\(\int \frac {2+x^4}{\sqrt [4]{x^2+x^4} (-1-x^4+2 x^8)} \, dx\) [1852]

Optimal result

Integrand size = 31, antiderivative size = 127 \[ \int \frac {2+x^4}{\sqrt [4]{x^2+x^4} \left (-1-x^4+2 x^8\right )} \, dx=-\frac {\left (x^2+x^4\right )^{3/4}}{x \left (1+x^2\right )}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}}+\frac {1}{4} \text {RootSum}\left [3-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{x^2+x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ] \]

[Out]

Unintegrable

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.42 (sec) , antiderivative size = 441, normalized size of antiderivative = 3.47, number of steps used = 19, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {2081, 6847, 6860, 1418, 390, 385, 218, 212, 209, 1443} \[ \int \frac {2+x^4}{\sqrt [4]{x^2+x^4} \left (-1-x^4+2 x^8\right )} \, dx=-\frac {\sqrt [4]{x^2+1} \sqrt {x} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \arctan \left (\frac {\sqrt [4]{\sqrt {2}-2 i} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{x^2+1}}\right )}{2^{7/8} \sqrt [4]{\sqrt {2}-2 i} \sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \arctan \left (\frac {\sqrt [4]{\sqrt {2}+2 i} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{x^2+1}}\right )}{2^{7/8} \sqrt [4]{\sqrt {2}+2 i} \sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \text {arctanh}\left (\frac {\sqrt [4]{\sqrt {2}-2 i} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{x^2+1}}\right )}{2^{7/8} \sqrt [4]{\sqrt {2}-2 i} \sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \text {arctanh}\left (\frac {\sqrt [4]{\sqrt {2}+2 i} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{x^2+1}}\right )}{2^{7/8} \sqrt [4]{\sqrt {2}+2 i} \sqrt [4]{x^4+x^2}}-\frac {x}{\sqrt [4]{x^4+x^2}} \]

[In]

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

[Out]

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

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 218

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] &&  !Gt
Q[a/b, 0]

Rule 385

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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 1418

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(d + e*x^n)^(p + q)*(a/d
+ (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rule 1443

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 2081

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 6847

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.26 \[ \int \frac {2+x^4}{\sqrt [4]{x^2+x^4} \left (-1-x^4+2 x^8\right )} \, dx=-\frac {\sqrt {x} \left (4 \sqrt {x}+2^{3/4} \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )+2^{3/4} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )-\sqrt [4]{1+x^2} \text {RootSum}\left [3-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt {x}\right )+\log \left (\sqrt [4]{1+x^2}-\sqrt {x} \text {$\#$1}\right )}{\text {$\#$1}}\&\right ]\right )}{4 \sqrt [4]{x^2+x^4}} \]

[In]

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

[Out]

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

Maple [F(-1)]

Timed out.

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 5.28 (sec) , antiderivative size = 2054, normalized size of antiderivative = 16.17 \[ \int \frac {2+x^4}{\sqrt [4]{x^2+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/48*(2*sqrt(3)*(x^3 + x)*sqrt(-sqrt(3)*sqrt(I*sqrt(2) + 1))*log(-(2*sqrt(3)*(2*x^4 - 22*x^2 + sqrt(2)*(22*I*
x^4 + I*x^2))*(x^4 + x^2)^(1/4)*sqrt(I*sqrt(2) + 1) + 6*(x^4 + x^2)^(3/4)*(14*x^2 + sqrt(2)*(-8*I*x^2 + 7*I) +
8) + (2*sqrt(x^4 + x^2)*(2*x^3 + sqrt(2)*(22*I*x^3 + I*x) - 22*x)*sqrt(I*sqrt(2) + 1) + sqrt(3)*(12*x^5 + 44*
x^3 - sqrt(2)*(30*I*x^5 + 2*I*x^3 - 7*I*x) + 8*x))*sqrt(-sqrt(3)*sqrt(I*sqrt(2) + 1)))/(2*x^5 + x)) - 2*sqrt(3
)*(x^3 + x)*sqrt(-sqrt(3)*sqrt(I*sqrt(2) + 1))*log(-(2*sqrt(3)*(2*x^4 - 22*x^2 + sqrt(2)*(22*I*x^4 + I*x^2))*(
x^4 + x^2)^(1/4)*sqrt(I*sqrt(2) + 1) + 6*(x^4 + x^2)^(3/4)*(14*x^2 + sqrt(2)*(-8*I*x^2 + 7*I) + 8) - (2*sqrt(x
^4 + x^2)*(2*x^3 - sqrt(2)*(-22*I*x^3 - I*x) - 22*x)*sqrt(I*sqrt(2) + 1) + sqrt(3)*(12*x^5 + 44*x^3 + sqrt(2)*
(-30*I*x^5 - 2*I*x^3 + 7*I*x) + 8*x))*sqrt(-sqrt(3)*sqrt(I*sqrt(2) + 1)))/(2*x^5 + x)) - 2*sqrt(3)*(x^3 + x)*s
qrt(sqrt(3)*sqrt(I*sqrt(2) + 1))*log((2*sqrt(3)*(2*x^4 - 22*x^2 - sqrt(2)*(-22*I*x^4 - I*x^2))*(x^4 + x^2)^(1/
4)*sqrt(I*sqrt(2) + 1) - 6*(x^4 + x^2)^(3/4)*(14*x^2 + sqrt(2)*(-8*I*x^2 + 7*I) + 8) - (2*sqrt(x^4 + x^2)*(2*x
^3 + sqrt(2)*(22*I*x^3 + I*x) - 22*x)*sqrt(I*sqrt(2) + 1) - sqrt(3)*(12*x^5 + 44*x^3 + sqrt(2)*(-30*I*x^5 - 2*
I*x^3 + 7*I*x) + 8*x))*sqrt(sqrt(3)*sqrt(I*sqrt(2) + 1)))/(2*x^5 + x)) + 2*sqrt(3)*(x^3 + x)*sqrt(sqrt(3)*sqrt
(I*sqrt(2) + 1))*log((2*sqrt(3)*(2*x^4 - 22*x^2 - sqrt(2)*(-22*I*x^4 - I*x^2))*(x^4 + x^2)^(1/4)*sqrt(I*sqrt(2
) + 1) - 6*(x^4 + x^2)^(3/4)*(14*x^2 + sqrt(2)*(-8*I*x^2 + 7*I) + 8) + (2*sqrt(x^4 + x^2)*(2*x^3 - sqrt(2)*(-2
2*I*x^3 - I*x) - 22*x)*sqrt(I*sqrt(2) + 1) - sqrt(3)*(12*x^5 + 44*x^3 - sqrt(2)*(30*I*x^5 + 2*I*x^3 - 7*I*x) +
8*x))*sqrt(sqrt(3)*sqrt(I*sqrt(2) + 1)))/(2*x^5 + x)) + 2*sqrt(3)*(x^3 + x)*sqrt(sqrt(3)*sqrt(-I*sqrt(2) + 1)
)*log((2*sqrt(3)*(2*x^4 - 22*x^2 - sqrt(2)*(22*I*x^4 + I*x^2))*(x^4 + x^2)^(1/4)*sqrt(-I*sqrt(2) + 1) - 6*(x^4
+ x^2)^(3/4)*(14*x^2 + sqrt(2)*(8*I*x^2 - 7*I) + 8) + (2*sqrt(x^4 + x^2)*(2*x^3 - sqrt(2)*(22*I*x^3 + I*x) -
22*x)*sqrt(-I*sqrt(2) + 1) - sqrt(3)*(12*x^5 + 44*x^3 - sqrt(2)*(-30*I*x^5 - 2*I*x^3 + 7*I*x) + 8*x))*sqrt(sqr
t(3)*sqrt(-I*sqrt(2) + 1)))/(2*x^5 + x)) - 2*sqrt(3)*(x^3 + x)*sqrt(sqrt(3)*sqrt(-I*sqrt(2) + 1))*log((2*sqrt(
3)*(2*x^4 - 22*x^2 - sqrt(2)*(22*I*x^4 + I*x^2))*(x^4 + x^2)^(1/4)*sqrt(-I*sqrt(2) + 1) - 6*(x^4 + x^2)^(3/4)*
(14*x^2 + sqrt(2)*(8*I*x^2 - 7*I) + 8) - (2*sqrt(x^4 + x^2)*(2*x^3 + sqrt(2)*(-22*I*x^3 - I*x) - 22*x)*sqrt(-I
*sqrt(2) + 1) - sqrt(3)*(12*x^5 + 44*x^3 + sqrt(2)*(30*I*x^5 + 2*I*x^3 - 7*I*x) + 8*x))*sqrt(sqrt(3)*sqrt(-I*s
qrt(2) + 1)))/(2*x^5 + x)) - 2*sqrt(3)*(x^3 + x)*sqrt(-sqrt(3)*sqrt(-I*sqrt(2) + 1))*log(-(2*sqrt(3)*(2*x^4 -
22*x^2 + sqrt(2)*(-22*I*x^4 - I*x^2))*(x^4 + x^2)^(1/4)*sqrt(-I*sqrt(2) + 1) + 6*(x^4 + x^2)^(3/4)*(14*x^2 + s
qrt(2)*(8*I*x^2 - 7*I) + 8) - (2*sqrt(x^4 + x^2)*(2*x^3 - sqrt(2)*(22*I*x^3 + I*x) - 22*x)*sqrt(-I*sqrt(2) + 1
) + sqrt(3)*(12*x^5 + 44*x^3 + sqrt(2)*(30*I*x^5 + 2*I*x^3 - 7*I*x) + 8*x))*sqrt(-sqrt(3)*sqrt(-I*sqrt(2) + 1)
))/(2*x^5 + x)) + 2*sqrt(3)*(x^3 + x)*sqrt(-sqrt(3)*sqrt(-I*sqrt(2) + 1))*log(-(2*sqrt(3)*(2*x^4 - 22*x^2 + sq
rt(2)*(-22*I*x^4 - I*x^2))*(x^4 + x^2)^(1/4)*sqrt(-I*sqrt(2) + 1) + 6*(x^4 + x^2)^(3/4)*(14*x^2 + sqrt(2)*(8*I
*x^2 - 7*I) + 8) + (2*sqrt(x^4 + x^2)*(2*x^3 + sqrt(2)*(-22*I*x^3 - I*x) - 22*x)*sqrt(-I*sqrt(2) + 1) + sqrt(3
)*(12*x^5 + 44*x^3 - sqrt(2)*(-30*I*x^5 - 2*I*x^3 + 7*I*x) + 8*x))*sqrt(-sqrt(3)*sqrt(-I*sqrt(2) + 1)))/(2*x^5
+ x)) + 3*2^(3/4)*(x^3 + x)*log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 + 2^(3/4)*(3*x^3 + x) + 4*2^(1/4)*sqrt(x^4 +
x^2)*x + 4*(x^4 + x^2)^(3/4))/(x^3 - x)) - 3*2^(3/4)*(x^3 + x)*log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)
*(3*x^3 + x) - 4*2^(1/4)*sqrt(x^4 + x^2)*x + 4*(x^4 + x^2)^(3/4))/(x^3 - x)) + 3*2^(3/4)*(I*x^3 + I*x)*log(-(4
*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(3*I*x^3 + I*x) + 4*I*2^(1/4)*sqrt(x^4 + x^2)*x - 4*(x^4 + x^2)^(3/4)
)/(x^3 - x)) + 3*2^(3/4)*(-I*x^3 - I*x)*log(-(4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(-3*I*x^3 - I*x) - 4*I
*2^(1/4)*sqrt(x^4 + x^2)*x - 4*(x^4 + x^2)^(3/4))/(x^3 - x)) + 48*(x^4 + x^2)^(3/4))/(x^3 + x)

Sympy [N/A]

Not integrable

Time = 4.78 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.28 \[ \int \frac {2+x^4}{\sqrt [4]{x^2+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\int \frac {x^{4} + 2}{\sqrt [4]{x^{2} \left (x^{2} + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (2 x^{4} + 1\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.33 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.21 \[ \int \frac {2+x^4}{\sqrt [4]{x^2+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\int { \frac {x^{4} + 2}{{\left (2 \, x^{8} - x^{4} - 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

2/231*(64*x^9 + 16*x^7 + 11*(4*x^5 + x^3 - 3*x)*x^4 - 28*x^5 - 2*x^3 - 22*x)/((2*x^(17/2) - x^(9/2) - sqrt(x))
*(x^2 + 1)^(1/4)) + integrate(4/231*(256*x^8 + 64*x^6 + (128*x^8 + 32*x^6 + 164*x^4 + 51*x^2 - 209)*x^4 - 68*x
^4 + 3*x^2 - 121)/((4*x^(33/2) - 4*x^(25/2) - 3*x^(17/2) + 2*x^(9/2) + sqrt(x))*(x^2 + 1)^(1/4)), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {2+x^4}{\sqrt [4]{x^2+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Invalid _EXT in replace_ext Error: Bad Argument ValueDone

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.24 \[ \int \frac {2+x^4}{\sqrt [4]{x^2+x^4} \left (-1-x^4+2 x^8\right )} \, dx=\int -\frac {x^4+2}{{\left (x^4+x^2\right )}^{1/4}\,\left (-2\,x^8+x^4+1\right )} \,d x \]

[In]

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

[Out]

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