\(\int \frac {x^4}{\sqrt [4]{-1+x^4} (-1+2 x^4+x^8)} \, dx\) [2100]

Optimal result

Integrand size = 25, antiderivative size = 152 \[ \int \frac {x^4}{\sqrt [4]{-1+x^4} \left (-1+2 x^4+x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [8]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4\ 2^{5/8}}+\frac {\arctan \left (\frac {2^{5/8} x \sqrt [4]{-1+x^4}}{\sqrt [4]{2} x^2-\sqrt {-1+x^4}}\right )}{8 \sqrt [8]{2}}+\frac {\text {arctanh}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4\ 2^{5/8}}-\frac {\text {arctanh}\left (\frac {2\ 2^{3/8} x \sqrt [4]{-1+x^4}}{2 x^2+2^{3/4} \sqrt {-1+x^4}}\right )}{8 \sqrt [8]{2}} \]

[Out]

1/8*arctan(2^(1/8)*x/(x^4-1)^(1/4))*2^(3/8)+1/16*arctan(2^(5/8)*x*(x^4-1)^(1/4)/(x^2*2^(1/4)-(x^4-1)^(1/2)))*2
^(7/8)+1/8*arctanh(2^(1/8)*x/(x^4-1)^(1/4))*2^(3/8)-1/16*arctanh(2*2^(3/8)*x*(x^4-1)^(1/4)/(2*x^2+2^(3/4)*(x^4
-1)^(1/2)))*2^(7/8)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.86, number of steps used = 16, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {1542, 385, 217, 1179, 642, 1176, 631, 210, 218, 212, 209} \[ \int \frac {x^4}{\sqrt [4]{-1+x^4} \left (-1+2 x^4+x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [8]{2} x}{\sqrt [4]{x^4-1}}\right )}{4\ 2^{5/8}}-\frac {\left (1-\sqrt {2}\right ) \arctan \left (1-\frac {2^{5/8} x}{\sqrt [4]{x^4-1}}\right )}{4\ 2^{5/8} \left (2-\sqrt {2}\right )}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {2^{5/8} x}{\sqrt [4]{x^4-1}}+1\right )}{4\ 2^{5/8} \left (2-\sqrt {2}\right )}+\frac {\text {arctanh}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{x^4-1}}\right )}{4\ 2^{5/8}}-\frac {\left (1-\sqrt {2}\right ) \log \left (-\frac {2 x}{\sqrt [4]{x^4-1}}+\frac {2^{5/8} x^2}{\sqrt {x^4-1}}+2^{3/8}\right )}{8\ 2^{5/8} \left (2-\sqrt {2}\right )}+\frac {\left (1-\sqrt {2}\right ) \log \left (\frac {2^{5/8} x}{\sqrt [4]{x^4-1}}+\frac {\sqrt [4]{2} x^2}{\sqrt {x^4-1}}+1\right )}{8\ 2^{5/8} \left (2-\sqrt {2}\right )} \]

[In]

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

[Out]

ArcTan[(2^(1/8)*x)/(-1 + x^4)^(1/4)]/(4*2^(5/8)) - ((1 - Sqrt[2])*ArcTan[1 - (2^(5/8)*x)/(-1 + x^4)^(1/4)])/(4
*2^(5/8)*(2 - Sqrt[2])) + ((1 - Sqrt[2])*ArcTan[1 + (2^(5/8)*x)/(-1 + x^4)^(1/4)])/(4*2^(5/8)*(2 - Sqrt[2])) +
ArcTanh[(2^(1/8)*x)/(-1 + x^4)^(1/4)]/(4*2^(5/8)) - ((1 - Sqrt[2])*Log[2^(3/8) + (2^(5/8)*x^2)/Sqrt[-1 + x^4]
- (2*x)/(-1 + x^4)^(1/4)])/(8*2^(5/8)*(2 - Sqrt[2])) + ((1 - Sqrt[2])*Log[1 + (2^(1/4)*x^2)/Sqrt[-1 + x^4] +
(2^(5/8)*x)/(-1 + x^4)^(1/4)])/(8*2^(5/8)*(2 - 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 1542

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

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

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.93 \[ \int \frac {x^4}{\sqrt [4]{-1+x^4} \left (-1+2 x^4+x^8\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [8]{2} x}{\sqrt [4]{-1+x^4}}\right )+\sqrt {2} \arctan \left (\frac {2^{5/8} x \sqrt [4]{-1+x^4}}{\sqrt [4]{2} x^2-\sqrt {-1+x^4}}\right )+2 \text {arctanh}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{-1+x^4}}\right )-\sqrt {2} \text {arctanh}\left (\frac {2\ 2^{3/8} x \sqrt [4]{-1+x^4}}{2 x^2+2^{3/4} \sqrt {-1+x^4}}\right )}{8\ 2^{5/8}} \]

[In]

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

[Out]

(2*ArcTan[(2^(1/8)*x)/(-1 + x^4)^(1/4)] + Sqrt[2]*ArcTan[(2^(5/8)*x*(-1 + x^4)^(1/4))/(2^(1/4)*x^2 - Sqrt[-1 +
x^4])] + 2*ArcTanh[(2^(1/8)*x)/(-1 + x^4)^(1/4)] - Sqrt[2]*ArcTanh[(2*2^(3/8)*x*(-1 + x^4)^(1/4))/(2*x^2 + 2^
(3/4)*Sqrt[-1 + x^4])])/(8*2^(5/8))

Maple [C] (verified)

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

Time = 10.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.22

method	result	size
pseudoelliptic	\(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{5}}\right )}{8}\)	\(33\)
trager	\(\text {Expression too large to display}\)	\(1111\)

[In]

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

[Out]

-1/8*sum(ln((-_R*x+(x^4-1)^(1/4))/x)/_R^5,_R=RootOf(_Z^8-2))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 8.18 (sec) , antiderivative size = 1100, normalized size of antiderivative = 7.24 \[ \int \frac {x^4}{\sqrt [4]{-1+x^4} \left (-1+2 x^4+x^8\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

(1/1024*I + 1/1024)*32^(7/8)*sqrt(2)*log((32^(7/8)*sqrt(2)*((I + 1)*x^8 + I + 1) - 4*32^(3/8)*sqrt(2)*((I + 1)
*x^8 - (2*I + 2)*x^4 - I - 1) + 64*(x^5 + sqrt(2)*x + x)*(x^4 - 1)^(3/4) - 4*sqrt(x^4 - 1)*(-(8*I - 8)*32^(1/8
)*sqrt(2)*x^2 + 32^(5/8)*sqrt(2)*(-(I - 1)*x^6 - (I - 1)*x^2)) - 64*(x^4 - 1)^(1/4)*(I*2^(3/4)*x^3 + 2^(1/4)*(
I*x^7 + I*x^3)))/(x^8 + 2*x^4 - 1)) - (1/1024*I - 1/1024)*32^(7/8)*sqrt(2)*log((32^(7/8)*sqrt(2)*(-(I - 1)*x^8
- I + 1) - 4*32^(3/8)*sqrt(2)*(-(I - 1)*x^8 + (2*I - 2)*x^4 + I - 1) + 64*(x^5 + sqrt(2)*x + x)*(x^4 - 1)^(3/
4) - 4*sqrt(x^4 - 1)*((8*I + 8)*32^(1/8)*sqrt(2)*x^2 + 32^(5/8)*sqrt(2)*((I + 1)*x^6 + (I + 1)*x^2)) - 64*(x^4
- 1)^(1/4)*(-I*2^(3/4)*x^3 + 2^(1/4)*(-I*x^7 - I*x^3)))/(x^8 + 2*x^4 - 1)) + (1/1024*I - 1/1024)*32^(7/8)*sqr
t(2)*log((32^(7/8)*sqrt(2)*((I - 1)*x^8 + I - 1) - 4*32^(3/8)*sqrt(2)*((I - 1)*x^8 - (2*I - 2)*x^4 - I + 1) +
64*(x^5 + sqrt(2)*x + x)*(x^4 - 1)^(3/4) - 4*sqrt(x^4 - 1)*(-(8*I + 8)*32^(1/8)*sqrt(2)*x^2 + 32^(5/8)*sqrt(2)
*(-(I + 1)*x^6 - (I + 1)*x^2)) - 64*(x^4 - 1)^(1/4)*(-I*2^(3/4)*x^3 + 2^(1/4)*(-I*x^7 - I*x^3)))/(x^8 + 2*x^4
- 1)) - (1/1024*I + 1/1024)*32^(7/8)*sqrt(2)*log((32^(7/8)*sqrt(2)*(-(I + 1)*x^8 - I - 1) - 4*32^(3/8)*sqrt(2)
*(-(I + 1)*x^8 + (2*I + 2)*x^4 + I + 1) + 64*(x^5 + sqrt(2)*x + x)*(x^4 - 1)^(3/4) - 4*sqrt(x^4 - 1)*((8*I - 8
)*32^(1/8)*sqrt(2)*x^2 + 32^(5/8)*sqrt(2)*((I - 1)*x^6 + (I - 1)*x^2)) - 64*(x^4 - 1)^(1/4)*(I*2^(3/4)*x^3 + 2
^(1/4)*(I*x^7 + I*x^3)))/(x^8 + 2*x^4 - 1)) + 1/512*32^(7/8)*log((32^(7/8)*(x^8 + 1) + 32*(x^5 - sqrt(2)*x + x
)*(x^4 - 1)^(3/4) - 4*sqrt(x^4 - 1)*(8*32^(1/8)*x^2 - 32^(5/8)*(x^6 + x^2)) + 4*32^(3/8)*(x^8 - 2*x^4 - 1) - 3
2*(x^4 - 1)^(1/4)*(2^(3/4)*x^3 - 2^(1/4)*(x^7 + x^3)))/(x^8 + 2*x^4 - 1)) - 1/512*32^(7/8)*log(-(32^(7/8)*(x^8
+ 1) - 32*(x^5 - sqrt(2)*x + x)*(x^4 - 1)^(3/4) - 4*sqrt(x^4 - 1)*(8*32^(1/8)*x^2 - 32^(5/8)*(x^6 + x^2)) + 4
*32^(3/8)*(x^8 - 2*x^4 - 1) + 32*(x^4 - 1)^(1/4)*(2^(3/4)*x^3 - 2^(1/4)*(x^7 + x^3)))/(x^8 + 2*x^4 - 1)) + 1/5
12*I*32^(7/8)*log((32^(7/8)*(I*x^8 + I) + 32*(x^5 - sqrt(2)*x + x)*(x^4 - 1)^(3/4) - 4*sqrt(x^4 - 1)*(-8*I*32^
(1/8)*x^2 + 32^(5/8)*(I*x^6 + I*x^2)) - 4*32^(3/8)*(-I*x^8 + 2*I*x^4 + I) + 32*(x^4 - 1)^(1/4)*(2^(3/4)*x^3 -
2^(1/4)*(x^7 + x^3)))/(x^8 + 2*x^4 - 1)) - 1/512*I*32^(7/8)*log((32^(7/8)*(-I*x^8 - I) + 32*(x^5 - sqrt(2)*x +
x)*(x^4 - 1)^(3/4) - 4*sqrt(x^4 - 1)*(8*I*32^(1/8)*x^2 + 32^(5/8)*(-I*x^6 - I*x^2)) - 4*32^(3/8)*(I*x^8 - 2*I
*x^4 - I) + 32*(x^4 - 1)^(1/4)*(2^(3/4)*x^3 - 2^(1/4)*(x^7 + x^3)))/(x^8 + 2*x^4 - 1))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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