\(\int \frac {\sqrt [4]{1+x^4} (2+x^4)}{x^2 (-1+2 x^4+x^8)} \, dx\) [1429]

Optimal result

Integrand size = 30, antiderivative size = 101 \[ \int \frac {\sqrt [4]{1+x^4} \left (2+x^4\right )}{x^2 \left (-1+2 x^4+x^8\right )} \, dx=\frac {2 \sqrt [4]{1+x^4}}{x}-\frac {1}{8} \text {RootSum}\left [2-4 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-4 \log (x)+4 \log \left (\sqrt [4]{1+x^4}-x \text {$\#$1}\right )+7 \log (x) \text {$\#$1}^4-7 \log \left (\sqrt [4]{1+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-2 \text {$\#$1}^3+\text {$\#$1}^7}\&\right ] \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(425\) vs. \(2(101)=202\).

Time = 0.83 (sec) , antiderivative size = 425, normalized size of antiderivative = 4.21, number of steps used = 40, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6860, 283, 338, 304, 209, 212, 1542, 524, 1532, 508} \[ \int \frac {\sqrt [4]{1+x^4} \left (2+x^4\right )}{x^2 \left (-1+2 x^4+x^8\right )} \, dx=-\frac {5}{12} \left (2-\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},-x^4,\left (1-\sqrt {2}\right ) x^4\right )-\frac {5}{12} \left (2+\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\left (1+\sqrt {2}\right ) x^4,-x^4\right )+\frac {1}{4} \sqrt [4]{10+7 \sqrt {2}} \arctan \left (\frac {\sqrt [4]{2-\sqrt {2}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{4} \sqrt [4]{2-\sqrt {2}} \arctan \left (\frac {\sqrt [4]{2-\sqrt {2}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{4} \sqrt [4]{2+\sqrt {2}} \arctan \left (\frac {\sqrt [4]{2+\sqrt {2}} x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{4} \sqrt [4]{10-7 \sqrt {2}} \arctan \left (\frac {\sqrt [4]{2+\sqrt {2}} x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{4} \sqrt [4]{10+7 \sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2-\sqrt {2}} x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{4} \sqrt [4]{2-\sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2-\sqrt {2}} x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{4} \sqrt [4]{2+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2+\sqrt {2}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{4} \sqrt [4]{10-7 \sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2+\sqrt {2}} x}{\sqrt [4]{x^4+1}}\right )+\frac {2 \sqrt [4]{x^4+1}}{x} \]

[In]

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

[Out]

(2*(1 + x^4)^(1/4))/x - (5*(2 - Sqrt[2])*x^3*AppellF1[3/4, -1/4, 1, 7/4, -x^4, (1 - Sqrt[2])*x^4])/12 - (5*(2
+ Sqrt[2])*x^3*AppellF1[3/4, 1, -1/4, 7/4, (1 + Sqrt[2])*x^4, -x^4])/12 + ((2 - Sqrt[2])^(1/4)*ArcTan[((2 - Sq
rt[2])^(1/4)*x)/(1 + x^4)^(1/4)])/4 + ((10 + 7*Sqrt[2])^(1/4)*ArcTan[((2 - Sqrt[2])^(1/4)*x)/(1 + x^4)^(1/4)])
/4 - ((10 - 7*Sqrt[2])^(1/4)*ArcTan[((2 + Sqrt[2])^(1/4)*x)/(1 + x^4)^(1/4)])/4 + ((2 + Sqrt[2])^(1/4)*ArcTan[
((2 + Sqrt[2])^(1/4)*x)/(1 + x^4)^(1/4)])/4 - ((2 - Sqrt[2])^(1/4)*ArcTanh[((2 - Sqrt[2])^(1/4)*x)/(1 + x^4)^(
1/4)])/4 - ((10 + 7*Sqrt[2])^(1/4)*ArcTanh[((2 - Sqrt[2])^(1/4)*x)/(1 + x^4)^(1/4)])/4 + ((10 - 7*Sqrt[2])^(1/
4)*ArcTanh[((2 + Sqrt[2])^(1/4)*x)/(1 + x^4)^(1/4)])/4 - ((2 + Sqrt[2])^(1/4)*ArcTanh[((2 + Sqrt[2])^(1/4)*x)/
(1 + x^4)^(1/4)])/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 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 1532

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), 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 - b*e)*x^n, x]/(a + b*x^n + c*x^(2*n))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && E
qQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] &&  !IntegerQ[q] && GtQ[q, 0] && GtQ[m, n - 1] && LeQ[m, 2*n
- 1]

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]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 \sqrt [4]{1+x^4}}{x^2}+\frac {x^2 \sqrt [4]{1+x^4} \left (5+2 x^4\right )}{-1+2 x^4+x^8}\right ) \, dx \\ & = -\left (2 \int \frac {\sqrt [4]{1+x^4}}{x^2} \, dx\right )+\int \frac {x^2 \sqrt [4]{1+x^4} \left (5+2 x^4\right )}{-1+2 x^4+x^8} \, dx \\ & = \frac {2 \sqrt [4]{1+x^4}}{x}-2 \int \frac {x^2}{\left (1+x^4\right )^{3/4}} \, dx+\int \left (\frac {5 x^2 \sqrt [4]{1+x^4}}{-1+2 x^4+x^8}+\frac {2 x^6 \sqrt [4]{1+x^4}}{-1+2 x^4+x^8}\right ) \, dx \\ & = \frac {2 \sqrt [4]{1+x^4}}{x}+2 \int \frac {x^6 \sqrt [4]{1+x^4}}{-1+2 x^4+x^8} \, dx-2 \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+5 \int \frac {x^2 \sqrt [4]{1+x^4}}{-1+2 x^4+x^8} \, dx \\ & = \frac {2 \sqrt [4]{1+x^4}}{x}+2 \int \frac {x^2}{\left (1+x^4\right )^{3/4}} \, dx-2 \int \frac {x^2 \left (-1+x^4\right )}{\left (1+x^4\right )^{3/4} \left (-1+2 x^4+x^8\right )} \, dx+5 \int \left (-\frac {x^2 \sqrt [4]{1+x^4}}{\sqrt {2} \left (-2+2 \sqrt {2}-2 x^4\right )}-\frac {x^2 \sqrt [4]{1+x^4}}{\sqrt {2} \left (2+2 \sqrt {2}+2 x^4\right )}\right ) \, dx-\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right ) \\ & = \frac {2 \sqrt [4]{1+x^4}}{x}+\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-2 \int \left (-\frac {x^2}{\left (1+x^4\right )^{3/4} \left (-1+2 x^4+x^8\right )}+\frac {x^6}{\left (1+x^4\right )^{3/4} \left (-1+2 x^4+x^8\right )}\right ) \, dx+2 \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \int \frac {x^2 \sqrt [4]{1+x^4}}{-2+2 \sqrt {2}-2 x^4} \, dx}{\sqrt {2}}-\frac {5 \int \frac {x^2 \sqrt [4]{1+x^4}}{2+2 \sqrt {2}+2 x^4} \, dx}{\sqrt {2}} \\ & = \frac {2 \sqrt [4]{1+x^4}}{x}-\frac {5}{12} \left (2-\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},-x^4,\left (1-\sqrt {2}\right ) x^4\right )-\frac {5}{12} \left (2+\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\left (1+\sqrt {2}\right ) x^4,-x^4\right )+\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+2 \int \frac {x^2}{\left (1+x^4\right )^{3/4} \left (-1+2 x^4+x^8\right )} \, dx-2 \int \frac {x^6}{\left (1+x^4\right )^{3/4} \left (-1+2 x^4+x^8\right )} \, dx+\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right ) \\ & = \frac {2 \sqrt [4]{1+x^4}}{x}-\frac {5}{12} \left (2-\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},-x^4,\left (1-\sqrt {2}\right ) x^4\right )-\frac {5}{12} \left (2+\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\left (1+\sqrt {2}\right ) x^4,-x^4\right )+2 \int \left (-\frac {x^2}{\sqrt {2} \left (-2+2 \sqrt {2}-2 x^4\right ) \left (1+x^4\right )^{3/4}}-\frac {x^2}{\sqrt {2} \left (1+x^4\right )^{3/4} \left (2+2 \sqrt {2}+2 x^4\right )}\right ) \, dx-2 \int \left (-\frac {\left (-2+2 \sqrt {2}\right ) x^2}{2 \sqrt {2} \left (-2+2 \sqrt {2}-2 x^4\right ) \left (1+x^4\right )^{3/4}}+\frac {\left (2+2 \sqrt {2}\right ) x^2}{2 \sqrt {2} \left (1+x^4\right )^{3/4} \left (2+2 \sqrt {2}+2 x^4\right )}\right ) \, dx \\ & = \frac {2 \sqrt [4]{1+x^4}}{x}-\frac {5}{12} \left (2-\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},-x^4,\left (1-\sqrt {2}\right ) x^4\right )-\frac {5}{12} \left (2+\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\left (1+\sqrt {2}\right ) x^4,-x^4\right )-\sqrt {2} \int \frac {x^2}{\left (-2+2 \sqrt {2}-2 x^4\right ) \left (1+x^4\right )^{3/4}} \, dx-\sqrt {2} \int \frac {x^2}{\left (1+x^4\right )^{3/4} \left (2+2 \sqrt {2}+2 x^4\right )} \, dx-\left (-2+\sqrt {2}\right ) \int \frac {x^2}{\left (-2+2 \sqrt {2}-2 x^4\right ) \left (1+x^4\right )^{3/4}} \, dx-\left (2+\sqrt {2}\right ) \int \frac {x^2}{\left (1+x^4\right )^{3/4} \left (2+2 \sqrt {2}+2 x^4\right )} \, dx \\ & = \frac {2 \sqrt [4]{1+x^4}}{x}-\frac {5}{12} \left (2-\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},-x^4,\left (1-\sqrt {2}\right ) x^4\right )-\frac {5}{12} \left (2+\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\left (1+\sqrt {2}\right ) x^4,-x^4\right )-\sqrt {2} \text {Subst}\left (\int \frac {x^2}{-2+2 \sqrt {2}-2 \sqrt {2} x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\sqrt {2} \text {Subst}\left (\int \frac {x^2}{2+2 \sqrt {2}-2 \sqrt {2} x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\left (-2+\sqrt {2}\right ) \text {Subst}\left (\int \frac {x^2}{-2+2 \sqrt {2}-2 \sqrt {2} x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\left (2+\sqrt {2}\right ) \text {Subst}\left (\int \frac {x^2}{2+2 \sqrt {2}-2 \sqrt {2} x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right ) \\ & = \frac {2 \sqrt [4]{1+x^4}}{x}-\frac {5}{12} \left (2-\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},-x^4,\left (1-\sqrt {2}\right ) x^4\right )-\frac {5}{12} \left (2+\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\left (1+\sqrt {2}\right ) x^4,-x^4\right )-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2-\sqrt {2}}-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2+\sqrt {2}}-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2-\sqrt {2}}+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2+\sqrt {2}}+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {2}}-\frac {1}{4} \left (2-\sqrt {2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2-\sqrt {2}}+\sqrt {2} x^2} \, 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 {2-\sqrt {2}}-\sqrt {2} x^2} \, 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 {2+\sqrt {2}}-\sqrt {2} x^2} \, 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 {2+\sqrt {2}}+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right ) \\ & = \frac {2 \sqrt [4]{1+x^4}}{x}-\frac {5}{12} \left (2-\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},-x^4,\left (1-\sqrt {2}\right ) x^4\right )-\frac {5}{12} \left (2+\sqrt {2}\right ) x^3 \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\left (1+\sqrt {2}\right ) x^4,-x^4\right )+\frac {1}{4} \sqrt [4]{2-\sqrt {2}} \arctan \left (\frac {\sqrt [4]{2-\sqrt {2}} x}{\sqrt [4]{1+x^4}}\right )+\frac {\left (2+\sqrt {2}\right )^{3/4} \arctan \left (\frac {\sqrt [4]{2-\sqrt {2}} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\left (2-\sqrt {2}\right )^{3/4} \arctan \left (\frac {\sqrt [4]{2+\sqrt {2}} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}}+\frac {1}{4} \sqrt [4]{2+\sqrt {2}} \arctan \left (\frac {\sqrt [4]{2+\sqrt {2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{4} \sqrt [4]{2-\sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2-\sqrt {2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {\left (2+\sqrt {2}\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2-\sqrt {2}} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}}+\frac {1}{4} \sqrt [4]{10-7 \sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2+\sqrt {2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{4} \sqrt [4]{2+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2+\sqrt {2}} x}{\sqrt [4]{1+x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{1+x^4} \left (2+x^4\right )}{x^2 \left (-1+2 x^4+x^8\right )} \, dx=\frac {2 \sqrt [4]{1+x^4}}{x}-\frac {1}{8} \text {RootSum}\left [2-4 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-4 \log (x)+4 \log \left (\sqrt [4]{1+x^4}-x \text {$\#$1}\right )+7 \log (x) \text {$\#$1}^4-7 \log \left (\sqrt [4]{1+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-2 \text {$\#$1}^3+\text {$\#$1}^7}\&\right ] \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Time = 254.62 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66

method	result	size
pseudoelliptic	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (7 \textit {\_R}^{4}-4\right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (\textit {\_R}^{4}-2\right )}\right ) x +16 \left (x^{4}+1\right )^{\frac {1}{4}}}{8 x}\)	\(67\)
trager	\(\text {Expression too large to display}\)	\(3650\)
risch	\(\text {Expression too large to display}\)	\(5218\)

[In]

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

[Out]

1/8*(sum((7*_R^4-4)*ln((-_R*x+(x^4+1)^(1/4))/x)/_R^3/(_R^4-2),_R=RootOf(_Z^8-4*_Z^4+2))*x+16*(x^4+1)^(1/4))/x

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 7.65 (sec) , antiderivative size = 1579, normalized size of antiderivative = 15.63 \[ \int \frac {\sqrt [4]{1+x^4} \left (2+x^4\right )}{x^2 \left (-1+2 x^4+x^8\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/32*(sqrt(2)*x*sqrt(-sqrt(-1393*sqrt(2) + 1970))*log((4*(11*x^5 + sqrt(2)*(6*x^5 - 5*x) - x)*(x^4 + 1)^(3/4)
- 4*(331*x^7 - 137*x^3 + sqrt(2)*(234*x^7 - 97*x^3))*(x^4 + 1)^(1/4)*sqrt(-1393*sqrt(2) + 1970) + (4*(57*x^6 -
23*x^2 + sqrt(2)*(40*x^6 - 17*x^2))*sqrt(x^4 + 1) - (10444*x^8 + 2260*x^4 + sqrt(2)*(7385*x^8 + 1598*x^4 - 19
29) - 2728)*sqrt(-1393*sqrt(2) + 1970))*sqrt(-sqrt(-1393*sqrt(2) + 1970)))/(x^8 + 2*x^4 - 1)) - sqrt(2)*x*sqrt
(-sqrt(-1393*sqrt(2) + 1970))*log((4*(11*x^5 + sqrt(2)*(6*x^5 - 5*x) - x)*(x^4 + 1)^(3/4) - 4*(331*x^7 - 137*x
^3 + sqrt(2)*(234*x^7 - 97*x^3))*(x^4 + 1)^(1/4)*sqrt(-1393*sqrt(2) + 1970) - (4*(57*x^6 - 23*x^2 + sqrt(2)*(4
0*x^6 - 17*x^2))*sqrt(x^4 + 1) - (10444*x^8 + 2260*x^4 + sqrt(2)*(7385*x^8 + 1598*x^4 - 1929) - 2728)*sqrt(-13
93*sqrt(2) + 1970))*sqrt(-sqrt(-1393*sqrt(2) + 1970)))/(x^8 + 2*x^4 - 1)) - sqrt(2)*x*sqrt(-sqrt(1393*sqrt(2)
+ 1970))*log((4*(11*x^5 - sqrt(2)*(6*x^5 - 5*x) - x)*(x^4 + 1)^(3/4) - 4*(331*x^7 - 137*x^3 - sqrt(2)*(234*x^7
- 97*x^3))*(x^4 + 1)^(1/4)*sqrt(1393*sqrt(2) + 1970) + (4*(57*x^6 - 23*x^2 - sqrt(2)*(40*x^6 - 17*x^2))*sqrt(
x^4 + 1) - (10444*x^8 + 2260*x^4 - sqrt(2)*(7385*x^8 + 1598*x^4 - 1929) - 2728)*sqrt(1393*sqrt(2) + 1970))*sqr
t(-sqrt(1393*sqrt(2) + 1970)))/(x^8 + 2*x^4 - 1)) + sqrt(2)*x*sqrt(-sqrt(1393*sqrt(2) + 1970))*log((4*(11*x^5
- sqrt(2)*(6*x^5 - 5*x) - x)*(x^4 + 1)^(3/4) - 4*(331*x^7 - 137*x^3 - sqrt(2)*(234*x^7 - 97*x^3))*(x^4 + 1)^(1
/4)*sqrt(1393*sqrt(2) + 1970) - (4*(57*x^6 - 23*x^2 - sqrt(2)*(40*x^6 - 17*x^2))*sqrt(x^4 + 1) - (10444*x^8 +
2260*x^4 - sqrt(2)*(7385*x^8 + 1598*x^4 - 1929) - 2728)*sqrt(1393*sqrt(2) + 1970))*sqrt(-sqrt(1393*sqrt(2) + 1
970)))/(x^8 + 2*x^4 - 1)) + sqrt(2)*x*(-1393*sqrt(2) + 1970)^(1/4)*log((4*(11*x^5 + sqrt(2)*(6*x^5 - 5*x) - x)
*(x^4 + 1)^(3/4) + 4*(331*x^7 - 137*x^3 + sqrt(2)*(234*x^7 - 97*x^3))*(x^4 + 1)^(1/4)*sqrt(-1393*sqrt(2) + 197
0) + (4*(57*x^6 - 23*x^2 + sqrt(2)*(40*x^6 - 17*x^2))*sqrt(x^4 + 1) + (10444*x^8 + 2260*x^4 + sqrt(2)*(7385*x^
8 + 1598*x^4 - 1929) - 2728)*sqrt(-1393*sqrt(2) + 1970))*(-1393*sqrt(2) + 1970)^(1/4))/(x^8 + 2*x^4 - 1)) - sq
rt(2)*x*(-1393*sqrt(2) + 1970)^(1/4)*log((4*(11*x^5 + sqrt(2)*(6*x^5 - 5*x) - x)*(x^4 + 1)^(3/4) + 4*(331*x^7
- 137*x^3 + sqrt(2)*(234*x^7 - 97*x^3))*(x^4 + 1)^(1/4)*sqrt(-1393*sqrt(2) + 1970) - (4*(57*x^6 - 23*x^2 + sqr
t(2)*(40*x^6 - 17*x^2))*sqrt(x^4 + 1) + (10444*x^8 + 2260*x^4 + sqrt(2)*(7385*x^8 + 1598*x^4 - 1929) - 2728)*s
qrt(-1393*sqrt(2) + 1970))*(-1393*sqrt(2) + 1970)^(1/4))/(x^8 + 2*x^4 - 1)) - sqrt(2)*x*(1393*sqrt(2) + 1970)^
(1/4)*log((4*(11*x^5 - sqrt(2)*(6*x^5 - 5*x) - x)*(x^4 + 1)^(3/4) + 4*(331*x^7 - 137*x^3 - sqrt(2)*(234*x^7 -
97*x^3))*(x^4 + 1)^(1/4)*sqrt(1393*sqrt(2) + 1970) + (4*(57*x^6 - 23*x^2 - sqrt(2)*(40*x^6 - 17*x^2))*sqrt(x^4
+ 1) + (10444*x^8 + 2260*x^4 - sqrt(2)*(7385*x^8 + 1598*x^4 - 1929) - 2728)*sqrt(1393*sqrt(2) + 1970))*(1393*
sqrt(2) + 1970)^(1/4))/(x^8 + 2*x^4 - 1)) + sqrt(2)*x*(1393*sqrt(2) + 1970)^(1/4)*log((4*(11*x^5 - sqrt(2)*(6*
x^5 - 5*x) - x)*(x^4 + 1)^(3/4) + 4*(331*x^7 - 137*x^3 - sqrt(2)*(234*x^7 - 97*x^3))*(x^4 + 1)^(1/4)*sqrt(1393
*sqrt(2) + 1970) - (4*(57*x^6 - 23*x^2 - sqrt(2)*(40*x^6 - 17*x^2))*sqrt(x^4 + 1) + (10444*x^8 + 2260*x^4 - sq
rt(2)*(7385*x^8 + 1598*x^4 - 1929) - 2728)*sqrt(1393*sqrt(2) + 1970))*(1393*sqrt(2) + 1970)^(1/4))/(x^8 + 2*x^
4 - 1)) + 64*(x^4 + 1)^(1/4))/x

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [N/A]

Not integrable

Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.30 \[ \int \frac {\sqrt [4]{1+x^4} \left (2+x^4\right )}{x^2 \left (-1+2 x^4+x^8\right )} \, dx=\int { \frac {{\left (x^{4} + 2\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{{\left (x^{8} + 2 \, x^{4} - 1\right )} x^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.30 \[ \int \frac {\sqrt [4]{1+x^4} \left (2+x^4\right )}{x^2 \left (-1+2 x^4+x^8\right )} \, dx=\int { \frac {{\left (x^{4} + 2\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{{\left (x^{8} + 2 \, x^{4} - 1\right )} x^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.30 \[ \int \frac {\sqrt [4]{1+x^4} \left (2+x^4\right )}{x^2 \left (-1+2 x^4+x^8\right )} \, dx=\int \frac {{\left (x^4+1\right )}^{1/4}\,\left (x^4+2\right )}{x^2\,\left (x^8+2\,x^4-1\right )} \,d x \]

[In]

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

[Out]

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