\(\int \frac {1}{x^8 (1+x^4+x^8)} \, dx\) [346]

Optimal result

Integrand size = 14, antiderivative size = 154 \[ \int \frac {1}{x^8 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{7 x^7}+\frac {1}{3 x^3}+\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \arctan \left (\sqrt {3}-2 x\right )-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{4} \arctan \left (\sqrt {3}+2 x\right )-\frac {1}{8} \log \left (1-x+x^2\right )+\frac {1}{8} \log \left (1+x+x^2\right )+\frac {\log \left (1-\sqrt {3} x+x^2\right )}{8 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x+x^2\right )}{8 \sqrt {3}} \]

[Out]

-1/7/x^7+1/3/x^3+1/4*arctan(2*x-3^(1/2))+1/4*arctan(2*x+3^(1/2))-1/8*ln(x^2-x+1)+1/8*ln(x^2+x+1)+1/12*arctan(1
/3*(1-2*x)*3^(1/2))*3^(1/2)-1/12*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)+1/24*ln(1+x^2-x*3^(1/2))*3^(1/2)-1/24*ln(
1+x^2+x*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {1382, 1518, 12, 1387, 1141, 1175, 632, 210, 1178, 642} \[ \int \frac {1}{x^8 \left (1+x^4+x^8\right )} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \arctan \left (\sqrt {3}-2 x\right )-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{4} \arctan \left (2 x+\sqrt {3}\right )-\frac {1}{7 x^7}+\frac {1}{3 x^3}-\frac {1}{8} \log \left (x^2-x+1\right )+\frac {1}{8} \log \left (x^2+x+1\right )+\frac {\log \left (x^2-\sqrt {3} x+1\right )}{8 \sqrt {3}}-\frac {\log \left (x^2+\sqrt {3} x+1\right )}{8 \sqrt {3}} \]

[In]

Int[1/(x^8*(1 + x^4 + x^8)),x]

[Out]

-1/7*1/x^7 + 1/(3*x^3) + ArcTan[(1 - 2*x)/Sqrt[3]]/(4*Sqrt[3]) - ArcTan[Sqrt[3] - 2*x]/4 - ArcTan[(1 + 2*x)/Sq
rt[3]]/(4*Sqrt[3]) + ArcTan[Sqrt[3] + 2*x]/4 - Log[1 - x + x^2]/8 + Log[1 + x + x^2]/8 + Log[1 - Sqrt[3]*x + x
^2]/(8*Sqrt[3]) - Log[1 + Sqrt[3]*x + x^2]/(8*Sqrt[3])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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

Rule 632

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

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1141

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

Rule 1175

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

Rule 1178

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

Rule 1382

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

Rule 1387

Int[(x_)^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[
2*q - b/c, 2]}, Dist[1/(2*c*r), Int[x^(m - n/2)/(q - r*x^(n/2) + x^n), x], x] - Dist[1/(2*c*r), Int[x^(m - n/2
)/(q + r*x^(n/2) + x^n), x], x]]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n/2, 0
] && IGtQ[m, 0] && GeQ[m, n/2] && LtQ[m, 3*(n/2)] && NegQ[b^2 - 4*a*c]

Rule 1518

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :>
Simp[d*(f*x)^(m + 1)*((a + b*x^n + c*x^(2*n))^(p + 1)/(a*f*(m + 1))), x] + Dist[1/(a*f^n*(m + 1)), Int[(f*x)^
(m + n)*(a + b*x^n + c*x^(2*n))^p*Simp[a*e*(m + 1) - b*d*(m + n*(p + 1) + 1) - c*d*(m + 2*n*(p + 1) + 1)*x^n,
x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -
1] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{7 x^7}+\frac {1}{7} \int \frac {-7-7 x^4}{x^4 \left (1+x^4+x^8\right )} \, dx \\ & = -\frac {1}{7 x^7}+\frac {1}{3 x^3}-\frac {1}{21} \int -\frac {21 x^4}{1+x^4+x^8} \, dx \\ & = -\frac {1}{7 x^7}+\frac {1}{3 x^3}+\int \frac {x^4}{1+x^4+x^8} \, dx \\ & = -\frac {1}{7 x^7}+\frac {1}{3 x^3}+\frac {1}{2} \int \frac {x^2}{1-x^2+x^4} \, dx-\frac {1}{2} \int \frac {x^2}{1+x^2+x^4} \, dx \\ & = -\frac {1}{7 x^7}+\frac {1}{3 x^3}-\frac {1}{4} \int \frac {1-x^2}{1-x^2+x^4} \, dx+\frac {1}{4} \int \frac {1+x^2}{1-x^2+x^4} \, dx+\frac {1}{4} \int \frac {1-x^2}{1+x^2+x^4} \, dx-\frac {1}{4} \int \frac {1+x^2}{1+x^2+x^4} \, dx \\ & = -\frac {1}{7 x^7}+\frac {1}{3 x^3}-\frac {1}{8} \int \frac {1+2 x}{-1-x-x^2} \, dx-\frac {1}{8} \int \frac {1-2 x}{-1+x-x^2} \, dx-\frac {1}{8} \int \frac {1}{1-x+x^2} \, dx-\frac {1}{8} \int \frac {1}{1+x+x^2} \, dx+\frac {1}{8} \int \frac {1}{1-\sqrt {3} x+x^2} \, dx+\frac {1}{8} \int \frac {1}{1+\sqrt {3} x+x^2} \, dx+\frac {\int \frac {\sqrt {3}+2 x}{-1-\sqrt {3} x-x^2} \, dx}{8 \sqrt {3}}+\frac {\int \frac {\sqrt {3}-2 x}{-1+\sqrt {3} x-x^2} \, dx}{8 \sqrt {3}} \\ & = -\frac {1}{7 x^7}+\frac {1}{3 x^3}-\frac {1}{8} \log \left (1-x+x^2\right )+\frac {1}{8} \log \left (1+x+x^2\right )+\frac {\log \left (1-\sqrt {3} x+x^2\right )}{8 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x+x^2\right )}{8 \sqrt {3}}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 x\right ) \\ & = -\frac {1}{7 x^7}+\frac {1}{3 x^3}+\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \tan ^{-1}\left (\sqrt {3}-2 x\right )-\frac {\tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{4} \tan ^{-1}\left (\sqrt {3}+2 x\right )-\frac {1}{8} \log \left (1-x+x^2\right )+\frac {1}{8} \log \left (1+x+x^2\right )+\frac {\log \left (1-\sqrt {3} x+x^2\right )}{8 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x+x^2\right )}{8 \sqrt {3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^8 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{7 x^7}+\frac {1}{3 x^3}+\frac {\left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )}{2 \sqrt {-6+6 i \sqrt {3}}}+\frac {\left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )}{2 \sqrt {-6-6 i \sqrt {3}}}-\frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{8} \log \left (1-x+x^2\right )+\frac {1}{8} \log \left (1+x+x^2\right ) \]

[In]

Integrate[1/(x^8*(1 + x^4 + x^8)),x]

[Out]

-1/7*1/x^7 + 1/(3*x^3) + ((I + Sqrt[3])*ArcTan[((1 - I*Sqrt[3])*x)/2])/(2*Sqrt[-6 + (6*I)*Sqrt[3]]) + ((-I + S
qrt[3])*ArcTan[((1 + I*Sqrt[3])*x)/2])/(2*Sqrt[-6 - (6*I)*Sqrt[3]]) - ArcTan[(-1 + 2*x)/Sqrt[3]]/(4*Sqrt[3]) -
ArcTan[(1 + 2*x)/Sqrt[3]]/(4*Sqrt[3]) - Log[1 - x + x^2]/8 + Log[1 + x + x^2]/8

Maple [C] (verified)

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

Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {\frac {x^{4}}{3}-\frac {1}{7}}{x^{7}}+\frac {\ln \left (4 x^{2}+4 x +4\right )}{8}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (6 \textit {\_R}^{3}+\textit {\_R} +x \right )\right )}{4}-\frac {\ln \left (x^{2}-x +1\right )}{8}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{12}\)	\(96\)
default	\(-\frac {1}{7 x^{7}}+\frac {1}{3 x^{3}}-\frac {\ln \left (x^{2}-x +1\right )}{8}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (x^{2}+x +1\right )}{8}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\sqrt {3}\, \left (-\frac {\ln \left (1+x^{2}-x \sqrt {3}\right )}{2}-\sqrt {3}\, \arctan \left (2 x -\sqrt {3}\right )\right )}{12}-\frac {\sqrt {3}\, \left (\frac {\ln \left (1+x^{2}+x \sqrt {3}\right )}{2}-\sqrt {3}\, \arctan \left (2 x +\sqrt {3}\right )\right )}{12}\)	\(131\)

[In]

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

[Out]

(1/3*x^4-1/7)/x^7+1/8*ln(4*x^2+4*x+4)-1/12*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)+1/4*sum(_R*ln(6*_R^3+_R+x),_R=R
ootOf(9*_Z^4+3*_Z^2+1))-1/8*ln(x^2-x+1)-1/12*3^(1/2)*arctan(2/3*(x-1/2)*3^(1/2))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.51 \[ \int \frac {1}{x^8 \left (1+x^4+x^8\right )} \, dx=\frac {7 \, \sqrt {6} x^{7} \sqrt {i \, \sqrt {3} - 1} \log \left (i \, \sqrt {6} \sqrt {3} \sqrt {i \, \sqrt {3} - 1} + 6 \, x\right ) - 7 \, \sqrt {6} x^{7} \sqrt {i \, \sqrt {3} - 1} \log \left (-i \, \sqrt {6} \sqrt {3} \sqrt {i \, \sqrt {3} - 1} + 6 \, x\right ) - 7 \, \sqrt {6} x^{7} \sqrt {-i \, \sqrt {3} - 1} \log \left (i \, \sqrt {6} \sqrt {3} \sqrt {-i \, \sqrt {3} - 1} + 6 \, x\right ) + 7 \, \sqrt {6} x^{7} \sqrt {-i \, \sqrt {3} - 1} \log \left (-i \, \sqrt {6} \sqrt {3} \sqrt {-i \, \sqrt {3} - 1} + 6 \, x\right ) - 14 \, \sqrt {3} x^{7} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 14 \, \sqrt {3} x^{7} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 21 \, x^{7} \log \left (x^{2} + x + 1\right ) - 21 \, x^{7} \log \left (x^{2} - x + 1\right ) + 56 \, x^{4} - 24}{168 \, x^{7}} \]

[In]

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

[Out]

1/168*(7*sqrt(6)*x^7*sqrt(I*sqrt(3) - 1)*log(I*sqrt(6)*sqrt(3)*sqrt(I*sqrt(3) - 1) + 6*x) - 7*sqrt(6)*x^7*sqrt
(I*sqrt(3) - 1)*log(-I*sqrt(6)*sqrt(3)*sqrt(I*sqrt(3) - 1) + 6*x) - 7*sqrt(6)*x^7*sqrt(-I*sqrt(3) - 1)*log(I*s
qrt(6)*sqrt(3)*sqrt(-I*sqrt(3) - 1) + 6*x) + 7*sqrt(6)*x^7*sqrt(-I*sqrt(3) - 1)*log(-I*sqrt(6)*sqrt(3)*sqrt(-I
*sqrt(3) - 1) + 6*x) - 14*sqrt(3)*x^7*arctan(1/3*sqrt(3)*(2*x + 1)) - 14*sqrt(3)*x^7*arctan(1/3*sqrt(3)*(2*x -
1)) + 21*x^7*log(x^2 + x + 1) - 21*x^7*log(x^2 - x + 1) + 56*x^4 - 24)/x^7

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.36 \[ \int \frac {1}{x^8 \left (1+x^4+x^8\right )} \, dx=\left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x - \frac {1}{2} + \frac {\sqrt {3} i}{6} - 18432 \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{5} \right )} + \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x - \frac {1}{2} - 18432 \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{5} - \frac {\sqrt {3} i}{6} \right )} + \left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x + \frac {1}{2} + \frac {\sqrt {3} i}{6} - 18432 \left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{5} \right )} + \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x + \frac {1}{2} - 18432 \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{5} - \frac {\sqrt {3} i}{6} \right )} + \operatorname {RootSum} {\left (2304 t^{4} + 48 t^{2} + 1, \left ( t \mapsto t \log {\left (- 18432 t^{5} - 4 t + x \right )} \right )\right )} + \frac {7 x^{4} - 3}{21 x^{7}} \]

[In]

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

[Out]

(1/8 - sqrt(3)*I/24)*log(x - 1/2 + sqrt(3)*I/6 - 18432*(1/8 - sqrt(3)*I/24)**5) + (1/8 + sqrt(3)*I/24)*log(x -
1/2 - 18432*(1/8 + sqrt(3)*I/24)**5 - sqrt(3)*I/6) + (-1/8 - sqrt(3)*I/24)*log(x + 1/2 + sqrt(3)*I/6 - 18432*
(-1/8 - sqrt(3)*I/24)**5) + (-1/8 + sqrt(3)*I/24)*log(x + 1/2 - 18432*(-1/8 + sqrt(3)*I/24)**5 - sqrt(3)*I/6)
+ RootSum(2304*_t**4 + 48*_t**2 + 1, Lambda(_t, _t*log(-18432*_t**5 - 4*_t + x))) + (7*x**4 - 3)/(21*x**7)

Maxima [F]

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

[In]

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

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/21*(7*x^4 - 3)/x^
7 + 1/2*integrate(x^2/(x^4 - x^2 + 1), x) + 1/8*log(x^2 + x + 1) - 1/8*log(x^2 - x + 1)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^8 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {7 \, x^{4} - 3}{21 \, x^{7}} + \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{4} \, \arctan \left (2 \, x - \sqrt {3}\right ) + \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \]

[In]

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

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/24*sqrt(3)*log(x^
2 + sqrt(3)*x + 1) + 1/24*sqrt(3)*log(x^2 - sqrt(3)*x + 1) + 1/21*(7*x^4 - 3)/x^7 + 1/4*arctan(2*x + sqrt(3))
+ 1/4*arctan(2*x - sqrt(3)) + 1/8*log(x^2 + x + 1) - 1/8*log(x^2 - x + 1)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^8 \left (1+x^4+x^8\right )} \, dx=\frac {\frac {x^4}{3}-\frac {1}{7}}{x^7}-\mathrm {atan}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}-\frac {1}{4}{}\mathrm {i}\right )-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}+\frac {1}{4}{}\mathrm {i}\right )-\mathrm {atan}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \]

[In]

int(1/(x^8*(x^4 + x^8 + 1)),x)

[Out]

(x^4/3 - 1/7)/x^7 - atan((2*x)/(3^(1/2)*1i + 1))*((3^(1/2)*1i)/12 - 1/4) - atan((x*2i)/(3^(1/2)*1i - 1))*(3^(1
/2)/12 - 1i/4) - atan((x*2i)/(3^(1/2)*1i + 1))*(3^(1/2)/12 + 1i/4) - atan((2*x)/(3^(1/2)*1i - 1))*((3^(1/2)*1i
)/12 + 1/4)