 2*atanh((160*x^2*(- 5^(1/2)/160 - 3/160)^(1/2))/(8*5^(1/2) + 1
8) + (72*5^(1/2)*x^2*(- 5^(1/2)/160 - 3/160)^(1/2))/(8*5^(1/2) + 18))*(- 5^(1/2)/160 - 3/160)^(1/2)

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