3.4.71 \(\int \frac {x^5}{1+3 x^4+x^8} \, dx\) [371]

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 = 0.06, 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 = {1373, 1144, 209} \begin {gather*} \frac {1}{2} \sqrt {\frac {1}{10} \left (3+\sqrt {5}\right )} \text {ArcTan}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )-\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 ) \end {gather*}

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 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 1144

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2/2)*(b/q + 1), Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2/2)*(b/q - 1), 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 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^5}{1+3 x^4+x^8} \, dx &=\frac {1}{2} \text {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 ) \text {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 ) \text {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 = 0.03, size = 75, normalized size = 0.93 \begin {gather*} \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}}} \end {gather*}

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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Maple [A]
time = 0.04, size = 70, normalized size = 0.86

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\RootOf \left (25 \textit {\_Z}^{4}+15 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-10 \textit {\_R}^{3}+x^{2}-3 \textit {\_R} \right )\right )}{4}\)	\(34\)
default	\(\frac {\left (3+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{2 \sqrt {5}+2}\right )}{10+10 \sqrt {5}}+\frac {\sqrt {5}\, \left (\sqrt {5}-3\right ) \arctan \left (\frac {4 x^{2}}{2 \sqrt {5}-2}\right )}{-10+10 \sqrt {5}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(3+5^(1/2))*5^(1/2)/(2*5^(1/2)+2)*arctan(4*x^2/(2*5^(1/2)+2))+1/5*5^(1/2)*(5^(1/2)-3)/(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^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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (41) = 82\).
time = 0.37, size = 149, normalized size = 1.84 \begin {gather*} -\frac {1}{10} \, \sqrt {10} \sqrt {\sqrt {5} + 3} \arctan \left (\frac {1}{40} \, \sqrt {10} \sqrt {2} \sqrt {2 \, x^{4} + \sqrt {5} + 3} {\left (3 \, \sqrt {5} - 5\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} \, {\left (\sqrt {10} \sqrt {2} \sqrt {2 \, x^{4} - \sqrt {5} + 3} {\left (3 \, \sqrt {5} + 5\right )} - 2 \, \sqrt {10} {\left (3 \, \sqrt {5} x^{2} + 5 \, x^{2}\right )}\right )} \sqrt {-\sqrt {5} + 3}\right ) \end {gather*}

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

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Sympy [A]
time = 0.09, size = 49, normalized size = 0.60 \begin {gather*} - 2 \cdot \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 )} \end {gather*}

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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Giac [A]
time = 2.89, size = 47, normalized size = 0.58 \begin {gather*} \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 ) \end {gather*}

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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Mupad [B]
time = 0.12, size = 117, normalized size = 1.44 \begin {gather*} 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}} \end {gather*}

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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