\(\int \frac {1}{x^3 (1+3 x^4+x^8)} \, dx\) [375]

Optimal result

Integrand size = 16, antiderivative size = 89 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}+\frac {1}{2} \sqrt {\frac {1}{5} \left (9-4 \sqrt {5}\right )} \arctan \left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )-\frac {\left (3+\sqrt {5}\right )^{3/2} \arctan \left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )}{4 \sqrt {10}} \]

[Out]

-1/2/x^2-1/40*arctan(x^2*(1/2+1/2*5^(1/2)))*(3+5^(1/2))^(3/2)*10^(1/2)+1/2*arctan(x^2*2^(1/2)/(3+5^(1/2))^(1/2
))*(1-2/5*5^(1/2))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1373, 1137, 1180, 209} \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=\frac {1}{2} \sqrt {\frac {1}{5} \left (9-4 \sqrt {5}\right )} \arctan \left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )-\frac {\left (3+\sqrt {5}\right )^{3/2} \arctan \left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )}{4 \sqrt {10}}-\frac {1}{2 x^2} \]

[In]

Int[1/(x^3*(1 + 3*x^4 + x^8)),x]

[Out]

-1/2*1/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 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 1137

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 1180

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 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*} \text {integral}& = \frac {1}{2} \text {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} \text {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 ) \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 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] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {1}{4} \text {RootSum}\left [1+3 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {3 \log (x-\text {$\#$1})+\log (x-\text {$\#$1}) \text {$\#$1}^4}{3 \text {$\#$1}^2+2 \text {$\#$1}^6}\&\right ] \]

[In]

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

[Out]

-1/2*1/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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.47

method	result	size
risch	\(-\frac {1}{2 x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+90 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (35 \textit {\_R}^{3}+8 x^{2}+123 \textit {\_R} \right )\right )}{4}\)	\(42\)
default	\(-\frac {1}{2 x^{2}}-\frac {\left (\sqrt {5}-3\right ) \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{2 \sqrt {5}+2}\right )}{5 \left (2 \sqrt {5}+2\right )}-\frac {\left (3+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{2 \sqrt {5}-2}\right )}{5 \left (2 \sqrt {5}-2\right )}\)	\(75\)

[In]

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

[Out]

-1/2/x^2+1/4*sum(_R*ln(35*_R^3+8*x^2+123*_R),_R=RootOf(25*_Z^4+90*_Z^2+1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (53) = 106\).

Time = 0.25 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.92 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=\frac {\sqrt {5} x^{2} \sqrt {4 \, \sqrt {5} - 9} \log \left (2 \, x^{2} + \sqrt {4 \, \sqrt {5} - 9} {\left (3 \, \sqrt {5} + 7\right )}\right ) - \sqrt {5} x^{2} \sqrt {4 \, \sqrt {5} - 9} \log \left (2 \, x^{2} - \sqrt {4 \, \sqrt {5} - 9} {\left (3 \, \sqrt {5} + 7\right )}\right ) + \sqrt {5} x^{2} \sqrt {-4 \, \sqrt {5} - 9} \log \left (2 \, x^{2} + {\left (3 \, \sqrt {5} - 7\right )} \sqrt {-4 \, \sqrt {5} - 9}\right ) - \sqrt {5} x^{2} \sqrt {-4 \, \sqrt {5} - 9} \log \left (2 \, x^{2} - {\left (3 \, \sqrt {5} - 7\right )} \sqrt {-4 \, \sqrt {5} - 9}\right ) - 10}{20 \, x^{2}} \]

[In]

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

[Out]

1/20*(sqrt(5)*x^2*sqrt(4*sqrt(5) - 9)*log(2*x^2 + sqrt(4*sqrt(5) - 9)*(3*sqrt(5) + 7)) - sqrt(5)*x^2*sqrt(4*sq
rt(5) - 9)*log(2*x^2 - sqrt(4*sqrt(5) - 9)*(3*sqrt(5) + 7)) + sqrt(5)*x^2*sqrt(-4*sqrt(5) - 9)*log(2*x^2 + (3*
sqrt(5) - 7)*sqrt(-4*sqrt(5) - 9)) - sqrt(5)*x^2*sqrt(-4*sqrt(5) - 9)*log(2*x^2 - (3*sqrt(5) - 7)*sqrt(-4*sqrt
(5) - 9)) - 10)/x^2

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=- 2 \left (\frac {\sqrt {5}}{10} + \frac {1}{4}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{-1 + \sqrt {5}} \right )} + 2 \cdot \left (\frac {1}{4} - \frac {\sqrt {5}}{10}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{1 + \sqrt {5}} \right )} - \frac {1}{2 x^{2}} \]

[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)

Maxima [F]

\[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} + 3 \, x^{4} + 1\right )} x^{3}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=-\frac {1}{20} \, {\left (x^{4} {\left (\sqrt {5} - 5\right )} + 3 \, \sqrt {5} - 15\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} + 1}\right ) - \frac {1}{20} \, {\left (x^{4} {\left (\sqrt {5} + 5\right )} + 3 \, \sqrt {5} + 15\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} - 1}\right ) - \frac {1}{2 \, x^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.46 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=2\,\mathrm {atanh}\left (\frac {26880\,x^2\,\sqrt {-\frac {\sqrt {5}}{20}-\frac {9}{80}}}{3520\,\sqrt {5}+7872}+\frac {12032\,\sqrt {5}\,x^2\,\sqrt {-\frac {\sqrt {5}}{20}-\frac {9}{80}}}{3520\,\sqrt {5}+7872}\right )\,\sqrt {-\frac {\sqrt {5}}{20}-\frac {9}{80}}-2\,\mathrm {atanh}\left (\frac {26880\,x^2\,\sqrt {\frac {\sqrt {5}}{20}-\frac {9}{80}}}{3520\,\sqrt {5}-7872}-\frac {12032\,\sqrt {5}\,x^2\,\sqrt {\frac {\sqrt {5}}{20}-\frac {9}{80}}}{3520\,\sqrt {5}-7872}\right )\,\sqrt {\frac {\sqrt {5}}{20}-\frac {9}{80}}-\frac {1}{2\,x^2} \]

[In]

int(1/(x^3*(3*x^4 + x^8 + 1)),x)

[Out]

2*atanh((26880*x^2*(- 5^(1/2)/20 - 9/80)^(1/2))/(3520*5^(1/2) + 7872) + (12032*5^(1/2)*x^2*(- 5^(1/2)/20 - 9/8
0)^(1/2))/(3520*5^(1/2) + 7872))*(- 5^(1/2)/20 - 9/80)^(1/2) - 2*atanh((26880*x^2*(5^(1/2)/20 - 9/80)^(1/2))/(
3520*5^(1/2) - 7872) - (12032*5^(1/2)*x^2*(5^(1/2)/20 - 9/80)^(1/2))/(3520*5^(1/2) - 7872))*(5^(1/2)/20 - 9/80
)^(1/2) - 1/(2*x^2)