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

Optimal. Leaf size=225 \[ -\frac {1}{8} \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{x^4-1}}{\sqrt {x^4-1}-x^2}\right )+\frac {1}{8} \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{x^4-1}}{\sqrt {x^4-1}-x^2}\right )-\frac {1}{8} \sqrt {2+\sqrt {2}} \tanh ^{-1}\left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{x^4-1}}{\sqrt {x^4-1}+x^2}\right )+\frac {1}{8} \sqrt {2-\sqrt {2}} \tanh ^{-1}\left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{x^4-1}}{\sqrt {x^4-1}+x^2}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 242, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 11, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {1528, 377, 211, 1165, 628, 1162, 617, 204, 212, 206, 203} \begin {gather*} \frac {1}{4} (-1)^{5/8} \tan ^{-1}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{x^4-1}}\right )+\frac {(-1)^{5/8} \tan ^{-1}\left (1-\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{x^4-1}}\right )}{4 \sqrt {2}}-\frac {(-1)^{5/8} \tan ^{-1}\left (\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{x^4-1}}+1\right )}{4 \sqrt {2}}+\frac {1}{4} (-1)^{5/8} \tanh ^{-1}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{x^4-1}}\right )+\frac {(-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 )}{8 \sqrt {2}}-\frac {(-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 )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

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

Rule 204

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

Rule 206

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

Rule 211

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 212

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] &&
!GtQ[a/b, 0]

Rule 377

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 617

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 628

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 1162

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 1165

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 1528

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*} \int \frac {x^4}{\sqrt [4]{-1+x^4} \left (1-2 x^4+2 x^8\right )} \, dx &=\int \left (\frac {1-i}{\sqrt [4]{-1+x^4} \left ((-2-2 i)+4 x^4\right )}+\frac {1+i}{\sqrt [4]{-1+x^4} \left ((-2+2 i)+4 x^4\right )}\right ) \, dx\\ &=(1-i) \int \frac {1}{\sqrt [4]{-1+x^4} \left ((-2-2 i)+4 x^4\right )} \, dx+(1+i) \int \frac {1}{\sqrt [4]{-1+x^4} \left ((-2+2 i)+4 x^4\right )} \, dx\\ &=(1-i) \operatorname {Subst}\left (\int \frac {1}{(-2-2 i)-(2-2 i) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+(1+i) \operatorname {Subst}\left (\int \frac {1}{(-2+2 i)-(2+2 i) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=-\frac {i \operatorname {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 )}{\sqrt {2}}-\frac {i \operatorname {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 )}{\sqrt {2}}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1}-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1}+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}\\ &=\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (-1)^{7/8} \tan ^{-1}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (-1)^{7/8} \tanh ^{-1}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}-\left (\left (\frac {1}{16}-\frac {i}{16}\right ) (-1)^{7/8}\right ) \operatorname {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 )-\left (\left (\frac {1}{16}-\frac {i}{16}\right ) (-1)^{7/8}\right ) \operatorname {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 )-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \operatorname {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 )}{\sqrt {2}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \operatorname {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 )}{\sqrt {2}}\\ &=\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (-1)^{7/8} \tan ^{-1}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (-1)^{7/8} \tanh ^{-1}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}+\left (\frac {1}{16}-\frac {i}{16}\right ) (-1)^{7/8} \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 )-\left (\frac {1}{16}-\frac {i}{16}\right ) (-1)^{7/8} \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 )-\left (\left (-\frac {1}{8}+\frac {i}{8}\right ) (-1)^{7/8}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )-\left (\left (\frac {1}{8}-\frac {i}{8}\right ) (-1)^{7/8}\right ) \operatorname {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 {\left (\frac {1}{4}-\frac {i}{4}\right ) (-1)^{7/8} \tan ^{-1}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}+\left (\frac {1}{8}-\frac {i}{8}\right ) (-1)^{7/8} \tan ^{-1}\left (1-\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )-\left (\frac {1}{8}-\frac {i}{8}\right ) (-1)^{7/8} \tan ^{-1}\left (1+\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (-1)^{7/8} \tanh ^{-1}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}+\left (\frac {1}{16}-\frac {i}{16}\right ) (-1)^{7/8} \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 )-\left (\frac {1}{16}-\frac {i}{16}\right ) (-1)^{7/8} \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 )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.18, size = 201, normalized size = 0.89 \begin {gather*} \left (\frac {1}{16}-\frac {i}{16}\right ) (-1)^{5/8} \left ((2+2 i) \tan ^{-1}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{x^4-1}}\right )+(2+2 i) \tanh ^{-1}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{x^4-1}}\right )+\sqrt [4]{-1} \left (2 \tan ^{-1}\left (1-\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{x^4-1}}\right )-2 \tan ^{-1}\left (\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{x^4-1}}+1\right )-\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 )+\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 )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.76, size = 225, normalized size = 1.00 \begin {gather*} -\frac {1}{8} \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{-1+x^4}}{-x^2+\sqrt {-1+x^4}}\right )+\frac {1}{8} \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{-1+x^4}}{-x^2+\sqrt {-1+x^4}}\right )-\frac {1}{8} \sqrt {2+\sqrt {2}} \tanh ^{-1}\left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{-1+x^4}}{x^2+\sqrt {-1+x^4}}\right )+\frac {1}{8} \sqrt {2-\sqrt {2}} \tanh ^{-1}\left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{-1+x^4}}{x^2+\sqrt {-1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{{\left (2 \, x^{8} - 2 \, x^{4} + 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 6.38, size = 466, normalized size = 2.07

method	result	size
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{8}+1\right )^{7} \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{8}+1\right )^{7} x^{4}+2 \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{8}+1\right )^{5} x^{2}-\RootOf \left (\textit {\_Z}^{8}+1\right )^{3} x^{4}-2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{8}+1\right )^{2} x^{3}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{8}+1\right )^{3}}{\RootOf \left (\textit {\_Z}^{8}+1\right )^{4} x^{4}-x^{4}+1}\right )}{8}+\frac {\RootOf \left (\textit {\_Z}^{8}+1\right )^{5} \ln \left (-\frac {2 \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{8}+1\right )^{7} x^{2}-2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{8}+1\right )^{6} x^{3}+\RootOf \left (\textit {\_Z}^{8}+1\right )^{5} x^{4}-\RootOf \left (\textit {\_Z}^{8}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{8}+1\right )}{\RootOf \left (\textit {\_Z}^{8}+1\right )^{4} x^{4}+x^{4}-1}\right )}{8}+\frac {\RootOf \left (\textit {\_Z}^{8}+1\right ) \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{8}+1\right )^{9} x^{4}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{8}+1\right )^{6} x^{3}+\RootOf \left (\textit {\_Z}^{8}+1\right )^{5} x^{4}-2 \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{8}+1\right )^{3} x^{2}-\RootOf \left (\textit {\_Z}^{8}+1\right )^{5}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x}{\RootOf \left (\textit {\_Z}^{8}+1\right )^{4} x^{4}+x^{4}-1}\right )}{8}+\frac {\RootOf \left (\textit {\_Z}^{8}+1\right )^{3} \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{8}+1\right )^{11} x^{4}-\RootOf \left (\textit {\_Z}^{8}+1\right )^{7} x^{4}+\RootOf \left (\textit {\_Z}^{8}+1\right )^{7}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{8}+1\right )^{2} x^{3}+2 \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{8}+1\right ) x^{2}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x}{\RootOf \left (\textit {\_Z}^{8}+1\right )^{4} x^{4}-x^{4}+1}\right )}{8}\)	\(466\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*RootOf(_Z^8+1)^7*ln(-(-RootOf(_Z^8+1)^7*x^4+2*(x^4-1)^(1/2)*RootOf(_Z^8+1)^5*x^2-RootOf(_Z^8+1)^3*x^4-2*(
x^4-1)^(1/4)*RootOf(_Z^8+1)^2*x^3+2*(x^4-1)^(3/4)*x+RootOf(_Z^8+1)^3)/(RootOf(_Z^8+1)^4*x^4-x^4+1))+1/8*RootOf
(_Z^8+1)^5*ln(-(2*(x^4-1)^(1/2)*RootOf(_Z^8+1)^7*x^2-2*(x^4-1)^(1/4)*RootOf(_Z^8+1)^6*x^3+RootOf(_Z^8+1)^5*x^4
-RootOf(_Z^8+1)*x^4+2*(x^4-1)^(3/4)*x+RootOf(_Z^8+1))/(RootOf(_Z^8+1)^4*x^4+x^4-1))+1/8*RootOf(_Z^8+1)*ln(-(-R
ootOf(_Z^8+1)^9*x^4+2*(x^4-1)^(1/4)*RootOf(_Z^8+1)^6*x^3+RootOf(_Z^8+1)^5*x^4-2*(x^4-1)^(1/2)*RootOf(_Z^8+1)^3
*x^2-RootOf(_Z^8+1)^5+2*(x^4-1)^(3/4)*x)/(RootOf(_Z^8+1)^4*x^4+x^4-1))+1/8*RootOf(_Z^8+1)^3*ln(-(-RootOf(_Z^8+
1)^11*x^4-RootOf(_Z^8+1)^7*x^4+RootOf(_Z^8+1)^7+2*(x^4-1)^(1/4)*RootOf(_Z^8+1)^2*x^3+2*(x^4-1)^(1/2)*RootOf(_Z
^8+1)*x^2+2*(x^4-1)^(3/4)*x)/(RootOf(_Z^8+1)^4*x^4-x^4+1))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{{\left (2 \, x^{8} - 2 \, x^{4} + 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4}{{\left (x^4-1\right )}^{1/4}\,\left (2\,x^8-2\,x^4+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________