3.11.40 \(\int \frac {\sqrt [4]{-1+x^4} (1-x^4+x^8)}{x^6 (1+2 x^8)} \, dx\) [1040]

Optimal. Leaf size=78 \[ \frac {\sqrt [4]{-1+x^4} \left (-1+6 x^4\right )}{5 x^5}-\frac {3}{8} \text {RootSum}\left [3-2 \text {$\#$1}^4+\text {$\#$1}^8\& ,\frac {-\log (x)+\log \left (\sqrt [4]{-1+x^4}-x \text {$\#$1}\right )}{-\text {$\#$1}^3+\text {$\#$1}^7}\& \right ] \]

[Out]

Unintegrable

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Rubi [C] Result contains complex when optimal does not.
time = 1.02, antiderivative size = 560, normalized size of antiderivative = 7.18, number of steps used = 43, number of rules used = 12, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6857, 270, 283, 338, 304, 209, 212, 1543, 525, 524, 1533, 508} \begin {gather*} -\frac {\sqrt [4]{x^4-1} x^3 F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,-i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}-\frac {\sqrt [4]{x^4-1} x^3 F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}-\frac {i \text {ArcTan}\left (\frac {\sqrt [4]{\sqrt {2}-2 i} x}{\sqrt [8]{2} \sqrt [4]{x^4-1}}\right )}{2 \sqrt [8]{2} \left (\sqrt {2}-2 i\right )^{3/4}}+\frac {\text {ArcTan}\left (\frac {\sqrt [4]{\sqrt {2}-2 i} x}{\sqrt [8]{2} \sqrt [4]{x^4-1}}\right )}{2\ 2^{5/8} \left (\sqrt {2}-2 i\right )^{3/4}}+\frac {i \text {ArcTan}\left (\frac {\sqrt [4]{\sqrt {2}+2 i} x}{\sqrt [8]{2} \sqrt [4]{x^4-1}}\right )}{2 \sqrt [8]{2} \left (\sqrt {2}+2 i\right )^{3/4}}+\frac {\text {ArcTan}\left (\frac {\sqrt [4]{\sqrt {2}+2 i} x}{\sqrt [8]{2} \sqrt [4]{x^4-1}}\right )}{2\ 2^{5/8} \left (\sqrt {2}+2 i\right )^{3/4}}+\frac {\sqrt [4]{x^4-1}}{x}+\frac {i \tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {2}-2 i} x}{\sqrt [8]{2} \sqrt [4]{x^4-1}}\right )}{2 \sqrt [8]{2} \left (\sqrt {2}-2 i\right )^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {2}-2 i} x}{\sqrt [8]{2} \sqrt [4]{x^4-1}}\right )}{2\ 2^{5/8} \left (\sqrt {2}-2 i\right )^{3/4}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {2}+2 i} x}{\sqrt [8]{2} \sqrt [4]{x^4-1}}\right )}{2 \sqrt [8]{2} \left (\sqrt {2}+2 i\right )^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {2}+2 i} x}{\sqrt [8]{2} \sqrt [4]{x^4-1}}\right )}{2\ 2^{5/8} \left (\sqrt {2}+2 i\right )^{3/4}}+\frac {\left (x^4-1\right )^{5/4}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1 + x^4)^(1/4)/x + (-1 + x^4)^(5/4)/(5*x^5) - (x^3*(-1 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, x^4, (-I)*Sq
rt[2]*x^4])/(6*(1 - x^4)^(1/4)) - (x^3*(-1 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, x^4, I*Sqrt[2]*x^4])/(6*(1
- x^4)^(1/4)) + ArcTan[((-2*I + Sqrt[2])^(1/4)*x)/(2^(1/8)*(-1 + x^4)^(1/4))]/(2*2^(5/8)*(-2*I + Sqrt[2])^(3/
4)) - ((I/2)*ArcTan[((-2*I + Sqrt[2])^(1/4)*x)/(2^(1/8)*(-1 + x^4)^(1/4))])/(2^(1/8)*(-2*I + Sqrt[2])^(3/4)) +
ArcTan[((2*I + Sqrt[2])^(1/4)*x)/(2^(1/8)*(-1 + x^4)^(1/4))]/(2*2^(5/8)*(2*I + Sqrt[2])^(3/4)) + ((I/2)*ArcTa
n[((2*I + Sqrt[2])^(1/4)*x)/(2^(1/8)*(-1 + x^4)^(1/4))])/(2^(1/8)*(2*I + Sqrt[2])^(3/4)) - ArcTanh[((-2*I + Sq
rt[2])^(1/4)*x)/(2^(1/8)*(-1 + x^4)^(1/4))]/(2*2^(5/8)*(-2*I + Sqrt[2])^(3/4)) + ((I/2)*ArcTanh[((-2*I + Sqrt[
2])^(1/4)*x)/(2^(1/8)*(-1 + x^4)^(1/4))])/(2^(1/8)*(-2*I + Sqrt[2])^(3/4)) - ArcTanh[((2*I + Sqrt[2])^(1/4)*x)
/(2^(1/8)*(-1 + x^4)^(1/4))]/(2*2^(5/8)*(2*I + Sqrt[2])^(3/4)) - ((I/2)*ArcTanh[((2*I + Sqrt[2])^(1/4)*x)/(2^(
1/8)*(-1 + x^4)^(1/4))])/(2^(1/8)*(2*I + Sqrt[2])^(3/4))

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 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 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 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 508

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[k*(a^(p + (m + 1)/n)/n), Subst[Int[x^(k*((m + 1)/n) - 1)*((c - (b*c - a*d)*x^k)^q/(1 - b*x^k)^(p +
q + (m + 1)/n + 1)), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 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 1533

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.)), x_Symbol] :> Dist[e*(f^n/c
), Int[(f*x)^(m - n)*(d + e*x^n)^(q - 1), x], x] - Dist[f^n/c, Int[(f*x)^(m - n)*(d + e*x^n)^(q - 1)*(Simp[a*e
- c*d*x^n, x]/(a + c*x^(2*n))), x], x] /; FreeQ[{a, c, d, e, f}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] &&  !Intege
rQ[q] && GtQ[q, 0] && GtQ[m, n - 1] && LeQ[m, 2*n - 1]

Rule 1543

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

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*} \int \frac {\sqrt [4]{-1+x^4} \left (1-x^4+x^8\right )}{x^6 \left (1+2 x^8\right )} \, dx &=\int \left (\frac {\sqrt [4]{-1+x^4}}{x^6}-\frac {\sqrt [4]{-1+x^4}}{x^2}+\frac {x^2 \sqrt [4]{-1+x^4} \left (-1+2 x^4\right )}{1+2 x^8}\right ) \, dx\\ &=\int \frac {\sqrt [4]{-1+x^4}}{x^6} \, dx-\int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx+\int \frac {x^2 \sqrt [4]{-1+x^4} \left (-1+2 x^4\right )}{1+2 x^8} \, dx\\ &=\frac {\sqrt [4]{-1+x^4}}{x}+\frac {\left (-1+x^4\right )^{5/4}}{5 x^5}-\int \frac {x^2}{\left (-1+x^4\right )^{3/4}} \, dx+\int \left (-\frac {x^2 \sqrt [4]{-1+x^4}}{1+2 x^8}+\frac {2 x^6 \sqrt [4]{-1+x^4}}{1+2 x^8}\right ) \, dx\\ &=\frac {\sqrt [4]{-1+x^4}}{x}+\frac {\left (-1+x^4\right )^{5/4}}{5 x^5}+2 \int \frac {x^6 \sqrt [4]{-1+x^4}}{1+2 x^8} \, dx-\int \frac {x^2 \sqrt [4]{-1+x^4}}{1+2 x^8} \, dx-\text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=\frac {\sqrt [4]{-1+x^4}}{x}+\frac {\left (-1+x^4\right )^{5/4}}{5 x^5}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\int \frac {x^2}{\left (-1+x^4\right )^{3/4}} \, dx-\int \frac {x^2 \left (1+2 x^4\right )}{\left (-1+x^4\right )^{3/4} \left (1+2 x^8\right )} \, dx-\int \left (-\frac {i x^2 \sqrt [4]{-1+x^4}}{\sqrt {2} \left (-i \sqrt {2}+2 x^4\right )}+\frac {i x^2 \sqrt [4]{-1+x^4}}{\sqrt {2} \left (i \sqrt {2}+2 x^4\right )}\right ) \, dx\\ &=\frac {\sqrt [4]{-1+x^4}}{x}+\frac {\left (-1+x^4\right )^{5/4}}{5 x^5}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {i \int \frac {x^2 \sqrt [4]{-1+x^4}}{-i \sqrt {2}+2 x^4} \, dx}{\sqrt {2}}-\frac {i \int \frac {x^2 \sqrt [4]{-1+x^4}}{i \sqrt {2}+2 x^4} \, dx}{\sqrt {2}}-\int \left (\frac {x^2}{\left (-1+x^4\right )^{3/4} \left (1+2 x^8\right )}+\frac {2 x^6}{\left (-1+x^4\right )^{3/4} \left (1+2 x^8\right )}\right ) \, dx+\text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=\frac {\sqrt [4]{-1+x^4}}{x}+\frac {\left (-1+x^4\right )^{5/4}}{5 x^5}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )-2 \int \frac {x^6}{\left (-1+x^4\right )^{3/4} \left (1+2 x^8\right )} \, dx+\frac {\left (i \sqrt [4]{-1+x^4}\right ) \int \frac {x^2 \sqrt [4]{1-x^4}}{-i \sqrt {2}+2 x^4} \, dx}{\sqrt {2} \sqrt [4]{1-x^4}}-\frac {\left (i \sqrt [4]{-1+x^4}\right ) \int \frac {x^2 \sqrt [4]{1-x^4}}{i \sqrt {2}+2 x^4} \, dx}{\sqrt {2} \sqrt [4]{1-x^4}}-\int \frac {x^2}{\left (-1+x^4\right )^{3/4} \left (1+2 x^8\right )} \, dx\\ &=\frac {\sqrt [4]{-1+x^4}}{x}+\frac {\left (-1+x^4\right )^{5/4}}{5 x^5}-\frac {x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,-i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}-\frac {x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}-2 \int \left (\frac {x^2}{2 \left (-1+x^4\right )^{3/4} \left (-i \sqrt {2}+2 x^4\right )}+\frac {x^2}{2 \left (-1+x^4\right )^{3/4} \left (i \sqrt {2}+2 x^4\right )}\right ) \, dx-\int \left (-\frac {i x^2}{\sqrt {2} \left (-1+x^4\right )^{3/4} \left (-i \sqrt {2}+2 x^4\right )}+\frac {i x^2}{\sqrt {2} \left (-1+x^4\right )^{3/4} \left (i \sqrt {2}+2 x^4\right )}\right ) \, dx\\ &=\frac {\sqrt [4]{-1+x^4}}{x}+\frac {\left (-1+x^4\right )^{5/4}}{5 x^5}-\frac {x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,-i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}-\frac {x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}+\frac {i \int \frac {x^2}{\left (-1+x^4\right )^{3/4} \left (-i \sqrt {2}+2 x^4\right )} \, dx}{\sqrt {2}}-\frac {i \int \frac {x^2}{\left (-1+x^4\right )^{3/4} \left (i \sqrt {2}+2 x^4\right )} \, dx}{\sqrt {2}}-\int \frac {x^2}{\left (-1+x^4\right )^{3/4} \left (-i \sqrt {2}+2 x^4\right )} \, dx-\int \frac {x^2}{\left (-1+x^4\right )^{3/4} \left (i \sqrt {2}+2 x^4\right )} \, dx\\ &=\frac {\sqrt [4]{-1+x^4}}{x}+\frac {\left (-1+x^4\right )^{5/4}}{5 x^5}-\frac {x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,-i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}-\frac {x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}+\frac {i \text {Subst}\left (\int \frac {x^2}{-i \sqrt {2}-\left (2-i \sqrt {2}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}-\frac {i \text {Subst}\left (\int \frac {x^2}{i \sqrt {2}-\left (2+i \sqrt {2}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {2}}-\text {Subst}\left (\int \frac {x^2}{-i \sqrt {2}-\left (2-i \sqrt {2}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )-\text {Subst}\left (\int \frac {x^2}{i \sqrt {2}-\left (2+i \sqrt {2}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=\frac {\sqrt [4]{-1+x^4}}{x}+\frac {\left (-1+x^4\right )^{5/4}}{5 x^5}-\frac {x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,-i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}-\frac {x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}+\frac {i \text {Subst}\left (\int \frac {1}{\sqrt [4]{2}-\sqrt {-2 i+\sqrt {2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {-2 i+\sqrt {2}}}-\frac {i \text {Subst}\left (\int \frac {1}{\sqrt [4]{2}+\sqrt {-2 i+\sqrt {2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {-2 i+\sqrt {2}}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt [4]{2}-\sqrt {-2 i+\sqrt {2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {2 \left (-2 i+\sqrt {2}\right )}}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt [4]{2}+\sqrt {-2 i+\sqrt {2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {2 \left (-2 i+\sqrt {2}\right )}}-\frac {i \text {Subst}\left (\int \frac {1}{\sqrt [4]{2}-\sqrt {2 i+\sqrt {2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {2 i+\sqrt {2}}}+\frac {i \text {Subst}\left (\int \frac {1}{\sqrt [4]{2}+\sqrt {2 i+\sqrt {2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {2 i+\sqrt {2}}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt [4]{2}-\sqrt {2 i+\sqrt {2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {2 \left (2 i+\sqrt {2}\right )}}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt [4]{2}+\sqrt {2 i+\sqrt {2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {2 \left (2 i+\sqrt {2}\right )}}\\ &=\frac {\sqrt [4]{-1+x^4}}{x}+\frac {\left (-1+x^4\right )^{5/4}}{5 x^5}-\frac {x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,-i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}-\frac {x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,i \sqrt {2} x^4\right )}{6 \sqrt [4]{1-x^4}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{-2 i+\sqrt {2}} x}{\sqrt [8]{2} \sqrt [4]{-1+x^4}}\right )}{2\ 2^{5/8} \left (-2 i+\sqrt {2}\right )^{3/4}}-\frac {i \tan ^{-1}\left (\frac {\sqrt [4]{-2 i+\sqrt {2}} x}{\sqrt [8]{2} \sqrt [4]{-1+x^4}}\right )}{2 \sqrt [8]{2} \left (-2 i+\sqrt {2}\right )^{3/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2 i+\sqrt {2}} x}{\sqrt [8]{2} \sqrt [4]{-1+x^4}}\right )}{2\ 2^{5/8} \left (2 i+\sqrt {2}\right )^{3/4}}+\frac {i \tan ^{-1}\left (\frac {\sqrt [4]{2 i+\sqrt {2}} x}{\sqrt [8]{2} \sqrt [4]{-1+x^4}}\right )}{2 \sqrt [8]{2} \left (2 i+\sqrt {2}\right )^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{-2 i+\sqrt {2}} x}{\sqrt [8]{2} \sqrt [4]{-1+x^4}}\right )}{2\ 2^{5/8} \left (-2 i+\sqrt {2}\right )^{3/4}}+\frac {i \tanh ^{-1}\left (\frac {\sqrt [4]{-2 i+\sqrt {2}} x}{\sqrt [8]{2} \sqrt [4]{-1+x^4}}\right )}{2 \sqrt [8]{2} \left (-2 i+\sqrt {2}\right )^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2 i+\sqrt {2}} x}{\sqrt [8]{2} \sqrt [4]{-1+x^4}}\right )}{2\ 2^{5/8} \left (2 i+\sqrt {2}\right )^{3/4}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt [4]{2 i+\sqrt {2}} x}{\sqrt [8]{2} \sqrt [4]{-1+x^4}}\right )}{2 \sqrt [8]{2} \left (2 i+\sqrt {2}\right )^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 78, normalized size = 1.00 \begin {gather*} \frac {\sqrt [4]{-1+x^4} \left (-1+6 x^4\right )}{5 x^5}-\frac {3}{8} \text {RootSum}\left [3-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{-1+x^4}-x \text {$\#$1}\right )}{-\text {$\#$1}^3+\text {$\#$1}^7}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1 + x^4)^(1/4)*(-1 + 6*x^4))/(5*x^5) - (3*RootSum[3 - 2*#1^4 + #1^8 & , (-Log[x] + Log[(-1 + x^4)^(1/4) - x
*#1])/(-#1^3 + #1^7) & ])/8

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (x^{4}-1\right )^{\frac {1}{4}} \left (x^{8}-x^{4}+1\right )}{x^{6} \left (2 x^{8}+1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
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^4-1)^(1/4)*(x^8-x^4+1)/x^6/(2*x^8+1),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (x^{8} - x^{4} + 1\right )}{x^{6} \cdot \left (2 x^{8} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^4-1\right )}^{1/4}\,\left (x^8-x^4+1\right )}{x^6\,\left (2\,x^8+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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