\(\int \frac {1-2 x^4+x^8}{\sqrt [4]{-1+x^4} (1-2 x^4+2 x^8)} \, dx\) [2849]

Optimal result

Integrand size = 34, antiderivative size = 293 \[ \int \frac {1-2 x^4+x^8}{\sqrt [4]{-1+x^4} \left (1-2 x^4+2 x^8\right )} \, dx=\frac {1}{4} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{8} \sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{-1+x^4}}{-x^2+\sqrt {-1+x^4}}\right )-\frac {1}{8} \sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )} \arctan \left (\frac {\left (\sqrt {\frac {2}{2-\sqrt {2}}}-\frac {2}{\sqrt {2-\sqrt {2}}}\right ) x \sqrt [4]{-1+x^4}}{-x^2+\sqrt {-1+x^4}}\right )+\frac {1}{4} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{8} \sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )} \text {arctanh}\left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{-1+x^4}}{x^2+\sqrt {-1+x^4}}\right )+\frac {1}{8} \sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )} \text {arctanh}\left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{-1+x^4}}{x^2+\sqrt {-1+x^4}}\right ) \]

[Out]

1/4*arctan(x/(x^4-1)^(1/4))+1/16*(4-2*2^(1/2))^(1/2)*arctan((2+2^(1/2))^(1/2)*x*(x^4-1)^(1/4)/(-x^2+(x^4-1)^(1
/2)))-1/16*(4+2*2^(1/2))^(1/2)*arctan((2^(1/2)/(2-2^(1/2))^(1/2)-2/(2-2^(1/2))^(1/2))*x*(x^4-1)^(1/4)/(-x^2+(x
^4-1)^(1/2)))+1/4*arctanh(x/(x^4-1)^(1/4))+1/16*(4+2*2^(1/2))^(1/2)*arctanh((2-2^(1/2))^(1/2)*x*(x^4-1)^(1/4)/
(x^2+(x^4-1)^(1/2)))+1/16*(4-2*2^(1/2))^(1/2)*arctanh((2+2^(1/2))^(1/2)*x*(x^4-1)^(1/4)/(x^2+(x^4-1)^(1/2)))

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.02, number of steps used = 28, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.441, Rules used = {28, 1442, 427, 544, 246, 218, 212, 209, 385, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1-2 x^4+x^8}{\sqrt [4]{-1+x^4} \left (1-2 x^4+2 x^8\right )} \, dx=\frac {1}{4} \arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\left (\frac {1}{8}+\frac {i}{8}\right ) \sqrt [8]{-1} \arctan \left (\frac {(-1)^{7/8} x}{\sqrt [4]{x^4-1}}\right )-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (-1)^{5/8} \arctan \left (1-\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{x^4-1}}\right )}{\sqrt {2}}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (-1)^{5/8} \arctan \left (\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{x^4-1}}+1\right )}{\sqrt {2}}+\frac {1}{4} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\left (\frac {1}{8}+\frac {i}{8}\right ) \sqrt [8]{-1} \text {arctanh}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{x^4-1}}\right )-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (-1)^{5/8} \log \left (\frac {\sqrt [8]{-1} \sqrt {2} x}{\sqrt [4]{x^4-1}}+\frac {x^2}{\sqrt {x^4-1}}+\sqrt [4]{-1}\right )}{\sqrt {2}}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (-1)^{5/8} \log \left (\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{x^4-1}}-\frac {(-1)^{3/4} x^2}{\sqrt {x^4-1}}+1\right )}{\sqrt {2}} \]

[In]

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

[Out]

ArcTan[x/(-1 + x^4)^(1/4)]/4 - (1/8 + I/8)*(-1)^(1/8)*ArcTan[((-1)^(7/8)*x)/(-1 + x^4)^(1/4)] - ((1/8 + I/8)*(
-1)^(5/8)*ArcTan[1 - ((-1)^(7/8)*Sqrt[2]*x)/(-1 + x^4)^(1/4)])/Sqrt[2] + ((1/8 + I/8)*(-1)^(5/8)*ArcTan[1 + ((
-1)^(7/8)*Sqrt[2]*x)/(-1 + x^4)^(1/4)])/Sqrt[2] + ArcTanh[x/(-1 + x^4)^(1/4)]/4 - (1/8 + I/8)*(-1)^(1/8)*ArcTa
nh[((-1)^(7/8)*x)/(-1 + x^4)^(1/4)] - ((1/16 + I/16)*(-1)^(5/8)*Log[(-1)^(1/4) + x^2/Sqrt[-1 + x^4] + ((-1)^(1
/8)*Sqrt[2]*x)/(-1 + x^4)^(1/4)])/Sqrt[2] + ((1/16 + I/16)*(-1)^(5/8)*Log[1 - ((-1)^(3/4)*x^2)/Sqrt[-1 + x^4]
+ ((-1)^(7/8)*Sqrt[2]*x)/(-1 + x^4)^(1/4)])/Sqrt[2]

Rule 28

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

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

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 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 544

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[f/d,
Int[(a + b*x^n)^p, x], x] + Dist[(d*e - c*f)/d, Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, p, n}, x]

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 1442

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[b^2 -
4*a*c, 2]}, Dist[2*(c/r), Int[(d + e*x^n)^q/(b - r + 2*c*x^n), x], x] - Dist[2*(c/r), Int[(d + e*x^n)^q/(b +
r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] &&  !IntegerQ[q]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-1+x^4\right )^{7/4}}{1-2 x^4+2 x^8} \, dx \\ & = -\left (2 i \int \frac {\left (-1+x^4\right )^{7/4}}{(-2-2 i)+4 x^4} \, dx\right )+2 i \int \frac {\left (-1+x^4\right )^{7/4}}{(-2+2 i)+4 x^4} \, dx \\ & = -\left (\frac {1}{8} i \int \frac {(14-2 i)-(20-8 i) x^4}{\sqrt [4]{-1+x^4} \left ((-2-2 i)+4 x^4\right )} \, dx\right )+\frac {1}{8} i \int \frac {(14+2 i)-(20+8 i) x^4}{\sqrt [4]{-1+x^4} \left ((-2+2 i)+4 x^4\right )} \, dx \\ & = -\left (\left (-\frac {1}{4}-\frac {5 i}{8}\right ) \int \frac {1}{\sqrt [4]{-1+x^4}} \, dx\right )+\left (\frac {1}{4}-\frac {5 i}{8}\right ) \int \frac {1}{\sqrt [4]{-1+x^4}} \, dx-\int \frac {1}{\sqrt [4]{-1+x^4} \left ((-2-2 i)+4 x^4\right )} \, dx-\int \frac {1}{\sqrt [4]{-1+x^4} \left ((-2+2 i)+4 x^4\right )} \, dx \\ & = -\left (\left (-\frac {1}{4}-\frac {5 i}{8}\right ) \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\right )+\left (\frac {1}{4}-\frac {5 i}{8}\right ) \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )-\text {Subst}\left (\int \frac {1}{(-2+2 i)-(2+2 i) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )-\text {Subst}\left (\int \frac {1}{(-2-2 i)-(2-2 i) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right ) \\ & = -\left (\left (-\frac {1}{8}-\frac {5 i}{16}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\right )-\left (-\frac {1}{8}-\frac {5 i}{16}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\left (\frac {1}{8}-\frac {5 i}{16}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\left (\frac {1}{8}-\frac {5 i}{16}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{2} (-1)^{3/4} \text {Subst}\left (\int \frac {\sqrt [4]{-1}-x^2}{(-2-2 i)-(2-2 i) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{2} (-1)^{3/4} \text {Subst}\left (\int \frac {\sqrt [4]{-1}+x^2}{(-2-2 i)-(2-2 i) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {i \text {Subst}\left (\int \frac {1}{\sqrt [4]{-1}-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt {2}}+\frac {i \text {Subst}\left (\int \frac {1}{\sqrt [4]{-1}+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt {2}} \\ & = \frac {1}{4} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {(-1)^{3/8} \arctan \left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt {2}}+\frac {1}{4} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {(-1)^{3/8} \text {arctanh}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt [4]{-1}-\sqrt [8]{-1} \sqrt {2} x+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{8 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt [4]{-1}+\sqrt [8]{-1} \sqrt {2} x+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{8 \sqrt {2}}+\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) (-1)^{5/8}\right ) \text {Subst}\left (\int \frac {\sqrt [8]{-1} \sqrt {2}+2 x}{-\sqrt [4]{-1}-\sqrt [8]{-1} \sqrt {2} x-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}+\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) (-1)^{5/8}\right ) \text {Subst}\left (\int \frac {\sqrt [8]{-1} \sqrt {2}-2 x}{-\sqrt [4]{-1}+\sqrt [8]{-1} \sqrt {2} x-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}} \\ & = \frac {1}{4} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {(-1)^{3/8} \arctan \left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt {2}}+\frac {1}{4} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {(-1)^{3/8} \text {arctanh}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt {2}}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (-1)^{5/8} \log \left (\sqrt [4]{-1}+\frac {x^2}{\sqrt {-1+x^4}}+\frac {\sqrt [8]{-1} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (-1)^{5/8} \log \left (1-\frac {(-1)^{3/4} x^2}{\sqrt {-1+x^4}}+\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}+\frac {1}{8} (-1)^{7/8} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{8} (-1)^{7/8} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right ) \\ & = \frac {1}{4} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {(-1)^{3/8} \arctan \left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt {2}}-\frac {1}{8} (-1)^{7/8} \arctan \left (1-\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{8} (-1)^{7/8} \arctan \left (1+\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{4} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {(-1)^{3/8} \text {arctanh}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt {2}}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (-1)^{5/8} \log \left (\sqrt [4]{-1}+\frac {x^2}{\sqrt {-1+x^4}}+\frac {\sqrt [8]{-1} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (-1)^{5/8} \log \left (1-\frac {(-1)^{3/4} x^2}{\sqrt {-1+x^4}}+\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.85 \[ \int \frac {1-2 x^4+x^8}{\sqrt [4]{-1+x^4} \left (1-2 x^4+2 x^8\right )} \, dx=\frac {1}{16} \left (4 \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\sqrt {2 \left (2+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{-1+x^4}}{-x^2+\sqrt {-1+x^4}}\right )+\sqrt {4-2 \sqrt {2}} \arctan \left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{-1+x^4}}{-x^2+\sqrt {-1+x^4}}\right )+4 \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\sqrt {2 \left (2+\sqrt {2}\right )} \text {arctanh}\left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{-1+x^4}}{x^2+\sqrt {-1+x^4}}\right )+\sqrt {4-2 \sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{-1+x^4}}{x^2+\sqrt {-1+x^4}}\right )\right ) \]

[In]

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

[Out]

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

Maple [C] (verified)

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

Time = 7.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.30

method	result	size
pseudoelliptic	\(-\frac {\arctan \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )}{4}+\frac {\ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}+x}{x}\right )}{8}-\frac {\ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}-x}{x}\right )}{8}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\sum }\frac {\left (\textit {\_R}^{4}+1\right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{5}}\right )}{16}\)	\(87\)
trager	\(\text {Expression too large to display}\)	\(1039\)

[In]

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

[Out]

-1/4*arctan((x^4-1)^(1/4)/x)+1/8*ln(((x^4-1)^(1/4)+x)/x)-1/8*ln(((x^4-1)^(1/4)-x)/x)-1/16*sum(1/_R^5*(_R^4+1)*
ln((-_R*x+(x^4-1)^(1/4))/x),_R=RootOf(_Z^8+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.17 \[ \int \frac {1-2 x^4+x^8}{\sqrt [4]{-1+x^4} \left (1-2 x^4+2 x^8\right )} \, dx=-\frac {1}{16} \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (\frac {4 \, {\left (\sqrt {2} {\left (\left (-1\right )^{\frac {7}{8}} x + \left (-1\right )^{\frac {3}{8}} x\right )} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}}{x}\right ) + \frac {1}{16} \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (-\frac {4 \, {\left (\sqrt {2} {\left (\left (-1\right )^{\frac {7}{8}} x + \left (-1\right )^{\frac {3}{8}} x\right )} - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}}{x}\right ) - \frac {1}{16} i \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (-\frac {4 \, {\left (\sqrt {2} {\left (i \, \left (-1\right )^{\frac {7}{8}} x + i \, \left (-1\right )^{\frac {3}{8}} x\right )} - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}}{x}\right ) + \frac {1}{16} i \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (-\frac {4 \, {\left (\sqrt {2} {\left (-i \, \left (-1\right )^{\frac {7}{8}} x - i \, \left (-1\right )^{\frac {3}{8}} x\right )} - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}}{x}\right ) - \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \left (-1\right )^{\frac {1}{8}} \log \left (-\frac {8 \, {\left (\left (i + 1\right ) \, \left (-1\right )^{\frac {7}{8}} x - \left (i + 1\right ) \, \left (-1\right )^{\frac {3}{8}} x - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}}{x}\right ) + \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \left (-1\right )^{\frac {1}{8}} \log \left (-\frac {8 \, {\left (-\left (i - 1\right ) \, \left (-1\right )^{\frac {7}{8}} x + \left (i - 1\right ) \, \left (-1\right )^{\frac {3}{8}} x - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}}{x}\right ) - \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \left (-1\right )^{\frac {1}{8}} \log \left (-\frac {8 \, {\left (\left (i - 1\right ) \, \left (-1\right )^{\frac {7}{8}} x - \left (i - 1\right ) \, \left (-1\right )^{\frac {3}{8}} x - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}}{x}\right ) + \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \left (-1\right )^{\frac {1}{8}} \log \left (-\frac {8 \, {\left (-\left (i + 1\right ) \, \left (-1\right )^{\frac {7}{8}} x + \left (i + 1\right ) \, \left (-1\right )^{\frac {3}{8}} x - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}}{x}\right ) - \frac {1}{4} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{8} \, \log \left (\frac {x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{8} \, \log \left (-\frac {x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

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

[Out]

-1/16*sqrt(2)*(-1)^(1/8)*log(4*(sqrt(2)*((-1)^(7/8)*x + (-1)^(3/8)*x) + 2*(x^4 - 1)^(1/4))/x) + 1/16*sqrt(2)*(
-1)^(1/8)*log(-4*(sqrt(2)*((-1)^(7/8)*x + (-1)^(3/8)*x) - 2*(x^4 - 1)^(1/4))/x) - 1/16*I*sqrt(2)*(-1)^(1/8)*lo
g(-4*(sqrt(2)*(I*(-1)^(7/8)*x + I*(-1)^(3/8)*x) - 2*(x^4 - 1)^(1/4))/x) + 1/16*I*sqrt(2)*(-1)^(1/8)*log(-4*(sq
rt(2)*(-I*(-1)^(7/8)*x - I*(-1)^(3/8)*x) - 2*(x^4 - 1)^(1/4))/x) - (1/16*I - 1/16)*(-1)^(1/8)*log(-8*((I + 1)*
(-1)^(7/8)*x - (I + 1)*(-1)^(3/8)*x - 2*(x^4 - 1)^(1/4))/x) + (1/16*I + 1/16)*(-1)^(1/8)*log(-8*(-(I - 1)*(-1)
^(7/8)*x + (I - 1)*(-1)^(3/8)*x - 2*(x^4 - 1)^(1/4))/x) - (1/16*I + 1/16)*(-1)^(1/8)*log(-8*((I - 1)*(-1)^(7/8
)*x - (I - 1)*(-1)^(3/8)*x - 2*(x^4 - 1)^(1/4))/x) + (1/16*I - 1/16)*(-1)^(1/8)*log(-8*(-(I + 1)*(-1)^(7/8)*x
+ (I + 1)*(-1)^(3/8)*x - 2*(x^4 - 1)^(1/4))/x) - 1/4*arctan((x^4 - 1)^(1/4)/x) + 1/8*log((x + (x^4 - 1)^(1/4))
/x) - 1/8*log(-(x - (x^4 - 1)^(1/4))/x)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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