\(\int \frac {-1+x^4}{(1+x^2+x^4) \sqrt [4]{x^2+x^6}} \, dx\) [1335]

Optimal result

Integrand size = 27, antiderivative size = 96 \[ \int \frac {-1+x^4}{\left (1+x^2+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt [4]{x^2+x^6}}{-x^2+\sqrt {x^2+x^6}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt {2}}+\frac {\sqrt {x^2+x^6}}{\sqrt {2}}}{x \sqrt [4]{x^2+x^6}}\right )}{\sqrt {2}} \]

[Out]

-1/2*arctan(2^(1/2)*x*(x^6+x^2)^(1/4)/(-x^2+(x^6+x^2)^(1/2)))*2^(1/2)-1/2*arctanh((1/2*2^(1/2)*x^2+1/2*(x^6+x^
2)^(1/2)*2^(1/2))/x/(x^6+x^2)^(1/4))*2^(1/2)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.56 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.32, number of steps used = 15, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2081, 6847, 6860, 251, 1452, 440, 524} \[ \int \frac {-1+x^4}{\left (1+x^2+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=-\frac {2 \sqrt [4]{x^4+1} x \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},-\frac {2 x^4}{1-i \sqrt {3}},-x^4\right )}{\sqrt [4]{x^6+x^2}}-\frac {2 \sqrt [4]{x^4+1} x \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},-\frac {2 x^4}{1+i \sqrt {3}},-x^4\right )}{\sqrt [4]{x^6+x^2}}-\frac {2 \left (-\sqrt {3}+i\right ) \sqrt [4]{x^4+1} x^3 \operatorname {AppellF1}\left (\frac {5}{8},\frac {1}{4},1,\frac {13}{8},-x^4,-\frac {2 x^4}{1-i \sqrt {3}}\right )}{5 \left (\sqrt {3}+i\right ) \sqrt [4]{x^6+x^2}}-\frac {2 \left (\sqrt {3}+i\right ) \sqrt [4]{x^4+1} x^3 \operatorname {AppellF1}\left (\frac {5}{8},\frac {1}{4},1,\frac {13}{8},-x^4,-\frac {2 x^4}{1+i \sqrt {3}}\right )}{5 \left (-\sqrt {3}+i\right ) \sqrt [4]{x^6+x^2}}+\frac {2 \sqrt [4]{x^4+1} x \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^6+x^2}} \]

[In]

Int[(-1 + x^4)/((1 + x^2 + x^4)*(x^2 + x^6)^(1/4)),x]

[Out]

(-2*x*(1 + x^4)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, (-2*x^4)/(1 - I*Sqrt[3]), -x^4])/(x^2 + x^6)^(1/4) - (2*x*(1
+ x^4)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, (-2*x^4)/(1 + I*Sqrt[3]), -x^4])/(x^2 + x^6)^(1/4) - (2*(I - Sqrt[3])*
x^3*(1 + x^4)^(1/4)*AppellF1[5/8, 1/4, 1, 13/8, -x^4, (-2*x^4)/(1 - I*Sqrt[3])])/(5*(I + Sqrt[3])*(x^2 + x^6)^
(1/4)) - (2*(I + Sqrt[3])*x^3*(1 + x^4)^(1/4)*AppellF1[5/8, 1/4, 1, 13/8, -x^4, (-2*x^4)/(1 + I*Sqrt[3])])/(5*
(I - Sqrt[3])*(x^2 + x^6)^(1/4)) + (2*x*(1 + x^4)^(1/4)*Hypergeometric2F1[1/8, 1/4, 9/8, -x^4])/(x^2 + x^6)^(1
/4)

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1452

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^(2
*n))^p, (d/(d^2 - e^2*x^(2*n)) - e*(x^n/(d^2 - e^2*x^(2*n))))^(-q), x], x] /; FreeQ[{a, c, d, e, n, p}, x] &&
EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[q, 0]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{1+x^4}\right ) \int \frac {-1+x^4}{\sqrt {x} \sqrt [4]{1+x^4} \left (1+x^2+x^4\right )} \, dx}{\sqrt [4]{x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {-1+x^8}{\sqrt [4]{1+x^8} \left (1+x^4+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \left (\frac {1}{\sqrt [4]{1+x^8}}-\frac {2+x^4}{\sqrt [4]{1+x^8} \left (1+x^4+x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {2+x^4}{\sqrt [4]{1+x^8} \left (1+x^4+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {2 x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \left (\frac {1-i \sqrt {3}}{\left (1-i \sqrt {3}+2 x^4\right ) \sqrt [4]{1+x^8}}+\frac {1+i \sqrt {3}}{\left (1+i \sqrt {3}+2 x^4\right ) \sqrt [4]{1+x^8}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {2 x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (2 \left (1-i \sqrt {3}\right ) \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (1-i \sqrt {3}+2 x^4\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (2 \left (1+i \sqrt {3}\right ) \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (1+i \sqrt {3}+2 x^4\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {2 x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (2 \left (1-i \sqrt {3}\right ) \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \left (\frac {i+\sqrt {3}}{2 \left (-i+\sqrt {3}-2 i x^8\right ) \sqrt [4]{1+x^8}}+\frac {x^4}{\sqrt [4]{1+x^8} \left (1+i \sqrt {3}+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (2 \left (1+i \sqrt {3}\right ) \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \left (\frac {-i+\sqrt {3}}{2 \left (i+\sqrt {3}+2 i x^8\right ) \sqrt [4]{1+x^8}}+\frac {x^4}{\sqrt [4]{1+x^8} \left (1-i \sqrt {3}+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {2 x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (2 \left (1-i \sqrt {3}\right ) \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt [4]{1+x^8} \left (1+i \sqrt {3}+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (2 \left (1+i \sqrt {3}\right ) \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt [4]{1+x^8} \left (1-i \sqrt {3}+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (\left (1+i \sqrt {3}\right ) \left (-i+\sqrt {3}\right ) \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (i+\sqrt {3}+2 i x^8\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (\left (1-i \sqrt {3}\right ) \left (i+\sqrt {3}\right ) \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (-i+\sqrt {3}-2 i x^8\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = -\frac {2 x \sqrt [4]{1+x^4} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},-\frac {2 x^4}{1-i \sqrt {3}},-x^4\right )}{\sqrt [4]{x^2+x^6}}-\frac {2 x \sqrt [4]{1+x^4} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},-\frac {2 x^4}{1+i \sqrt {3}},-x^4\right )}{\sqrt [4]{x^2+x^6}}-\frac {2 \left (i-\sqrt {3}\right ) x^3 \sqrt [4]{1+x^4} \operatorname {AppellF1}\left (\frac {5}{8},\frac {1}{4},1,\frac {13}{8},-x^4,-\frac {2 x^4}{1-i \sqrt {3}}\right )}{5 \left (i+\sqrt {3}\right ) \sqrt [4]{x^2+x^6}}-\frac {2 \left (i+\sqrt {3}\right ) x^3 \sqrt [4]{1+x^4} \operatorname {AppellF1}\left (\frac {5}{8},\frac {1}{4},1,\frac {13}{8},-x^4,-\frac {2 x^4}{1+i \sqrt {3}}\right )}{5 \left (i-\sqrt {3}\right ) \sqrt [4]{x^2+x^6}}+\frac {2 x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^2+x^6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.07 \[ \int \frac {-1+x^4}{\left (1+x^2+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=-\frac {\sqrt {x} \sqrt [4]{1+x^4} \left (\arctan \left (\frac {\sqrt {2} \sqrt {x} \sqrt [4]{1+x^4}}{-x+\sqrt {1+x^4}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt {x} \sqrt [4]{1+x^4}}{x+\sqrt {1+x^4}}\right )\right )}{\sqrt {2} \sqrt [4]{x^2+x^6}} \]

[In]

Integrate[(-1 + x^4)/((1 + x^2 + x^4)*(x^2 + x^6)^(1/4)),x]

[Out]

-((Sqrt[x]*(1 + x^4)^(1/4)*(ArcTan[(Sqrt[2]*Sqrt[x]*(1 + x^4)^(1/4))/(-x + Sqrt[1 + x^4])] + ArcTanh[(Sqrt[2]*
Sqrt[x]*(1 + x^4)^(1/4))/(x + Sqrt[1 + x^4])]))/(Sqrt[2]*(x^2 + x^6)^(1/4)))

Maple [A] (verified)

Time = 0.00 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.29

method	result	size
pseudoelliptic	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}{\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}\right )+2 \arctan \left (\frac {\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} \sqrt {2}-x}{x}\right )\right )}{4}\)	\(124\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {2 \sqrt {x^{6}+x^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{5}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \left (x^{6}+x^{2}\right )^{\frac {1}{4}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{3}-2 \left (x^{6}+x^{2}\right )^{\frac {3}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x}{\left (x^{2}-x +1\right ) x \left (x^{2}+x +1\right )}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{5}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \left (x^{6}+x^{2}\right )^{\frac {1}{4}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x -2 \sqrt {x^{6}+x^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x -2 \left (x^{6}+x^{2}\right )^{\frac {3}{4}}}{\left (x^{2}-x +1\right ) x \left (x^{2}+x +1\right )}\right )}{2}\)	\(237\)

[In]

int((x^4-1)/(x^4+x^2+1)/(x^6+x^2)^(1/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 34.64 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.18 \[ \int \frac {-1+x^4}{\left (1+x^2+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (\frac {4 i \, {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} - \left (2 i - 2\right ) \, \sqrt {2} \sqrt {x^{6} + x^{2}} x + \sqrt {2} {\left (\left (i + 1\right ) \, x^{5} - \left (i + 1\right ) \, x^{3} + \left (i + 1\right ) \, x\right )} - 4 \, {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{x^{5} + x^{3} + x}\right ) - \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (\frac {4 i \, {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + \left (2 i - 2\right ) \, \sqrt {2} \sqrt {x^{6} + x^{2}} x + \sqrt {2} {\left (-\left (i + 1\right ) \, x^{5} + \left (i + 1\right ) \, x^{3} - \left (i + 1\right ) \, x\right )} - 4 \, {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{x^{5} + x^{3} + x}\right ) - \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (\frac {-4 i \, {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + \left (2 i + 2\right ) \, \sqrt {2} \sqrt {x^{6} + x^{2}} x + \sqrt {2} {\left (-\left (i - 1\right ) \, x^{5} + \left (i - 1\right ) \, x^{3} - \left (i - 1\right ) \, x\right )} - 4 \, {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{x^{5} + x^{3} + x}\right ) + \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (\frac {-4 i \, {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} - \left (2 i + 2\right ) \, \sqrt {2} \sqrt {x^{6} + x^{2}} x + \sqrt {2} {\left (\left (i - 1\right ) \, x^{5} - \left (i - 1\right ) \, x^{3} + \left (i - 1\right ) \, x\right )} - 4 \, {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{x^{5} + x^{3} + x}\right ) \]

[In]

integrate((x^4-1)/(x^4+x^2+1)/(x^6+x^2)^(1/4),x, algorithm="fricas")

[Out]

(1/8*I + 1/8)*sqrt(2)*log((4*I*(x^6 + x^2)^(1/4)*x^2 - (2*I - 2)*sqrt(2)*sqrt(x^6 + x^2)*x + sqrt(2)*((I + 1)*
x^5 - (I + 1)*x^3 + (I + 1)*x) - 4*(x^6 + x^2)^(3/4))/(x^5 + x^3 + x)) - (1/8*I + 1/8)*sqrt(2)*log((4*I*(x^6 +
x^2)^(1/4)*x^2 + (2*I - 2)*sqrt(2)*sqrt(x^6 + x^2)*x + sqrt(2)*(-(I + 1)*x^5 + (I + 1)*x^3 - (I + 1)*x) - 4*(
x^6 + x^2)^(3/4))/(x^5 + x^3 + x)) - (1/8*I - 1/8)*sqrt(2)*log((-4*I*(x^6 + x^2)^(1/4)*x^2 + (2*I + 2)*sqrt(2)
*sqrt(x^6 + x^2)*x + sqrt(2)*(-(I - 1)*x^5 + (I - 1)*x^3 - (I - 1)*x) - 4*(x^6 + x^2)^(3/4))/(x^5 + x^3 + x))
+ (1/8*I - 1/8)*sqrt(2)*log((-4*I*(x^6 + x^2)^(1/4)*x^2 - (2*I + 2)*sqrt(2)*sqrt(x^6 + x^2)*x + sqrt(2)*((I -
1)*x^5 - (I - 1)*x^3 + (I - 1)*x) - 4*(x^6 + x^2)^(3/4))/(x^5 + x^3 + x))

Sympy [F]

\[ \int \frac {-1+x^4}{\left (1+x^2+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}{\sqrt [4]{x^{2} \left (x^{4} + 1\right )} \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )}\, dx \]

[In]

integrate((x**4-1)/(x**4+x**2+1)/(x**6+x**2)**(1/4),x)

[Out]

Integral((x - 1)*(x + 1)*(x**2 + 1)/((x**2*(x**4 + 1))**(1/4)*(x**2 - x + 1)*(x**2 + x + 1)), x)

Maxima [F]

\[ \int \frac {-1+x^4}{\left (1+x^2+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\int { \frac {x^{4} - 1}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} + x^{2} + 1\right )}} \,d x } \]

[In]

integrate((x^4-1)/(x^4+x^2+1)/(x^6+x^2)^(1/4),x, algorithm="maxima")

[Out]

integrate((x^4 - 1)/((x^6 + x^2)^(1/4)*(x^4 + x^2 + 1)), x)

Giac [F]

\[ \int \frac {-1+x^4}{\left (1+x^2+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\int { \frac {x^{4} - 1}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} + x^{2} + 1\right )}} \,d x } \]

[In]

integrate((x^4-1)/(x^4+x^2+1)/(x^6+x^2)^(1/4),x, algorithm="giac")

[Out]

integrate((x^4 - 1)/((x^6 + x^2)^(1/4)*(x^4 + x^2 + 1)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {-1+x^4}{\left (1+x^2+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\int \frac {x^4-1}{{\left (x^6+x^2\right )}^{1/4}\,\left (x^4+x^2+1\right )} \,d x \]

[In]

int((x^4 - 1)/((x^2 + x^6)^(1/4)*(x^2 + x^4 + 1)),x)

[Out]

int((x^4 - 1)/((x^2 + x^6)^(1/4)*(x^2 + x^4 + 1)), x)