3.371 \(\int \frac {x^5}{1+3 x^4+x^8} \, dx\)

Optimal. Leaf size=81 \[ \frac {1}{2} \sqrt {\frac {1}{10} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (3-\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1359, 1130, 203} \[ \frac {1}{2} \sqrt {\frac {1}{10} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (3-\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^5/(1 + 3*x^4 + x^8),x]

[Out]

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

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 1130

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

Rule 1359

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^5}{1+3 x^4+x^8} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{1+3 x^2+x^4} \, dx,x,x^2\right )\\ &=\frac {1}{20} \left (5-3 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {3}{2}-\frac {\sqrt {5}}{2}+x^2} \, dx,x,x^2\right )+\frac {1}{20} \left (5+3 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {3}{2}+\frac {\sqrt {5}}{2}+x^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \sqrt {\frac {1}{10} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\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 = 0.05, size = 75, normalized size = 0.93 \[ \frac {2 \sqrt {5} \tan ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )+\left (5-3 \sqrt {5}\right ) \tan ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )}{10 \sqrt {6-2 \sqrt {5}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(1 + 3*x^4 + x^8),x]

[Out]

(2*Sqrt[5]*ArcTan[Sqrt[2/(3 + Sqrt[5])]*x^2] + (5 - 3*Sqrt[5])*ArcTan[Sqrt[(3 + Sqrt[5])/2]*x^2])/(10*Sqrt[6 -
2*Sqrt[5]])

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fricas [B]  time = 1.06, size = 165, normalized size = 2.04 \[ -\frac {1}{10} \, \sqrt {10} \sqrt {\sqrt {5} + 3} \arctan \left (\frac {1}{40} \, \sqrt {10} \sqrt {2 \, x^{4} + \sqrt {5} + 3} {\left (3 \, \sqrt {5} \sqrt {2} - 5 \, \sqrt {2}\right )} \sqrt {\sqrt {5} + 3} - \frac {1}{20} \, \sqrt {10} {\left (3 \, \sqrt {5} x^{2} - 5 \, x^{2}\right )} \sqrt {\sqrt {5} + 3}\right ) + \frac {1}{10} \, \sqrt {10} \sqrt {-\sqrt {5} + 3} \arctan \left (\frac {1}{40} \, \sqrt {10} \sqrt {2 \, x^{4} - \sqrt {5} + 3} {\left (3 \, \sqrt {5} \sqrt {2} + 5 \, \sqrt {2}\right )} \sqrt {-\sqrt {5} + 3} - \frac {1}{20} \, \sqrt {10} {\left (3 \, \sqrt {5} x^{2} + 5 \, x^{2}\right )} \sqrt {-\sqrt {5} + 3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^8+3*x^4+1),x, algorithm="fricas")

[Out]

-1/10*sqrt(10)*sqrt(sqrt(5) + 3)*arctan(1/40*sqrt(10)*sqrt(2*x^4 + sqrt(5) + 3)*(3*sqrt(5)*sqrt(2) - 5*sqrt(2)
)*sqrt(sqrt(5) + 3) - 1/20*sqrt(10)*(3*sqrt(5)*x^2 - 5*x^2)*sqrt(sqrt(5) + 3)) + 1/10*sqrt(10)*sqrt(-sqrt(5) +
3)*arctan(1/40*sqrt(10)*sqrt(2*x^4 - sqrt(5) + 3)*(3*sqrt(5)*sqrt(2) + 5*sqrt(2))*sqrt(-sqrt(5) + 3) - 1/20*s
qrt(10)*(3*sqrt(5)*x^2 + 5*x^2)*sqrt(-sqrt(5) + 3))

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giac [A]  time = 0.55, size = 47, normalized size = 0.58 \[ \frac {1}{20} \, x^{4} {\left (\sqrt {5} - 5\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} + 1}\right ) + \frac {1}{20} \, x^{4} {\left (\sqrt {5} + 5\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^8+3*x^4+1),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.02, size = 110, normalized size = 1.36 \[ \frac {\arctan \left (\frac {4 x^{2}}{-2+2 \sqrt {5}}\right )}{-2+2 \sqrt {5}}-\frac {3 \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{-2+2 \sqrt {5}}\right )}{5 \left (-2+2 \sqrt {5}\right )}+\frac {\arctan \left (\frac {4 x^{2}}{2+2 \sqrt {5}}\right )}{2+2 \sqrt {5}}+\frac {3 \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{2+2 \sqrt {5}}\right )}{5 \left (2+2 \sqrt {5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(x^8+3*x^4+1),x)

[Out]

1/(2+2*5^(1/2))*arctan(4/(2+2*5^(1/2))*x^2)+3/5*5^(1/2)/(2+2*5^(1/2))*arctan(4/(2+2*5^(1/2))*x^2)+1/(-2+2*5^(1
/2))*arctan(4/(-2+2*5^(1/2))*x^2)-3/5*5^(1/2)/(-2+2*5^(1/2))*arctan(4/(-2+2*5^(1/2))*x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{x^{8} + 3 \, x^{4} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^8+3*x^4+1),x, algorithm="maxima")

[Out]

integrate(x^5/(x^8 + 3*x^4 + 1), x)

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mupad [B]  time = 0.12, size = 117, normalized size = 1.44 \[ 2\,\mathrm {atanh}\left (\frac {60\,x^2\,\sqrt {\frac {\sqrt {5}}{160}-\frac {3}{160}}}{\sqrt {5}+3}+\frac {28\,\sqrt {5}\,x^2\,\sqrt {\frac {\sqrt {5}}{160}-\frac {3}{160}}}{\sqrt {5}+3}\right )\,\sqrt {\frac {\sqrt {5}}{160}-\frac {3}{160}}-2\,\mathrm {atanh}\left (\frac {60\,x^2\,\sqrt {-\frac {\sqrt {5}}{160}-\frac {3}{160}}}{\sqrt {5}-3}-\frac {28\,\sqrt {5}\,x^2\,\sqrt {-\frac {\sqrt {5}}{160}-\frac {3}{160}}}{\sqrt {5}-3}\right )\,\sqrt {-\frac {\sqrt {5}}{160}-\frac {3}{160}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(3*x^4 + x^8 + 1),x)

[Out]

2*atanh((60*x^2*(5^(1/2)/160 - 3/160)^(1/2))/(5^(1/2) + 3) + (28*5^(1/2)*x^2*(5^(1/2)/160 - 3/160)^(1/2))/(5^(
1/2) + 3))*(5^(1/2)/160 - 3/160)^(1/2) - 2*atanh((60*x^2*(- 5^(1/2)/160 - 3/160)^(1/2))/(5^(1/2) - 3) - (28*5^
(1/2)*x^2*(- 5^(1/2)/160 - 3/160)^(1/2))/(5^(1/2) - 3))*(- 5^(1/2)/160 - 3/160)^(1/2)

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sympy [A]  time = 0.20, size = 49, normalized size = 0.60 \[ - 2 \left (\frac {1}{8} - \frac {\sqrt {5}}{40}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{-1 + \sqrt {5}} \right )} + 2 \left (\frac {\sqrt {5}}{40} + \frac {1}{8}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{1 + \sqrt {5}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(x**8+3*x**4+1),x)

[Out]

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

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