\(\int \frac {-2+x^4}{\sqrt [4]{1+x^4} (-1-x^4+2 x^8)} \, dx\) [1952]

Optimal result

Integrand size = 29, antiderivative size = 137 \[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}+\frac {5 \arctan \left (\frac {\sqrt {2} x \sqrt [4]{1+x^4}}{-x^2+\sqrt {1+x^4}}\right )}{6 \sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}+\frac {5 \text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{1+x^4}}{x^2+\sqrt {1+x^4}}\right )}{6 \sqrt {2}} \]

[Out]

1/12*arctan(2^(1/4)*x/(x^4+1)^(1/4))*2^(3/4)+5/12*arctan(2^(1/2)*x*(x^4+1)^(1/4)/(-x^2+(x^4+1)^(1/2)))*2^(1/2)
+1/12*arctanh(2^(1/4)*x/(x^4+1)^(1/4))*2^(3/4)+5/12*arctanh(2^(1/2)*x*(x^4+1)^(1/4)/(x^2+(x^4+1)^(1/2)))*2^(1/
2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.41, number of steps used = 16, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {6860, 385, 218, 212, 209, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{6 \sqrt [4]{2}}-\frac {5 \arctan \left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}\right )}{6 \sqrt {2}}+\frac {5 \arctan \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}+1\right )}{6 \sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{6 \sqrt [4]{2}}-\frac {5 \log \left (-\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}+\frac {x^2}{\sqrt {x^4+1}}+1\right )}{12 \sqrt {2}}+\frac {5 \log \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}+\frac {x^2}{\sqrt {x^4+1}}+1\right )}{12 \sqrt {2}} \]

[In]

Int[(-2 + x^4)/((1 + x^4)^(1/4)*(-1 - x^4 + 2*x^8)),x]

[Out]

ArcTan[(2^(1/4)*x)/(1 + x^4)^(1/4)]/(6*2^(1/4)) - (5*ArcTan[1 - (Sqrt[2]*x)/(1 + x^4)^(1/4)])/(6*Sqrt[2]) + (5
*ArcTan[1 + (Sqrt[2]*x)/(1 + x^4)^(1/4)])/(6*Sqrt[2]) + ArcTanh[(2^(1/4)*x)/(1 + x^4)^(1/4)]/(6*2^(1/4)) - (5*
Log[1 + x^2/Sqrt[1 + x^4] - (Sqrt[2]*x)/(1 + x^4)^(1/4)])/(12*Sqrt[2]) + (5*Log[1 + x^2/Sqrt[1 + x^4] + (Sqrt[
2]*x)/(1 + x^4)^(1/4)])/(12*Sqrt[2])

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 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 212

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

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 385

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[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, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 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, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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}& = \int \left (-\frac {4}{3 \sqrt [4]{1+x^4} \left (-4+4 x^4\right )}+\frac {10}{3 \sqrt [4]{1+x^4} \left (2+4 x^4\right )}\right ) \, dx \\ & = -\left (\frac {4}{3} \int \frac {1}{\sqrt [4]{1+x^4} \left (-4+4 x^4\right )} \, dx\right )+\frac {10}{3} \int \frac {1}{\sqrt [4]{1+x^4} \left (2+4 x^4\right )} \, dx \\ & = -\left (\frac {4}{3} \text {Subst}\left (\int \frac {1}{-4+8 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\right )+\frac {10}{3} \text {Subst}\left (\int \frac {1}{2+2 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right ) \\ & = \frac {1}{6} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{6} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {5}{3} \text {Subst}\left (\int \frac {1-x^2}{2+2 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {5}{3} \text {Subst}\left (\int \frac {1+x^2}{2+2 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right ) \\ & = \frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}+\frac {5}{12} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {5}{12} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}}-\frac {5 \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}} \\ & = \frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}-\frac {5 \log \left (1+\frac {x^2}{\sqrt {1+x^4}}-\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}}+\frac {5 \log \left (1+\frac {x^2}{\sqrt {1+x^4}}+\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}}+\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}-\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}} \\ & = \frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}-\frac {5 \arctan \left (1-\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}+\frac {5 \arctan \left (1+\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}-\frac {5 \log \left (1+\frac {x^2}{\sqrt {1+x^4}}-\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}}+\frac {5 \log \left (1+\frac {x^2}{\sqrt {1+x^4}}+\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.92 \[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\frac {\sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )+5 \arctan \left (\frac {\sqrt {2} x \sqrt [4]{1+x^4}}{-x^2+\sqrt {1+x^4}}\right )+\sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )+5 \text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{1+x^4}}{x^2+\sqrt {1+x^4}}\right )}{6 \sqrt {2}} \]

[In]

Integrate[(-2 + x^4)/((1 + x^4)^(1/4)*(-1 - x^4 + 2*x^8)),x]

[Out]

(2^(1/4)*ArcTan[(2^(1/4)*x)/(1 + x^4)^(1/4)] + 5*ArcTan[(Sqrt[2]*x*(1 + x^4)^(1/4))/(-x^2 + Sqrt[1 + x^4])] +
2^(1/4)*ArcTanh[(2^(1/4)*x)/(1 + x^4)^(1/4)] + 5*ArcTanh[(Sqrt[2]*x*(1 + x^4)^(1/4))/(x^2 + Sqrt[1 + x^4])])/(
6*Sqrt[2])

Maple [A] (verified)

Time = 6.16 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.22

method	result	size
pseudoelliptic	\(-\frac {\arctan \left (\frac {2^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {3}{4}}}{12}+\frac {\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{4}+1\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{4}+1\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}}}{24}-\frac {5 \ln \left (\frac {-\left (x^{4}+1\right )^{\frac {1}{4}} x \sqrt {2}+x^{2}+\sqrt {x^{4}+1}}{\left (x^{4}+1\right )^{\frac {1}{4}} x \sqrt {2}+x^{2}+\sqrt {x^{4}+1}}\right ) \sqrt {2}}{24}-\frac {5 \arctan \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}} \sqrt {2}+x}{x}\right ) \sqrt {2}}{12}-\frac {5 \arctan \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}} \sqrt {2}-x}{x}\right ) \sqrt {2}}{12}\)	\(167\)
trager	\(\text {Expression too large to display}\)	\(642\)

[In]

int((x^4-2)/(x^4+1)^(1/4)/(2*x^8-x^4-1),x,method=_RETURNVERBOSE)

[Out]

-1/12*arctan(1/2*2^(3/4)/x*(x^4+1)^(1/4))*2^(3/4)+1/24*ln((-2^(1/4)*x-(x^4+1)^(1/4))/(2^(1/4)*x-(x^4+1)^(1/4))
)*2^(3/4)-5/24*ln((-(x^4+1)^(1/4)*x*2^(1/2)+x^2+(x^4+1)^(1/2))/((x^4+1)^(1/4)*x*2^(1/2)+x^2+(x^4+1)^(1/2)))*2^
(1/2)-5/12*arctan(((x^4+1)^(1/4)*2^(1/2)+x)/x)*2^(1/2)-5/12*arctan(((x^4+1)^(1/4)*2^(1/2)-x)/x)*2^(1/2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.18 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.66 \[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\frac {1}{48} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) + \frac {1}{48} i \cdot 2^{\frac {3}{4}} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 2^{\frac {3}{4}} {\left (3 i \, x^{4} + i\right )} - 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) - \frac {1}{48} i \cdot 2^{\frac {3}{4}} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 2^{\frac {3}{4}} {\left (-3 i \, x^{4} - i\right )} - 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) - \frac {1}{48} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) - \left (\frac {5}{48} i + \frac {5}{48}\right ) \, \sqrt {2} \log \left (\frac {\left (i + 1\right ) \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 2 i \, \sqrt {x^{4} + 1} x^{2} + \left (i - 1\right ) \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 1}{2 \, x^{4} + 1}\right ) + \left (\frac {5}{48} i - \frac {5}{48}\right ) \, \sqrt {2} \log \left (\frac {-\left (i - 1\right ) \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 2 i \, \sqrt {x^{4} + 1} x^{2} - \left (i + 1\right ) \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 1}{2 \, x^{4} + 1}\right ) - \left (\frac {5}{48} i - \frac {5}{48}\right ) \, \sqrt {2} \log \left (\frac {\left (i - 1\right ) \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 2 i \, \sqrt {x^{4} + 1} x^{2} + \left (i + 1\right ) \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 1}{2 \, x^{4} + 1}\right ) + \left (\frac {5}{48} i + \frac {5}{48}\right ) \, \sqrt {2} \log \left (\frac {-\left (i + 1\right ) \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 2 i \, \sqrt {x^{4} + 1} x^{2} - \left (i - 1\right ) \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 1}{2 \, x^{4} + 1}\right ) \]

[In]

integrate((x^4-2)/(x^4+1)^(1/4)/(2*x^8-x^4-1),x, algorithm="fricas")

[Out]

1/48*2^(3/4)*log((4*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 4*2^(1/4)*sqrt(x^4 + 1)*x^2 + 2^(3/4)*(3*x^4 + 1) + 4*(x^4 +
1)^(3/4)*x)/(x^4 - 1)) + 1/48*I*2^(3/4)*log(-(4*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 4*I*2^(1/4)*sqrt(x^4 + 1)*x^2 -
2^(3/4)*(3*I*x^4 + I) - 4*(x^4 + 1)^(3/4)*x)/(x^4 - 1)) - 1/48*I*2^(3/4)*log(-(4*sqrt(2)*(x^4 + 1)^(1/4)*x^3
- 4*I*2^(1/4)*sqrt(x^4 + 1)*x^2 - 2^(3/4)*(-3*I*x^4 - I) - 4*(x^4 + 1)^(3/4)*x)/(x^4 - 1)) - 1/48*2^(3/4)*log(
(4*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 4*2^(1/4)*sqrt(x^4 + 1)*x^2 - 2^(3/4)*(3*x^4 + 1) + 4*(x^4 + 1)^(3/4)*x)/(x^4
- 1)) - (5/48*I + 5/48)*sqrt(2)*log(((I + 1)*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 2*I*sqrt(x^4 + 1)*x^2 + (I - 1)*sq
rt(2)*(x^4 + 1)^(3/4)*x + 1)/(2*x^4 + 1)) + (5/48*I - 5/48)*sqrt(2)*log((-(I - 1)*sqrt(2)*(x^4 + 1)^(1/4)*x^3
+ 2*I*sqrt(x^4 + 1)*x^2 - (I + 1)*sqrt(2)*(x^4 + 1)^(3/4)*x + 1)/(2*x^4 + 1)) - (5/48*I - 5/48)*sqrt(2)*log(((
I - 1)*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 2*I*sqrt(x^4 + 1)*x^2 + (I + 1)*sqrt(2)*(x^4 + 1)^(3/4)*x + 1)/(2*x^4 + 1
)) + (5/48*I + 5/48)*sqrt(2)*log((-(I + 1)*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 2*I*sqrt(x^4 + 1)*x^2 - (I - 1)*sqrt(
2)*(x^4 + 1)^(3/4)*x + 1)/(2*x^4 + 1))

Sympy [F]

\[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\int \frac {x^{4} - 2}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \sqrt [4]{x^{4} + 1} \cdot \left (2 x^{4} + 1\right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\int { \frac {x^{4} - 2}{{\left (2 \, x^{8} - x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate((x^4-2)/(x^4+1)^(1/4)/(2*x^8-x^4-1),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\int { \frac {x^{4} - 2}{{\left (2 \, x^{8} - x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate((x^4-2)/(x^4+1)^(1/4)/(2*x^8-x^4-1),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1-x^4+2 x^8\right )} \, dx=-\int \frac {x^4-2}{{\left (x^4+1\right )}^{1/4}\,\left (-2\,x^8+x^4+1\right )} \,d x \]

[In]

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

[Out]

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