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

Optimal. Leaf size=89 \[ -\frac{1}{2 x^2}+\frac{1}{2} \sqrt{\frac{1}{5} \left (9-4 \sqrt{5}\right )} \tan ^{-1}\left (\sqrt{\frac{2}{3+\sqrt{5}}} x^2\right )-\frac{\left (3+\sqrt{5}\right )^{3/2} \tan ^{-1}\left (\sqrt{\frac{1}{2} \left (3+\sqrt{5}\right )} x^2\right )}{4 \sqrt{10}} \]

[Out]

-1/(2*x^2) + (Sqrt[(9 - 4*Sqrt[5])/5]*ArcTan[Sqrt[2/(3 + Sqrt[5])]*x^2])/2 - ((3 + Sqrt[5])^(3/2)*ArcTan[Sqrt[
(3 + Sqrt[5])/2]*x^2])/(4*Sqrt[10])

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Rubi [A]  time = 0.0724102, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1359, 1123, 1166, 203} \[ -\frac{1}{2 x^2}+\frac{1}{2} \sqrt{\frac{1}{5} \left (9-4 \sqrt{5}\right )} \tan ^{-1}\left (\sqrt{\frac{2}{3+\sqrt{5}}} x^2\right )-\frac{\left (3+\sqrt{5}\right )^{3/2} \tan ^{-1}\left (\sqrt{\frac{1}{2} \left (3+\sqrt{5}\right )} x^2\right )}{4 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 + 3*x^4 + x^8)),x]

[Out]

-1/(2*x^2) + (Sqrt[(9 - 4*Sqrt[5])/5]*ArcTan[Sqrt[2/(3 + Sqrt[5])]*x^2])/2 - ((3 + Sqrt[5])^(3/2)*ArcTan[Sqrt[
(3 + Sqrt[5])/2]*x^2])/(4*Sqrt[10])

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]

Rule 1123

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*x^2 +
c*x^4)^(p + 1))/(a*d*(m + 1)), x] - Dist[1/(a*d^2*(m + 1)), Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p +
5)*x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && In
tegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1166

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

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])

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (1+3 x^4+x^8\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+3 x^2+x^4\right )} \, dx,x,x^2\right )\\ &=-\frac{1}{2 x^2}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{-3-x^2}{1+3 x^2+x^4} \, dx,x,x^2\right )\\ &=-\frac{1}{2 x^2}+\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 x^2}+\frac{1}{10} \sqrt{45-20 \sqrt{5}} \tan ^{-1}\left (\sqrt{\frac{2}{3+\sqrt{5}}} x^2\right )-\frac{\left (3+\sqrt{5}\right )^{3/2} \tan ^{-1}\left (\sqrt{\frac{1}{2} \left (3+\sqrt{5}\right )} x^2\right )}{4 \sqrt{10}}\\ \end{align*}

Mathematica [C]  time = 0.0176194, size = 65, normalized size = 0.73 \[ -\frac{1}{4} \text{RootSum}\left [\text{$\#$1}^8+3 \text{$\#$1}^4+1\& ,\frac{\text{$\#$1}^4 \log (x-\text{$\#$1})+3 \log (x-\text{$\#$1})}{2 \text{$\#$1}^6+3 \text{$\#$1}^2}\& \right ]-\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(1 + 3*x^4 + x^8)),x]

[Out]

-1/(2*x^2) - RootSum[1 + 3*#1^4 + #1^8 & , (3*Log[x - #1] + Log[x - #1]*#1^4)/(3*#1^2 + 2*#1^6) & ]/4

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Maple [B]  time = 0.021, size = 117, normalized size = 1.3 \begin{align*} -{\frac{1}{-2+2\,\sqrt{5}}\arctan \left ( 4\,{\frac{{x}^{2}}{-2+2\,\sqrt{5}}} \right ) }-{\frac{3\,\sqrt{5}}{-10+10\,\sqrt{5}}\arctan \left ( 4\,{\frac{{x}^{2}}{-2+2\,\sqrt{5}}} \right ) }-{\frac{1}{2+2\,\sqrt{5}}\arctan \left ( 4\,{\frac{{x}^{2}}{2+2\,\sqrt{5}}} \right ) }+{\frac{3\,\sqrt{5}}{10+10\,\sqrt{5}}\arctan \left ( 4\,{\frac{{x}^{2}}{2+2\,\sqrt{5}}} \right ) }-{\frac{1}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2 \, x^{2}} - \int \frac{{\left (x^{4} + 3\right )} x}{x^{8} + 3 \, x^{4} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x^2 - integrate((x^4 + 3)*x/(x^8 + 3*x^4 + 1), x)

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Fricas [B]  time = 1.59149, size = 460, normalized size = 5.17 \begin{align*} -\frac{2 \, \sqrt{5} x^{2} \sqrt{-4 \, \sqrt{5} + 9} \arctan \left (\frac{1}{4} \, \sqrt{2 \, x^{4} + \sqrt{5} + 3}{\left (\sqrt{5} \sqrt{2} + 3 \, \sqrt{2}\right )} \sqrt{-4 \, \sqrt{5} + 9} - \frac{1}{2} \,{\left (\sqrt{5} x^{2} + 3 \, x^{2}\right )} \sqrt{-4 \, \sqrt{5} + 9}\right ) + 2 \, \sqrt{5} x^{2} \sqrt{4 \, \sqrt{5} + 9} \arctan \left (-\frac{1}{4} \,{\left (2 \, \sqrt{5} x^{2} - 6 \, x^{2} - \sqrt{2 \, x^{4} - \sqrt{5} + 3}{\left (\sqrt{5} \sqrt{2} - 3 \, \sqrt{2}\right )}\right )} \sqrt{4 \, \sqrt{5} + 9}\right ) + 5}{10 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(2*sqrt(5)*x^2*sqrt(-4*sqrt(5) + 9)*arctan(1/4*sqrt(2*x^4 + sqrt(5) + 3)*(sqrt(5)*sqrt(2) + 3*sqrt(2))*s
qrt(-4*sqrt(5) + 9) - 1/2*(sqrt(5)*x^2 + 3*x^2)*sqrt(-4*sqrt(5) + 9)) + 2*sqrt(5)*x^2*sqrt(4*sqrt(5) + 9)*arct
an(-1/4*(2*sqrt(5)*x^2 - 6*x^2 - sqrt(2*x^4 - sqrt(5) + 3)*(sqrt(5)*sqrt(2) - 3*sqrt(2)))*sqrt(4*sqrt(5) + 9))
+ 5)/x^2

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Sympy [A]  time = 0.230296, size = 56, normalized size = 0.63 \begin{align*} - 2 \left (\frac{\sqrt{5}}{10} + \frac{1}{4}\right ) \operatorname{atan}{\left (\frac{2 x^{2}}{-1 + \sqrt{5}} \right )} + 2 \left (\frac{1}{4} - \frac{\sqrt{5}}{10}\right ) \operatorname{atan}{\left (\frac{2 x^{2}}{1 + \sqrt{5}} \right )} - \frac{1}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(sqrt(5)/10 + 1/4)*atan(2*x**2/(-1 + sqrt(5))) + 2*(1/4 - sqrt(5)/10)*atan(2*x**2/(1 + sqrt(5))) - 1/(2*x**
2)

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Giac [A]  time = 1.25711, size = 92, normalized size = 1.03 \begin{align*} -\frac{1}{20} \,{\left (x^{4}{\left (\sqrt{5} - 5\right )} + 3 \, \sqrt{5} - 15\right )} \arctan \left (\frac{2 \, x^{2}}{\sqrt{5} + 1}\right ) - \frac{1}{20} \,{\left (x^{4}{\left (\sqrt{5} + 5\right )} + 3 \, \sqrt{5} + 15\right )} \arctan \left (\frac{2 \, x^{2}}{\sqrt{5} - 1}\right ) - \frac{1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(x^4*(sqrt(5) - 5) + 3*sqrt(5) - 15)*arctan(2*x^2/(sqrt(5) + 1)) - 1/20*(x^4*(sqrt(5) + 5) + 3*sqrt(5) +
15)*arctan(2*x^2/(sqrt(5) - 1)) - 1/2/x^2