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

Optimal. Leaf size=272 \[ \frac {1}{8} \text {RootSum}\left [\text {$\#$1}^8-2 \text {$\#$1}^4+2\& ,\frac {-3 \text {$\#$1}^4 \log \left (\sqrt [4]{x^4-x^2}-\text {$\#$1} x\right )+3 \text {$\#$1}^4 \log (x)-2 \log \left (\sqrt [4]{x^4-x^2}-\text {$\#$1} x\right )+2 \log (x)}{\text {$\#$1}^5-\text {$\#$1}}\& \right ]+\frac {5}{8} \text {RootSum}\left [\text {$\#$1}^8-2 \text {$\#$1}^4+2\& ,\frac {\text {$\#$1}^3 \log \left (\sqrt [4]{x^4-x^2}-\text {$\#$1} x\right )-\text {$\#$1}^3 \log (x)}{\text {$\#$1}^4-1}\& \right ]-\frac {\left (x^4-x^2\right )^{3/4}}{x \left (x^2-1\right )}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^2}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-x^2}}\right )}{2 \sqrt [4]{2}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^2}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-x^2}}\right )}{2 \sqrt [4]{2}} \]

________________________________________________________________________________________

Rubi [C]  time = 0.75, antiderivative size = 559, normalized size of antiderivative = 2.06, number of steps used = 27, number of rules used = 12, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2056, 6715, 6725, 240, 212, 206, 203, 2073, 1152, 380, 377, 1429} \begin {gather*} -\frac {x (x+1) \left (\frac {1-x}{x+1}\right )^{5/4} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {3}{2};\frac {2 x}{x+1}\right )}{2 (1-x) \sqrt [4]{x^4-x^2}}-\frac {x \sqrt [4]{\frac {1-x}{x+1}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {2 x}{x+1}\right )}{2 \sqrt [4]{x^4-x^2}}+\frac {\sqrt {x} \sqrt [4]{x^2-1} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}}-\frac {\sqrt {x} \sqrt [4]{x^2-1} \tan ^{-1}\left (\frac {\sqrt [4]{1-i} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{2 \sqrt [4]{1-i} \sqrt [4]{x^4-x^2}}-\frac {\sqrt {x} \sqrt [4]{x^2-1} \tan ^{-1}\left (\frac {\sqrt [4]{1+i} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{2 \sqrt [4]{1+i} \sqrt [4]{x^4-x^2}}-\frac {\sqrt {x} \sqrt [4]{x^2-1} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^4-x^2}}+\frac {\sqrt {x} \sqrt [4]{x^2-1} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}}-\frac {\sqrt {x} \sqrt [4]{x^2-1} \tanh ^{-1}\left (\frac {\sqrt [4]{1-i} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{2 \sqrt [4]{1-i} \sqrt [4]{x^4-x^2}}-\frac {\sqrt {x} \sqrt [4]{x^2-1} \tanh ^{-1}\left (\frac {\sqrt [4]{1+i} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{2 \sqrt [4]{1+i} \sqrt [4]{x^4-x^2}}-\frac {\sqrt {x} \sqrt [4]{x^2-1} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^4-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-1 + x^2)^(1/4)*ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)])/(-x^2 + x^4)^(1/4) - (Sqrt[x]*(-1 + x^2)^(1/4)*Arc
Tan[((1 - I)^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)])/(2*(1 - I)^(1/4)*(-x^2 + x^4)^(1/4)) - (Sqrt[x]*(-1 + x^2)^(1/4
)*ArcTan[((1 + I)^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)])/(2*(1 + I)^(1/4)*(-x^2 + x^4)^(1/4)) - (Sqrt[x]*(-1 + x^2)
^(1/4)*ArcTan[(2^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)])/(2*2^(1/4)*(-x^2 + x^4)^(1/4)) + (Sqrt[x]*(-1 + x^2)^(1/4)*
ArcTanh[Sqrt[x]/(-1 + x^2)^(1/4)])/(-x^2 + x^4)^(1/4) - (Sqrt[x]*(-1 + x^2)^(1/4)*ArcTanh[((1 - I)^(1/4)*Sqrt[
x])/(-1 + x^2)^(1/4)])/(2*(1 - I)^(1/4)*(-x^2 + x^4)^(1/4)) - (Sqrt[x]*(-1 + x^2)^(1/4)*ArcTanh[((1 + I)^(1/4)
*Sqrt[x])/(-1 + x^2)^(1/4)])/(2*(1 + I)^(1/4)*(-x^2 + x^4)^(1/4)) - (Sqrt[x]*(-1 + x^2)^(1/4)*ArcTanh[(2^(1/4)
*Sqrt[x])/(-1 + x^2)^(1/4)])/(2*2^(1/4)*(-x^2 + x^4)^(1/4)) - (x*((1 - x)/(1 + x))^(1/4)*Hypergeometric2F1[1/4
, 1/2, 3/2, (2*x)/(1 + x)])/(2*(-x^2 + x^4)^(1/4)) - (x*((1 - x)/(1 + x))^(5/4)*(1 + x)*Hypergeometric2F1[1/2,
5/4, 3/2, (2*x)/(1 + x)])/(2*(1 - x)*(-x^2 + x^4)^(1/4))

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 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 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 240

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 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 380

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(x*(a + b*x^n)^p*Hypergeome
tric2F1[1/n, -p, 1 + 1/n, -(((b*c - a*d)*x^n)/(a*(c + d*x^n)))])/(c*((c*(a + b*x^n))/(a*(c + d*x^n)))^p*(c + d
*x^n)^(1/n + p)), x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0]

Rule 1152

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

Rule 1429

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

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 2073

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(PP /. x ->
x^2)^p*Q^q, x], x] /;  !SumQ[NonfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x^2] && PolyQ[Q, x] && ILtQ[p,
0]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6725

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*} \int \frac {1+x^8}{\sqrt [4]{-x^2+x^4} \left (-1+x^8\right )} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \int \frac {1+x^8}{\sqrt {x} \sqrt [4]{-1+x^2} \left (-1+x^8\right )} \, dx}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1+x^{16}}{\sqrt [4]{-1+x^4} \left (-1+x^{16}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{\sqrt [4]{-1+x^4}}+\frac {2}{\sqrt [4]{-1+x^4} \left (-1+x^{16}\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (4 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4} \left (-1+x^{16}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (4 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{8 \left (-1+x^2\right ) \sqrt [4]{-1+x^4}}-\frac {1}{8 \left (1+x^2\right ) \sqrt [4]{-1+x^4}}-\frac {1}{4 \sqrt [4]{-1+x^4} \left (1+x^4\right )}-\frac {1}{2 \sqrt [4]{-1+x^4} \left (1+x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-1+x^2\right ) \sqrt [4]{-1+x^4}} \, dx,x,\sqrt {x}\right )}{2 \sqrt [4]{-x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt [4]{-1+x^4}} \, dx,x,\sqrt {x}\right )}{2 \sqrt [4]{-x^2+x^4}}+\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4} \left (1+x^4\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4} \left (1+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\left (\sqrt [4]{-1+x} \sqrt {x} \sqrt [4]{1+x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^2} \left (1+x^2\right )^{5/4}} \, dx,x,\sqrt {x}\right )}{2 \sqrt [4]{-x^2+x^4}}+\frac {\left (\sqrt [4]{-1+x} \sqrt {x} \sqrt [4]{1+x}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-1+x^2\right )^{5/4} \sqrt [4]{1+x^2}} \, dx,x,\sqrt {x}\right )}{2 \sqrt [4]{-x^2+x^4}}-\frac {\left (i \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (i-x^4\right ) \sqrt [4]{-1+x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\left (i \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4} \left (i+x^4\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-2 x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {x \sqrt [4]{\frac {1-x}{1+x}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {2 x}{1+x}\right )}{2 \sqrt [4]{-x^2+x^4}}-\frac {x \left (\frac {1-x}{1+x}\right )^{5/4} (1+x) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {3}{2};\frac {2 x}{1+x}\right )}{2 (1-x) \sqrt [4]{-x^2+x^4}}-\frac {\left (i \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{i-(1+i) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\left (i \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{i+(1-i) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{-x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{-x^2+x^4}}\\ &=\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{2} \sqrt [4]{-x^2+x^4}}+\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{2} \sqrt [4]{-x^2+x^4}}-\frac {x \sqrt [4]{\frac {1-x}{1+x}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {2 x}{1+x}\right )}{2 \sqrt [4]{-x^2+x^4}}-\frac {x \left (\frac {1-x}{1+x}\right )^{5/4} (1+x) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {3}{2};\frac {2 x}{1+x}\right )}{2 (1-x) \sqrt [4]{-x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {1-i} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{-x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {1-i} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{-x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {1+i} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{-x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {1+i} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{-x^2+x^4}}\\ &=\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{1-i} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{1-i} \sqrt [4]{-x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{1+i} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{1+i} \sqrt [4]{-x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{2} \sqrt [4]{-x^2+x^4}}+\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{1-i} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{1-i} \sqrt [4]{-x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{1+i} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{1+i} \sqrt [4]{-x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{2 \sqrt [4]{2} \sqrt [4]{-x^2+x^4}}-\frac {x \sqrt [4]{\frac {1-x}{1+x}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {2 x}{1+x}\right )}{2 \sqrt [4]{-x^2+x^4}}-\frac {x \left (\frac {1-x}{1+x}\right )^{5/4} (1+x) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {3}{2};\frac {2 x}{1+x}\right )}{2 (1-x) \sqrt [4]{-x^2+x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 0.69, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+x^8}{\sqrt [4]{-x^2+x^4} \left (-1+x^8\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.00, size = 167, normalized size = 0.61 \begin {gather*} -\frac {\left (-x^2+x^4\right )^{3/4}}{x \left (-1+x^2\right )}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^2+x^4}}\right )}{2 \sqrt [4]{2}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^2+x^4}}\right )}{2 \sqrt [4]{2}}+\frac {1}{4} \text {RootSum}\left [2-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{-x^2+x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((-x^2 + x^4)^(3/4)/(x*(-1 + x^2))) + ArcTan[x/(-x^2 + x^4)^(1/4)] - ArcTan[(2^(1/4)*x)/(-x^2 + x^4)^(1/4)]/(
2*2^(1/4)) + ArcTanh[x/(-x^2 + x^4)^(1/4)] - ArcTanh[(2^(1/4)*x)/(-x^2 + x^4)^(1/4)]/(2*2^(1/4)) + RootSum[2 -
2*#1^4 + #1^8 & , (-Log[x] + Log[(-x^2 + x^4)^(1/4) - x*#1])/#1 & ]/4

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

giac [B]  time = 0.40, size = 282, normalized size = 1.04 \begin {gather*} -\frac {1}{4} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{8} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{8} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} - {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + 4 i \, \left (-\frac {1}{131072} i + \frac {1}{131072}\right )^{\frac {1}{4}} \log \left (i \, \left (-2251799813685248 i + 2251799813685248\right )^{\frac {1}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 8192\right ) - 4 i \, \left (-\frac {1}{131072} i + \frac {1}{131072}\right )^{\frac {1}{4}} \log \left (-i \, \left (-2251799813685248 i + 2251799813685248\right )^{\frac {1}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 8192\right ) + i \, \left (\frac {1}{512} i + \frac {1}{512}\right )^{\frac {1}{4}} \log \left (\left (134217728 i + 134217728\right )^{\frac {1}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 128 i\right ) - \left (\frac {1}{512} i + \frac {1}{512}\right )^{\frac {1}{4}} \log \left (i \, \left (134217728 i + 134217728\right )^{\frac {1}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 128 i\right ) + \left (\frac {1}{512} i + \frac {1}{512}\right )^{\frac {1}{4}} \log \left (-i \, \left (134217728 i + 134217728\right )^{\frac {1}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 128 i\right ) - i \, \left (\frac {1}{512} i + \frac {1}{512}\right )^{\frac {1}{4}} \log \left (-\left (134217728 i + 134217728\right )^{\frac {1}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 128 i\right ) + \left (-\frac {1}{512} i + \frac {1}{512}\right )^{\frac {1}{4}} \log \left (\left (-134217728 i + 134217728\right )^{\frac {1}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 128\right ) - \left (-\frac {1}{512} i + \frac {1}{512}\right )^{\frac {1}{4}} \log \left (-\left (-134217728 i + 134217728\right )^{\frac {1}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 128\right ) + \frac {1}{{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}} + \arctan \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left (-{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*2^(3/4)*arctan(1/2*2^(3/4)*(-1/x^2 + 1)^(1/4)) + 1/8*2^(3/4)*log(2^(1/4) + (-1/x^2 + 1)^(1/4)) - 1/8*2^(3
/4)*log(2^(1/4) - (-1/x^2 + 1)^(1/4)) + 4*I*(-1/131072*I + 1/131072)^(1/4)*log(I*(-2251799813685248*I + 225179
9813685248)^(1/4)*(-1/x^2 + 1)^(1/4) + 8192) - 4*I*(-1/131072*I + 1/131072)^(1/4)*log(-I*(-2251799813685248*I
+ 2251799813685248)^(1/4)*(-1/x^2 + 1)^(1/4) + 8192) + I*(1/512*I + 1/512)^(1/4)*log((134217728*I + 134217728)
^(1/4)*(-1/x^2 + 1)^(1/4) - 128*I) - (1/512*I + 1/512)^(1/4)*log(I*(134217728*I + 134217728)^(1/4)*(-1/x^2 + 1
)^(1/4) - 128*I) + (1/512*I + 1/512)^(1/4)*log(-I*(134217728*I + 134217728)^(1/4)*(-1/x^2 + 1)^(1/4) - 128*I)
- I*(1/512*I + 1/512)^(1/4)*log(-(134217728*I + 134217728)^(1/4)*(-1/x^2 + 1)^(1/4) - 128*I) + (-1/512*I + 1/5
12)^(1/4)*log((-134217728*I + 134217728)^(1/4)*(-1/x^2 + 1)^(1/4) + 128) - (-1/512*I + 1/512)^(1/4)*log(-(-134
217728*I + 134217728)^(1/4)*(-1/x^2 + 1)^(1/4) + 128) + 1/(-1/x^2 + 1)^(1/4) + arctan((-1/x^2 + 1)^(1/4)) - 1/
2*log((-1/x^2 + 1)^(1/4) + 1) + 1/2*log(-(-1/x^2 + 1)^(1/4) + 1)

________________________________________________________________________________________

maple [B]  time = 135.70, size = 2036, normalized size = 7.49

method	result	size
risch	\(\text {Expression too large to display}\)	\(2036\)
trager	\(\text {Expression too large to display}\)	\(2042\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(x^2*(x^2-1))^(1/4)+1/2*RootOf(_Z^2+1)*ln((-2*(x^4-x^2)^(1/2)*RootOf(_Z^2+1)*x+2*RootOf(_Z^2+1)*x^3+2*(x^4-
x^2)^(3/4)-2*x^2*(x^4-x^2)^(1/4)-RootOf(_Z^2+1)*x)/x)+1/2*ln((2*(x^4-x^2)^(3/4)+2*(x^4-x^2)^(1/2)*x+2*x^2*(x^4
-x^2)^(1/4)+2*x^3-x)/x)+1/8*RootOf(_Z^2+1)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*ln(-(-367*(x^4-x^2)^(1/2)*RootOf(_Z
^2+1)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^3*x-1519*(x^4-x^2)^(1/2)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^3*x-3038*RootOf
(_Z^2+1)*(x^4-x^2)^(1/4)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^2*x^2+8*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1
)^2*x^3+734*(x^4-x^2)^(1/4)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^2*x^2-418*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*RootOf(_
Z^2+1)*x^3+7544*RootOf(_Z^2+1)*(x^4-x^2)^(3/4)-32*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)^2*x+4932*Root
Of(_Z^4+8*RootOf(_Z^2+1)-8)*x^3+4608*(x^4-x^2)^(3/4)+1152*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)*x-191
8*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*x)/x/(RootOf(_Z^2+1)*x^2+4*x^2-4*RootOf(_Z^2+1)+1))-1/8*RootOf(_Z^4+8*RootOf
(_Z^2+1)-8)*ln(-(1519*(x^4-x^2)^(1/2)*RootOf(_Z^2+1)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^3*x-367*(x^4-x^2)^(1/2)*R
ootOf(_Z^4+8*RootOf(_Z^2+1)-8)^3*x+3038*RootOf(_Z^2+1)*(x^4-x^2)^(1/4)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^2*x^2-2
74*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)^2*x^3-734*(x^4-x^2)^(1/4)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^2*
x^2+4924*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)*x^3+7544*RootOf(_Z^2+1)*(x^4-x^2)^(3/4)+1096*RootOf(_Z
^4+8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)^2*x+144*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*x^3+4608*(x^4-x^2)^(3/4)-1886*Ro
otOf(_Z^4+8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)*x-56*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*x)/x/(RootOf(_Z^2+1)*x^2+4*x
^2-4*RootOf(_Z^2+1)+1))-1/8*RootOf(_Z^2+1)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*ln((-367*(x^4-x^2)^(1/2)*RootOf(_Z^
2+1)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)^3*x+1519*(x^4-x^2)^(1/2)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)^3*x-3038*RootOf(
_Z^2+1)*(x^4-x^2)^(1/4)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)^2*x^2-8*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)
^2*x^3-734*(x^4-x^2)^(1/4)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)^2*x^2-418*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z
^2+1)*x^3+7544*RootOf(_Z^2+1)*(x^4-x^2)^(3/4)+32*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)^2*x-4932*RootO
f(_Z^4-8*RootOf(_Z^2+1)-8)*x^3-4608*(x^4-x^2)^(3/4)+1152*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)*x+1918
*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*x)/x/(RootOf(_Z^2+1)*x^2-4*x^2-4*RootOf(_Z^2+1)-1))+1/8*RootOf(_Z^4-8*RootOf(
_Z^2+1)-8)*ln((-1519*(x^4-x^2)^(1/2)*RootOf(_Z^2+1)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)^3*x-367*(x^4-x^2)^(1/2)*Ro
otOf(_Z^4-8*RootOf(_Z^2+1)-8)^3*x+3038*RootOf(_Z^2+1)*(x^4-x^2)^(1/4)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)^2*x^2-27
4*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)^2*x^3+734*(x^4-x^2)^(1/4)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)^2*x
^2-4924*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)*x^3+7544*RootOf(_Z^2+1)*(x^4-x^2)^(3/4)+1096*RootOf(_Z^
4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)^2*x+144*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*x^3-4608*(x^4-x^2)^(3/4)+1886*Roo
tOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)*x-56*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*x)/x/(RootOf(_Z^2+1)*x^2-4*x^
2-4*RootOf(_Z^2+1)-1))+1/16*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*ln((-(x^4-x^2)^(1/
2)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)^3*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^3*x+4*(x^4-x^2)^(1/4)*RootOf(_Z^4-8*RootO
f(_Z^2+1)-8)^2*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^2*x^2-12*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^4+8*RootOf(_
Z^2+1)-8)*x^3+32*(x^4-x^2)^(3/4)+4*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*x)/x/(x^2+1
))-1/16*RootOf(_Z^2+1)*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*ln((-(x^4-x^2)^(1/2)*Ro
otOf(_Z^4-8*RootOf(_Z^2+1)-8)^3*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^3*RootOf(_Z^2+1)*x-4*(x^4-x^2)^(1/4)*RootOf(_Z
^4-8*RootOf(_Z^2+1)-8)^2*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)^2*x^2+12*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^4+
8*RootOf(_Z^2+1)-8)*RootOf(_Z^2+1)*x^3-4*RootOf(_Z^4-8*RootOf(_Z^2+1)-8)*RootOf(_Z^4+8*RootOf(_Z^2+1)-8)*RootO
f(_Z^2+1)*x+32*(x^4-x^2)^(3/4))/x/(x^2+1))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________