\(\int \frac {1}{x^4 (1+x^8)} \, dx\) [1506]

Optimal result

Integrand size = 11, antiderivative size = 354 \[ \int \frac {1}{x^4 \left (1+x^8\right )} \, dx=-\frac {1}{3 x^3}-\frac {\arctan \left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}+\frac {\arctan \left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\arctan \left (\frac {\sqrt {2-\sqrt {2}}+2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}-\frac {\arctan \left (\frac {\sqrt {2+\sqrt {2}}+2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\log \left (1-\sqrt {2-\sqrt {2}} x+x^2\right )}{8 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\log \left (1+\sqrt {2-\sqrt {2}} x+x^2\right )}{8 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}}+\frac {\log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {331, 305, 1141, 1175, 632, 210, 1178, 642} \[ \int \frac {1}{x^4 \left (1+x^8\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}+\frac {\arctan \left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\arctan \left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}-\frac {\arctan \left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {1}{3 x^3}+\frac {\log \left (x^2-\sqrt {2-\sqrt {2}} x+1\right )}{8 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\log \left (x^2+\sqrt {2-\sqrt {2}} x+1\right )}{8 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\log \left (x^2-\sqrt {2+\sqrt {2}} x+1\right )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}}+\frac {\log \left (x^2+\sqrt {2+\sqrt {2}} x+1\right )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}} \]

[In]

Int[1/(x^4*(1 + x^8)),x]

[Out]

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

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 305

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Numerator[Rt[a/b, 4]], s = Denominator[Rt[a/b,
4]]}, Dist[s^3/(2*Sqrt[2]*b*r), Int[x^(m - n/4)/(r^2 - Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x] - Dist[s^3/
(2*Sqrt[2]*b*r), Int[x^(m - n/4)/(r^2 + Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x]] /; FreeQ[{a, b}, x] && IGt
Q[n/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && GtQ[a/b, 0]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^4 \left (1+x^8\right )} \, dx=-\frac {1}{3 x^3}-\frac {1}{4} \arctan \left (\left (x-\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right ) \cos \left (\frac {\pi }{8}\right )-\frac {1}{4} \arctan \left (\left (x+\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right ) \cos \left (\frac {\pi }{8}\right )+\frac {1}{8} \cos \left (\frac {\pi }{8}\right ) \log \left (1+x^2-2 x \sin \left (\frac {\pi }{8}\right )\right )-\frac {1}{8} \cos \left (\frac {\pi }{8}\right ) \log \left (1+x^2+2 x \sin \left (\frac {\pi }{8}\right )\right )+\frac {1}{4} \arctan \left (\sec \left (\frac {\pi }{8}\right ) \left (x-\sin \left (\frac {\pi }{8}\right )\right )\right ) \sin \left (\frac {\pi }{8}\right )+\frac {1}{4} \arctan \left (\sec \left (\frac {\pi }{8}\right ) \left (x+\sin \left (\frac {\pi }{8}\right )\right )\right ) \sin \left (\frac {\pi }{8}\right )-\frac {1}{8} \log \left (1+x^2-2 x \cos \left (\frac {\pi }{8}\right )\right ) \sin \left (\frac {\pi }{8}\right )+\frac {1}{8} \log \left (1+x^2+2 x \cos \left (\frac {\pi }{8}\right )\right ) \sin \left (\frac {\pi }{8}\right ) \]

[In]

Integrate[1/(x^4*(1 + x^8)),x]

[Out]

-1/3*1/x^3 - (ArcTan[(x - Cos[Pi/8])*Csc[Pi/8]]*Cos[Pi/8])/4 - (ArcTan[(x + Cos[Pi/8])*Csc[Pi/8]]*Cos[Pi/8])/4
+ (Cos[Pi/8]*Log[1 + x^2 - 2*x*Sin[Pi/8]])/8 - (Cos[Pi/8]*Log[1 + x^2 + 2*x*Sin[Pi/8]])/8 + (ArcTan[Sec[Pi/8]
*(x - Sin[Pi/8])]*Sin[Pi/8])/4 + (ArcTan[Sec[Pi/8]*(x + Sin[Pi/8])]*Sin[Pi/8])/4 - (Log[1 + x^2 - 2*x*Cos[Pi/8
]]*Sin[Pi/8])/8 + (Log[1 + x^2 + 2*x*Cos[Pi/8]]*Sin[Pi/8])/8

Maple [C] (verified)

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

Time = 3.41 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.07

method	result	size
risch	\(-\frac {1}{3 x^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\sum }\textit {\_R} \ln \left (\textit {\_R}^{5}+x \right )\right )}{8}\)	\(26\)
default	\(-\frac {1}{3 x^{3}}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{8}\)	\(28\)
meijerg	\(-\frac {1}{3 x^{3}}-\frac {x^{5} \left (\frac {\cos \left (\frac {3 \pi }{8}\right ) \ln \left (1-2 \cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{\left (x^{8}\right )^{\frac {5}{8}}}+\frac {2 \sin \left (\frac {3 \pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1-\cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{\left (x^{8}\right )^{\frac {5}{8}}}-\frac {\cos \left (\frac {\pi }{8}\right ) \ln \left (1-2 \cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{\left (x^{8}\right )^{\frac {5}{8}}}-\frac {2 \sin \left (\frac {\pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1-\cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{\left (x^{8}\right )^{\frac {5}{8}}}+\frac {\cos \left (\frac {\pi }{8}\right ) \ln \left (1+2 \cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{\left (x^{8}\right )^{\frac {5}{8}}}-\frac {2 \sin \left (\frac {\pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1+\cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{\left (x^{8}\right )^{\frac {5}{8}}}-\frac {\cos \left (\frac {3 \pi }{8}\right ) \ln \left (1+2 \cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{\left (x^{8}\right )^{\frac {5}{8}}}+\frac {2 \sin \left (\frac {3 \pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1+\cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{\left (x^{8}\right )^{\frac {5}{8}}}\right )}{8}\)	\(277\)

[In]

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

[Out]

-1/3/x^3+1/8*sum(_R*ln(_R^5+x),_R=RootOf(_Z^8+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^4 \left (1+x^8\right )} \, dx=\frac {-\left (3 i + 3\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} x^{3} \log \left (\left (i + 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {5}{8}} + 2 \, x\right ) + \left (3 i - 3\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} x^{3} \log \left (-\left (i - 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {5}{8}} + 2 \, x\right ) - \left (3 i - 3\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} x^{3} \log \left (\left (i - 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {5}{8}} + 2 \, x\right ) + \left (3 i + 3\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} x^{3} \log \left (-\left (i + 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {5}{8}} + 2 \, x\right ) + 6 \, \left (-1\right )^{\frac {1}{8}} x^{3} \log \left (x + \left (-1\right )^{\frac {5}{8}}\right ) + 6 i \, \left (-1\right )^{\frac {1}{8}} x^{3} \log \left (x + i \, \left (-1\right )^{\frac {5}{8}}\right ) - 6 i \, \left (-1\right )^{\frac {1}{8}} x^{3} \log \left (x - i \, \left (-1\right )^{\frac {5}{8}}\right ) - 6 \, \left (-1\right )^{\frac {1}{8}} x^{3} \log \left (x - \left (-1\right )^{\frac {5}{8}}\right ) - 16}{48 \, x^{3}} \]

[In]

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

[Out]

1/48*(-(3*I + 3)*sqrt(2)*(-1)^(1/8)*x^3*log((I + 1)*sqrt(2)*(-1)^(5/8) + 2*x) + (3*I - 3)*sqrt(2)*(-1)^(1/8)*x
^3*log(-(I - 1)*sqrt(2)*(-1)^(5/8) + 2*x) - (3*I - 3)*sqrt(2)*(-1)^(1/8)*x^3*log((I - 1)*sqrt(2)*(-1)^(5/8) +
2*x) + (3*I + 3)*sqrt(2)*(-1)^(1/8)*x^3*log(-(I + 1)*sqrt(2)*(-1)^(5/8) + 2*x) + 6*(-1)^(1/8)*x^3*log(x + (-1)
^(5/8)) + 6*I*(-1)^(1/8)*x^3*log(x + I*(-1)^(5/8)) - 6*I*(-1)^(1/8)*x^3*log(x - I*(-1)^(5/8)) - 6*(-1)^(1/8)*x
^3*log(x - (-1)^(5/8)) - 16)/x^3

Sympy [A] (verification not implemented)

Time = 1.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.06 \[ \int \frac {1}{x^4 \left (1+x^8\right )} \, dx=\operatorname {RootSum} {\left (16777216 t^{8} + 1, \left ( t \mapsto t \log {\left (32768 t^{5} + x \right )} \right )\right )} - \frac {1}{3 x^{3}} \]

[In]

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

[Out]

RootSum(16777216*_t**8 + 1, Lambda(_t, _t*log(32768*_t**5 + x))) - 1/(3*x**3)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^4 \left (1+x^8\right )} \, dx=\frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {2 \, x + \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {2 \, x - \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {2 \, x + \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {2 \, x - \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \log \left (x^{2} + x \sqrt {\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \log \left (x^{2} - x \sqrt {\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \log \left (x^{2} + x \sqrt {-\sqrt {2} + 2} + 1\right ) + \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \log \left (x^{2} - x \sqrt {-\sqrt {2} + 2} + 1\right ) - \frac {1}{3 \, x^{3}} \]

[In]

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

[Out]

1/8*sqrt(-sqrt(2) + 2)*arctan((2*x + sqrt(-sqrt(2) + 2))/sqrt(sqrt(2) + 2)) + 1/8*sqrt(-sqrt(2) + 2)*arctan((2
*x - sqrt(-sqrt(2) + 2))/sqrt(sqrt(2) + 2)) - 1/8*sqrt(sqrt(2) + 2)*arctan((2*x + sqrt(sqrt(2) + 2))/sqrt(-sqr
t(2) + 2)) - 1/8*sqrt(sqrt(2) + 2)*arctan((2*x - sqrt(sqrt(2) + 2))/sqrt(-sqrt(2) + 2)) + 1/16*sqrt(-sqrt(2) +
2)*log(x^2 + x*sqrt(sqrt(2) + 2) + 1) - 1/16*sqrt(-sqrt(2) + 2)*log(x^2 - x*sqrt(sqrt(2) + 2) + 1) - 1/16*sqr
t(sqrt(2) + 2)*log(x^2 + x*sqrt(-sqrt(2) + 2) + 1) + 1/16*sqrt(sqrt(2) + 2)*log(x^2 - x*sqrt(-sqrt(2) + 2) + 1
) - 1/3/x^3

Mupad [B] (verification not implemented)

Time = 5.92 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^4 \left (1+x^8\right )} \, dx=\mathrm {atan}\left (\frac {x\,\sqrt {-\sqrt {2}-2}\,1{}\mathrm {i}}{\sqrt {2-\sqrt {2}}\,\sqrt {-\sqrt {2}-2}-\sqrt {2}}-\frac {x\,\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{\sqrt {2-\sqrt {2}}\,\sqrt {-\sqrt {2}-2}-\sqrt {2}}\right )\,\left (\frac {\sqrt {-\sqrt {2}-2}\,1{}\mathrm {i}}{8}-\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )-\frac {1}{3\,x^3}+\mathrm {atan}\left (\frac {x\,\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{\sqrt {2}-\sqrt {\sqrt {2}-2}\,\sqrt {\sqrt {2}+2}}+\frac {x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{\sqrt {2}-\sqrt {\sqrt {2}-2}\,\sqrt {\sqrt {2}+2}}\right )\,\left (\frac {\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )-\mathrm {atan}\left (-\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+2}}{2}+x\,\sqrt {\sqrt {2}+2}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {2}}{16}-\frac {1}{16}+\frac {1}{16}{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,2{}\mathrm {i}-\mathrm {atan}\left (x\,\sqrt {\sqrt {2}+2}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {2}\,1{}\mathrm {i}}{16}-\frac {1}{16}-\frac {1}{16}{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,2{}\mathrm {i} \]

[In]

int(1/(x^4*(x^8 + 1)),x)

[Out]

atan((x*(- 2^(1/2) - 2)^(1/2)*1i)/((2 - 2^(1/2))^(1/2)*(- 2^(1/2) - 2)^(1/2) - 2^(1/2)) - (x*(2 - 2^(1/2))^(1/
2)*1i)/((2 - 2^(1/2))^(1/2)*(- 2^(1/2) - 2)^(1/2) - 2^(1/2)))*(((- 2^(1/2) - 2)^(1/2)*1i)/8 - ((2 - 2^(1/2))^(
1/2)*1i)/8) - 1/(3*x^3) + atan((x*(2^(1/2) - 2)^(1/2)*1i)/(2^(1/2) - (2^(1/2) - 2)^(1/2)*(2^(1/2) + 2)^(1/2))
+ (x*(2^(1/2) + 2)^(1/2)*1i)/(2^(1/2) - (2^(1/2) - 2)^(1/2)*(2^(1/2) + 2)^(1/2)))*(((2^(1/2) - 2)^(1/2)*1i)/8
+ ((2^(1/2) + 2)^(1/2)*1i)/8) - atan(x*(2^(1/2) + 2)^(1/2)*(1/2 + 1i/2) - (2^(1/2)*x*(2^(1/2) + 2)^(1/2))/2)*(
2^(1/2)/16 - (1/16 - 1i/16))*(2^(1/2) + 2)^(1/2)*2i - atan(x*(2^(1/2) + 2)^(1/2)*(1/2 - 1i/2) + (2^(1/2)*x*(2^
(1/2) + 2)^(1/2)*1i)/2)*((2^(1/2)*1i)/16 - (1/16 + 1i/16))*(2^(1/2) + 2)^(1/2)*2i