\(\int \frac {\sqrt [4]{-1+2 x^4} (-1+x^4+x^8)}{x^6 (-1+x^4)} \, dx\) [1545]

Optimal result

Integrand size = 30, antiderivative size = 106 \[ \int \frac {\sqrt [4]{-1+2 x^4} \left (-1+x^4+x^8\right )}{x^6 \left (-1+x^4\right )} \, dx=\frac {\left (-1+2 x^4\right )^{5/4}}{5 x^5}+\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{-1+2 x^4}}\right )-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+2 x^4}}\right )}{2^{3/4}}-\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+2 x^4}}\right )+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+2 x^4}}\right )}{2^{3/4}} \]

[Out]

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

Rubi [C] (verified)

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

Time = 0.53 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.63, number of steps used = 23, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6857, 270, 1254, 416, 418, 1227, 551, 525, 524} \[ \int \frac {\sqrt [4]{-1+2 x^4} \left (-1+x^4+x^8\right )}{x^6 \left (-1+x^4\right )} \, dx=\frac {\left (2 x^4-1\right )^{5/4}}{5 x^5}-\frac {x^3 \sqrt [4]{2 x^4-1} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},2 x^4,x^4\right )}{3 \sqrt [4]{1-2 x^4}} \]

[In]

Int[((-1 + 2*x^4)^(1/4)*(-1 + x^4 + x^8))/(x^6*(-1 + x^4)),x]

[Out]

(-1 + 2*x^4)^(5/4)/(5*x^5) - (x^3*(-1 + 2*x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, 2*x^4, x^4])/(3*(1 - 2*x^4)^(
1/4))

Rule 270

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 416

Int[((a_) + (b_.)*(x_)^4)^(1/4)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[Sqrt[a + b*x^4]*Sqrt[a/(a + b*x^4)],
Subst[Int[1/(Sqrt[1 - b*x^4]*(c - (b*c - a*d)*x^4)), x], x, x/(a + b*x^4)^(1/4)], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[b*c - a*d, 0]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-d/c, 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), x], x] /; FreeQ[{a,
b, c, d}, x] && NeQ[b*c - a*d, 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{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])

Rule 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-f/e, -d/c])

Rule 1227

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

Rule 1254

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^4)^p, (d/
(d^2 - e^2*x^4) - e*(x^2/(d^2 - e^2*x^4)))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&&  !IntegerQ[p] && ILtQ[q, 0]

Rule 6857

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\sqrt [4]{-1+2 x^4}}{x^6}+\frac {\sqrt [4]{-1+2 x^4}}{2 \left (-1+x^2\right )}+\frac {\sqrt [4]{-1+2 x^4}}{2 \left (1+x^2\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {\sqrt [4]{-1+2 x^4}}{-1+x^2} \, dx+\frac {1}{2} \int \frac {\sqrt [4]{-1+2 x^4}}{1+x^2} \, dx+\int \frac {\sqrt [4]{-1+2 x^4}}{x^6} \, dx \\ & = \frac {\left (-1+2 x^4\right )^{5/4}}{5 x^5}+\frac {1}{2} \int \left (\frac {\sqrt [4]{-1+2 x^4}}{1-x^4}+\frac {x^2 \sqrt [4]{-1+2 x^4}}{-1+x^4}\right ) \, dx+\frac {1}{2} \int \left (\frac {\sqrt [4]{-1+2 x^4}}{-1+x^4}+\frac {x^2 \sqrt [4]{-1+2 x^4}}{-1+x^4}\right ) \, dx \\ & = \frac {\left (-1+2 x^4\right )^{5/4}}{5 x^5}+\frac {1}{2} \int \frac {\sqrt [4]{-1+2 x^4}}{1-x^4} \, dx+\frac {1}{2} \int \frac {\sqrt [4]{-1+2 x^4}}{-1+x^4} \, dx+2 \left (\frac {1}{2} \int \frac {x^2 \sqrt [4]{-1+2 x^4}}{-1+x^4} \, dx\right ) \\ & = \frac {\left (-1+2 x^4\right )^{5/4}}{5 x^5}+2 \frac {\sqrt [4]{-1+2 x^4} \int \frac {x^2 \sqrt [4]{1-2 x^4}}{-1+x^4} \, dx}{2 \sqrt [4]{1-2 x^4}}+\frac {1}{2} \left (\sqrt {-\frac {1}{-1+2 x^4}} \sqrt {-1+2 x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-2 x^4} \left (1-x^4\right )} \, dx,x,\frac {x}{\sqrt [4]{-1+2 x^4}}\right )+\frac {1}{2} \left (\sqrt {-\frac {1}{-1+2 x^4}} \sqrt {-1+2 x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-2 x^4} \left (-1+x^4\right )} \, dx,x,\frac {x}{\sqrt [4]{-1+2 x^4}}\right ) \\ & = \frac {\left (-1+2 x^4\right )^{5/4}}{5 x^5}-\frac {x^3 \sqrt [4]{-1+2 x^4} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},2 x^4,x^4\right )}{3 \sqrt [4]{1-2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt [4]{-1+2 x^4} \left (-1+x^4+x^8\right )}{x^6 \left (-1+x^4\right )} \, dx=\frac {1}{10} \left (\frac {2 \left (-1+2 x^4\right )^{5/4}}{x^5}-5 \sqrt [4]{2} \arctan \left (\frac {x}{\sqrt [4]{-\frac {1}{2}+x^4}}\right )+5 \arctan \left (\frac {x}{\sqrt [4]{-1+2 x^4}}\right )+5 \sqrt [4]{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{-\frac {1}{2}+x^4}}\right )-5 \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+2 x^4}}\right )\right ) \]

[In]

Integrate[((-1 + 2*x^4)^(1/4)*(-1 + x^4 + x^8))/(x^6*(-1 + x^4)),x]

[Out]

((2*(-1 + 2*x^4)^(5/4))/x^5 - 5*2^(1/4)*ArcTan[x/(-1/2 + x^4)^(1/4)] + 5*ArcTan[x/(-1 + 2*x^4)^(1/4)] + 5*2^(1
/4)*ArcTanh[x/(-1/2 + x^4)^(1/4)] - 5*ArcTanh[x/(-1 + 2*x^4)^(1/4)])/10

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(167\) vs. \(2(83)=166\).

Time = 11.69 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.58

method	result	size
pseudoelliptic	\(\frac {5 \ln \left (\frac {-2^{\frac {1}{4}} x -\left (2 x^{4}-1\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (2 x^{4}-1\right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}} x^{5}+10 \arctan \left (\frac {\left (2 x^{4}-1\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right ) 2^{\frac {1}{4}} x^{5}-5 \ln \left (\frac {x +\left (2 x^{4}-1\right )^{\frac {1}{4}}}{x}\right ) x^{5}+5 \ln \left (\frac {\left (2 x^{4}-1\right )^{\frac {1}{4}}-x}{x}\right ) x^{5}-10 \arctan \left (\frac {\left (2 x^{4}-1\right )^{\frac {1}{4}}}{x}\right ) x^{5}+8 x^{4} \left (2 x^{4}-1\right )^{\frac {1}{4}}-4 \left (2 x^{4}-1\right )^{\frac {1}{4}}}{20 x^{5}}\)	\(168\)
trager	\(\frac {\left (2 x^{4}-1\right )^{\frac {5}{4}}}{5 x^{5}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{4}-4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (2 x^{4}-1\right )^{\frac {1}{4}} x^{3}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \sqrt {2 x^{4}-1}\, x^{2}+4 \left (2 x^{4}-1\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right )\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \ln \left (4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{4}+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (2 x^{4}-1\right )^{\frac {1}{4}} x^{3}+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \sqrt {2 x^{4}-1}\, x^{2}+4 \left (2 x^{4}-1\right )^{\frac {3}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3}\right )}{4}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} \sqrt {2 x^{4}-1}\, x^{2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{4}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right )+4 \left (2 x^{4}-1\right )^{\frac {3}{4}} x -4 \left (2 x^{4}-1\right )^{\frac {1}{4}} x^{3}}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{8}+\frac {\ln \left (\frac {2 \left (2 x^{4}-1\right )^{\frac {3}{4}} x -2 \sqrt {2 x^{4}-1}\, x^{2}+2 \left (2 x^{4}-1\right )^{\frac {1}{4}} x^{3}-3 x^{4}+1}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{4}\)	\(449\)

[In]

int((2*x^4-1)^(1/4)*(x^8+x^4-1)/x^6/(x^4-1),x,method=_RETURNVERBOSE)

[Out]

1/20*(5*ln((-2^(1/4)*x-(2*x^4-1)^(1/4))/(2^(1/4)*x-(2*x^4-1)^(1/4)))*2^(1/4)*x^5+10*arctan(1/2*(2*x^4-1)^(1/4)
/x*2^(3/4))*2^(1/4)*x^5-5*ln((x+(2*x^4-1)^(1/4))/x)*x^5+5*ln(((2*x^4-1)^(1/4)-x)/x)*x^5-10*arctan((2*x^4-1)^(1
/4)/x)*x^5+8*x^4*(2*x^4-1)^(1/4)-4*(2*x^4-1)^(1/4))/x^5

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 13.32 (sec) , antiderivative size = 388, normalized size of antiderivative = 3.66 \[ \int \frac {\sqrt [4]{-1+2 x^4} \left (-1+x^4+x^8\right )}{x^6 \left (-1+x^4\right )} \, dx=\frac {5 \cdot 2^{\frac {1}{4}} x^{5} \log \left (4 \, \sqrt {2} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} \sqrt {2 \, x^{4} - 1} x^{2} + 2^{\frac {3}{4}} {\left (4 \, x^{4} - 1\right )} + 4 \, {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x\right ) - 5 \cdot 2^{\frac {1}{4}} x^{5} \log \left (4 \, \sqrt {2} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - 4 \cdot 2^{\frac {1}{4}} \sqrt {2 \, x^{4} - 1} x^{2} - 2^{\frac {3}{4}} {\left (4 \, x^{4} - 1\right )} + 4 \, {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x\right ) + 5 i \cdot 2^{\frac {1}{4}} x^{5} \log \left (-4 \, \sqrt {2} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 4 i \cdot 2^{\frac {1}{4}} \sqrt {2 \, x^{4} - 1} x^{2} + 2^{\frac {3}{4}} {\left (-4 i \, x^{4} + i\right )} + 4 \, {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x\right ) - 5 i \cdot 2^{\frac {1}{4}} x^{5} \log \left (-4 \, \sqrt {2} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - 4 i \cdot 2^{\frac {1}{4}} \sqrt {2 \, x^{4} - 1} x^{2} + 2^{\frac {3}{4}} {\left (4 i \, x^{4} - i\right )} + 4 \, {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x\right ) + 10 \, x^{5} \arctan \left (\frac {2 \, {\left ({\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x\right )}}{x^{4} - 1}\right ) + 10 \, x^{5} \log \left (-\frac {3 \, x^{4} - 2 \, {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, \sqrt {2 \, x^{4} - 1} x^{2} - 2 \, {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x - 1}{x^{4} - 1}\right ) + 8 \, {\left (2 \, x^{4} - 1\right )}^{\frac {5}{4}}}{40 \, x^{5}} \]

[In]

integrate((2*x^4-1)^(1/4)*(x^8+x^4-1)/x^6/(x^4-1),x, algorithm="fricas")

[Out]

1/40*(5*2^(1/4)*x^5*log(4*sqrt(2)*(2*x^4 - 1)^(1/4)*x^3 + 4*2^(1/4)*sqrt(2*x^4 - 1)*x^2 + 2^(3/4)*(4*x^4 - 1)
+ 4*(2*x^4 - 1)^(3/4)*x) - 5*2^(1/4)*x^5*log(4*sqrt(2)*(2*x^4 - 1)^(1/4)*x^3 - 4*2^(1/4)*sqrt(2*x^4 - 1)*x^2 -
2^(3/4)*(4*x^4 - 1) + 4*(2*x^4 - 1)^(3/4)*x) + 5*I*2^(1/4)*x^5*log(-4*sqrt(2)*(2*x^4 - 1)^(1/4)*x^3 + 4*I*2^(
1/4)*sqrt(2*x^4 - 1)*x^2 + 2^(3/4)*(-4*I*x^4 + I) + 4*(2*x^4 - 1)^(3/4)*x) - 5*I*2^(1/4)*x^5*log(-4*sqrt(2)*(2
*x^4 - 1)^(1/4)*x^3 - 4*I*2^(1/4)*sqrt(2*x^4 - 1)*x^2 + 2^(3/4)*(4*I*x^4 - I) + 4*(2*x^4 - 1)^(3/4)*x) + 10*x^
5*arctan(2*((2*x^4 - 1)^(1/4)*x^3 + (2*x^4 - 1)^(3/4)*x)/(x^4 - 1)) + 10*x^5*log(-(3*x^4 - 2*(2*x^4 - 1)^(1/4)
*x^3 + 2*sqrt(2*x^4 - 1)*x^2 - 2*(2*x^4 - 1)^(3/4)*x - 1)/(x^4 - 1)) + 8*(2*x^4 - 1)^(5/4))/x^5

Sympy [F]

\[ \int \frac {\sqrt [4]{-1+2 x^4} \left (-1+x^4+x^8\right )}{x^6 \left (-1+x^4\right )} \, dx=\int \frac {\sqrt [4]{2 x^{4} - 1} \left (x^{8} + x^{4} - 1\right )}{x^{6} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}\, dx \]

[In]

integrate((2*x**4-1)**(1/4)*(x**8+x**4-1)/x**6/(x**4-1),x)

[Out]

Integral((2*x**4 - 1)**(1/4)*(x**8 + x**4 - 1)/(x**6*(x - 1)*(x + 1)*(x**2 + 1)), x)

Maxima [F]

\[ \int \frac {\sqrt [4]{-1+2 x^4} \left (-1+x^4+x^8\right )}{x^6 \left (-1+x^4\right )} \, dx=\int { \frac {{\left (x^{8} + x^{4} - 1\right )} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{{\left (x^{4} - 1\right )} x^{6}} \,d x } \]

[In]

integrate((2*x^4-1)^(1/4)*(x^8+x^4-1)/x^6/(x^4-1),x, algorithm="maxima")

[Out]

integrate((x^8 + x^4 - 1)*(2*x^4 - 1)^(1/4)/((x^4 - 1)*x^6), x)

Giac [A] (verification not implemented)

none

Time = 0.69 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt [4]{-1+2 x^4} \left (-1+x^4+x^8\right )}{x^6 \left (-1+x^4\right )} \, dx=\frac {1}{2} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{2 \, x}\right ) + \frac {1}{4} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + \frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} - \frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} {\left (\frac {1}{x^{4}} - 2\right )}}{5 \, x} - \frac {1}{2} \, \arctan \left (\frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \log \left (\frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {1}{4} \, \log \left ({\left | \frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{x} - 1 \right |}\right ) \]

[In]

integrate((2*x^4-1)^(1/4)*(x^8+x^4-1)/x^6/(x^4-1),x, algorithm="giac")

[Out]

1/2*2^(1/4)*arctan(1/2*2^(3/4)*(2*x^4 - 1)^(1/4)/x) + 1/4*2^(1/4)*log(2^(1/4) + (2*x^4 - 1)^(1/4)/x) - 1/4*2^(
1/4)*log(2^(1/4) - (2*x^4 - 1)^(1/4)/x) - 1/5*(2*x^4 - 1)^(1/4)*(1/x^4 - 2)/x - 1/2*arctan((2*x^4 - 1)^(1/4)/x
) - 1/4*log((2*x^4 - 1)^(1/4)/x + 1) + 1/4*log(abs((2*x^4 - 1)^(1/4)/x - 1))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{-1+2 x^4} \left (-1+x^4+x^8\right )}{x^6 \left (-1+x^4\right )} \, dx=\int \frac {{\left (2\,x^4-1\right )}^{1/4}\,\left (x^8+x^4-1\right )}{x^6\,\left (x^4-1\right )} \,d x \]

[In]

int(((2*x^4 - 1)^(1/4)*(x^4 + x^8 - 1))/(x^6*(x^4 - 1)),x)

[Out]

int(((2*x^4 - 1)^(1/4)*(x^4 + x^8 - 1))/(x^6*(x^4 - 1)), x)