\(\int \frac {1+x^{16}}{\sqrt {1+x^4} (-1+x^{16})} \, dx\) [1772]

Optimal result

Integrand size = 22, antiderivative size = 119 \[ \int \frac {1+x^{16}}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx=-\frac {x}{4 \sqrt {1+x^4}}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}} \]

[Out]

-1/4*x/(x^4+1)^(1/2)-1/8*arctan(2^(1/4)*x/(x^4+1)^(1/2))*2^(3/4)-1/16*arctan(2^(1/2)*x/(x^4+1)^(1/2))*2^(1/2)-
1/8*arctanh(2^(1/4)*x/(x^4+1)^(1/2))*2^(3/4)-1/16*arctanh(2^(1/2)*x/(x^4+1)^(1/2))*2^(1/2)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.61 (sec) , antiderivative size = 404, normalized size of antiderivative = 3.39, number of steps used = 37, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {6857, 226, 2098, 1236, 1193, 1210, 1225, 1713, 213, 1212, 209, 21, 1443, 418, 1231, 1721} \[ \int \frac {1+x^{16}}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}}+\frac {i \left (\sqrt {2}+(1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{16 \sqrt {x^4+1}}-\frac {i \left (\sqrt {2}+(1-i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{16 \sqrt {x^4+1}}+\frac {i \left (\sqrt {2}+(-1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{16 \sqrt {x^4+1}}-\frac {i \left (\sqrt {2}+(-1-i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{16 \sqrt {x^4+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}}-\frac {x \left (1-x^2\right )}{8 \sqrt {x^4+1}}-\frac {x \left (x^2+1\right )}{8 \sqrt {x^4+1}} \]

[In]

Int[(1 + x^16)/(Sqrt[1 + x^4]*(-1 + x^16)),x]

[Out]

-1/8*(x*(1 - x^2))/Sqrt[1 + x^4] - (x*(1 + x^2))/(8*Sqrt[1 + x^4]) - ArcTan[(2^(1/4)*x)/Sqrt[1 + x^4]]/(4*2^(1
/4)) - ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(8*Sqrt[2]) - ArcTanh[(2^(1/4)*x)/Sqrt[1 + x^4]]/(4*2^(1/4)) - ArcTan
h[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(8*Sqrt[2]) + ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])
/(4*Sqrt[1 + x^4]) - ((I/16)*((-1 - I) + Sqrt[2])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x],
1/2])/Sqrt[1 + x^4] + ((I/16)*((-1 + I) + Sqrt[2])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x
], 1/2])/Sqrt[1 + x^4] - ((I/16)*((1 - I) + Sqrt[2])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[
x], 1/2])/Sqrt[1 + x^4] + ((I/16)*((1 + I) + Sqrt[2])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan
[x], 1/2])/Sqrt[1 + x^4]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 213

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

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-d/c, 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), x], x] /; FreeQ[{a,
b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1193

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x)*(d + e*x^2)*((a + c*x^4)^(p + 1)/
(4*a*(p + 1))), x] + Dist[1/(4*a*(p + 1)), Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x],
x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rule 1225

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[a + c*x^4], x],
x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d
^2 + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1231

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(c*d + a*e*q
)/(c*d^2 - a*e^2), Int[1/Sqrt[a + c*x^4], x], x] - Dist[(a*e*(e + d*q))/(c*d^2 - a*e^2), Int[(1 + q*x^2)/((d +
e*x^2)*Sqrt[a + c*x^4]), x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0]
&& PosQ[c/a]

Rule 1236

Int[((a_) + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x^2)
*(a + c*x^4)^p, x], x] + Dist[e^2/(c*d^2 + a*e^2), Int[(a + c*x^4)^(p + 1)/(d + e*x^2), x], x] /; FreeQ[{a, c,
d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 0]

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 1713

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 1721

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, Simp[(-(B*d - A*e))*(ArcTan[Rt[c*(d/e) + a*(e/d), 2]*(x/Sqrt[a + c*x^4])]/(2*d*e*Rt[c*(d/e) + a*(e/d), 2]))
, x] + Simp[(B*d + A*e)*(A + B*x^2)*(Sqrt[A^2*((a + c*x^4)/(a*(A + B*x^2)^2))]/(4*d*e*A*q*Sqrt[a + c*x^4]))*El
lipticPi[Cancel[-(B*d - A*e)^2/(4*d*e*A*B)], 2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rule 2098

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 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*} \text {integral}& = \int \left (\frac {1}{\sqrt {1+x^4}}+\frac {2}{\sqrt {1+x^4} \left (-1+x^{16}\right )}\right ) \, dx \\ & = 2 \int \frac {1}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx+\int \frac {1}{\sqrt {1+x^4}} \, dx \\ & = \frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {1+x^4}}+2 \int \left (\frac {1}{4 \left (-1+x^2\right ) \left (1+x^4\right )^{3/2}}-\frac {1}{4 \left (1+x^2\right ) \left (1+x^4\right )^{3/2}}+\frac {-1-x^4}{2 \left (1+x^4\right )^{3/2} \left (1+x^8\right )}\right ) \, dx \\ & = \frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {1+x^4}}+\frac {1}{2} \int \frac {1}{\left (-1+x^2\right ) \left (1+x^4\right )^{3/2}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+x^2\right ) \left (1+x^4\right )^{3/2}} \, dx+\int \frac {-1-x^4}{\left (1+x^4\right )^{3/2} \left (1+x^8\right )} \, dx \\ & = \frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {1+x^4}}+\frac {1}{4} \int \frac {-1-x^2}{\left (1+x^4\right )^{3/2}} \, dx-\frac {1}{4} \int \frac {1-x^2}{\left (1+x^4\right )^{3/2}} \, dx+\frac {1}{4} \int \frac {1}{\left (-1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-\int \frac {1}{\sqrt {1+x^4} \left (1+x^8\right )} \, dx \\ & = -\frac {x \left (1-x^2\right )}{8 \sqrt {1+x^4}}-\frac {x \left (1+x^2\right )}{8 \sqrt {1+x^4}}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {1}{2} i \int \frac {1}{\left (i-x^4\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} i \int \frac {1}{\left (i+x^4\right ) \sqrt {1+x^4}} \, dx-2 \left (\frac {1}{8} \int \frac {1}{\sqrt {1+x^4}} \, dx\right )+\frac {1}{8} \int \frac {-1-x^2}{\sqrt {1+x^4}} \, dx-\frac {1}{8} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx-\frac {1}{8} \int \frac {-1-x^2}{\left (-1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{8} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx \\ & = -\frac {x \left (1-x^2\right )}{8 \sqrt {1+x^4}}-\frac {x \left (1+x^2\right )}{8 \sqrt {1+x^4}}+\frac {x \sqrt {1+x^4}}{8 \left (1+x^2\right )}-\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{8 \sqrt {1+x^4}}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{8 \sqrt {1+x^4}}+\frac {1}{8} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx+\frac {1}{8} \text {Subst}\left (\int \frac {1}{-1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )-\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )-\frac {1}{4} \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1}{\left (1-\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1}{\left (1+\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1}{\left (1-(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1}{\left (1+(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx \\ & = -\frac {x \left (1-x^2\right )}{8 \sqrt {1+x^4}}-\frac {x \left (1+x^2\right )}{8 \sqrt {1+x^4}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {1+x^4}}-\left (\left (\frac {1}{8}+\frac {i}{8}\right ) \left (1+\sqrt [4]{-1}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\left (\left (-\frac {1}{8}-\frac {i}{8}\right ) \sqrt [4]{-1} \left (1+\sqrt [4]{-1}\right )\right ) \int \frac {1+x^2}{\left (1+\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\left (\left (\frac {1}{8}-\frac {i}{8}\right ) (-1)^{3/4} \left (1-(-1)^{3/4}\right )\right ) \int \frac {1+x^2}{\left (1-(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\left (\left (\frac {1}{8}-\frac {i}{8}\right ) \left (1+(-1)^{3/4}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\left (\left (-\frac {1}{8}+\frac {i}{8}\right ) (-1)^{3/4} \left (1+(-1)^{3/4}\right )\right ) \int \frac {1+x^2}{\left (1+(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{8} \left ((1+i)-i \sqrt {2}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{8} \left (i \left ((-1-i)+\sqrt {2}\right )\right ) \int \frac {1+x^2}{\left (1-\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{8} \left (i \left ((1+i)+\sqrt {2}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx \\ & = -\frac {x \left (1-x^2\right )}{8 \sqrt {1+x^4}}-\frac {x \left (1+x^2\right )}{8 \sqrt {1+x^4}}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {1+x^4}}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) \left (1+\sqrt [4]{-1}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{\sqrt {1+x^4}}-\frac {\left ((1+i)-i \sqrt {2}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{16 \sqrt {1+x^4}}-\frac {i \left ((-1-i)+\sqrt {2}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{16 \sqrt {1+x^4}}+\frac {i \left ((1+i)+\sqrt {2}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{16 \sqrt {1+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.95 \[ \int \frac {1+x^{16}}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx=\frac {1}{16} \left (-\frac {4 x}{\sqrt {1+x^4}}-2\ 2^{3/4} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )-\sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )-2\ 2^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )\right ) \]

[In]

Integrate[(1 + x^16)/(Sqrt[1 + x^4]*(-1 + x^16)),x]

[Out]

((-4*x)/Sqrt[1 + x^4] - 2*2^(3/4)*ArcTan[(2^(1/4)*x)/Sqrt[1 + x^4]] - Sqrt[2]*ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]
] - 2*2^(3/4)*ArcTanh[(2^(1/4)*x)/Sqrt[1 + x^4]] - Sqrt[2]*ArcTanh[(Sqrt[2]*x)/Sqrt[1 + x^4]])/16

Maple [A] (verified)

Time = 8.95 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.17

method	result	size
risch	\(-\frac {x}{4 \sqrt {x^{4}+1}}-\frac {\arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right ) \sqrt {2}}{16}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )}{32}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )}{32}+\frac {\arctan \left (\frac {2^{\frac {3}{4}} \sqrt {x^{4}+1}}{2 x}\right ) 2^{\frac {3}{4}}}{8}-\frac {\ln \left (\frac {-2^{\frac {1}{4}} x -\sqrt {x^{4}+1}}{2^{\frac {1}{4}} x -\sqrt {x^{4}+1}}\right ) 2^{\frac {3}{4}}}{16}\)	\(139\)
elliptic	\(\frac {\left (-\frac {\sqrt {2}\, x}{4 \sqrt {x^{4}+1}}-\frac {\ln \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{16}+\frac {\ln \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{16}+\frac {2^{\frac {1}{4}} \left (2 \arctan \left (\frac {2^{\frac {3}{4}} \sqrt {x^{4}+1}}{2 x}\right )-\ln \left (\frac {\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}+\frac {2^{\frac {3}{4}}}{2}}{\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}-\frac {2^{\frac {3}{4}}}{2}}\right )\right )}{8}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{8}\right ) \sqrt {2}}{2}\)	\(150\)
default	\(\frac {-8 \sqrt {x^{4}+1}\, x -\left (x^{4}+1\right ) \left (\left (2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right ) \sqrt {2}-2 \,2^{\frac {3}{4}} \left (\arctan \left (\frac {\left (\sqrt {2-\sqrt {2}}\, x^{2}+\sqrt {2-\sqrt {2}}-2 x \right ) 2^{\frac {3}{4}}}{2 \sqrt {x^{4}+1}}\right )-\arctan \left (\frac {\left (\sqrt {2-\sqrt {2}}\, x^{2}+\sqrt {2-\sqrt {2}}+2 x \right ) 2^{\frac {3}{4}}}{2 \sqrt {x^{4}+1}}\right )-\operatorname {arctanh}\left (\frac {\left (\sqrt {2+\sqrt {2}}\, x^{2}+\sqrt {2+\sqrt {2}}-2 x \right ) 2^{\frac {3}{4}}}{2 \sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (\sqrt {2+\sqrt {2}}\, x^{2}+\sqrt {2+\sqrt {2}}+2 x \right ) 2^{\frac {3}{4}}}{2 \sqrt {x^{4}+1}}\right )\right )\right )}{32 \left (-x \sqrt {2}+x^{2}+1\right ) \left (x \sqrt {2}+x^{2}+1\right )}\)	\(265\)
pseudoelliptic	\(\frac {-8 \sqrt {x^{4}+1}\, x -\left (x^{4}+1\right ) \left (\left (2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right ) \sqrt {2}-2 \,2^{\frac {3}{4}} \left (\arctan \left (\frac {\left (\sqrt {2-\sqrt {2}}\, x^{2}+\sqrt {2-\sqrt {2}}-2 x \right ) 2^{\frac {3}{4}}}{2 \sqrt {x^{4}+1}}\right )-\arctan \left (\frac {\left (\sqrt {2-\sqrt {2}}\, x^{2}+\sqrt {2-\sqrt {2}}+2 x \right ) 2^{\frac {3}{4}}}{2 \sqrt {x^{4}+1}}\right )-\operatorname {arctanh}\left (\frac {\left (\sqrt {2+\sqrt {2}}\, x^{2}+\sqrt {2+\sqrt {2}}-2 x \right ) 2^{\frac {3}{4}}}{2 \sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (\sqrt {2+\sqrt {2}}\, x^{2}+\sqrt {2+\sqrt {2}}+2 x \right ) 2^{\frac {3}{4}}}{2 \sqrt {x^{4}+1}}\right )\right )\right )}{32 \left (-x \sqrt {2}+x^{2}+1\right ) \left (x \sqrt {2}+x^{2}+1\right )}\)	\(265\)
trager	\(-\frac {x}{4 \sqrt {x^{4}+1}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x -\sqrt {x^{4}+1}}{\left (1+x \right ) \left (-1+x \right )}\right )}{16}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{16}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right ) \ln \left (\frac {-x^{4} \operatorname {RootOf}\left (\textit {\_Z}^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right )+x^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )-4 \sqrt {x^{4}+1}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right )}{x^{4}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}+1}\right )}{16}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right ) x^{4}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right ) x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right )+8 \sqrt {x^{4}+1}\, x}{-x^{4}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}-1}\right )}{32}\)	\(331\)

[In]

int((x^16+1)/(x^4+1)^(1/2)/(x^16-1),x,method=_RETURNVERBOSE)

[Out]

-1/4*x/(x^4+1)^(1/2)-1/16*arctan(2^(1/2)*x/(x^4+1)^(1/2))*2^(1/2)-1/32*2^(1/2)*arctanh((x^2-x+1)*2^(1/2)/(x^4+
1)^(1/2))+1/32*2^(1/2)*arctanh((x^2+x+1)*2^(1/2)/(x^4+1)^(1/2))+1/8*arctan(1/2*2^(3/4)/x*(x^4+1)^(1/2))*2^(3/4
)-1/16*ln((-2^(1/4)*x-(x^4+1)^(1/2))/(2^(1/4)*x-(x^4+1)^(1/2)))*2^(3/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 376, normalized size of antiderivative = 3.16 \[ \int \frac {1+x^{16}}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx=-\frac {2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (-\frac {2^{\frac {3}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )} + 4 \, {\left (x^{5} + \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} + 1}\right ) - 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {2^{\frac {3}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )} - 4 \, {\left (x^{5} + \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} + 1}\right ) - 2^{\frac {3}{4}} {\left (i \, x^{4} + i\right )} \log \left (\frac {2^{\frac {3}{4}} {\left (i \, x^{8} + 4 i \, x^{4} + i\right )} - 4 \, {\left (x^{5} - \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} - 4 \cdot 2^{\frac {1}{4}} {\left (i \, x^{6} + i \, x^{2}\right )}}{x^{8} + 1}\right ) - 2^{\frac {3}{4}} {\left (-i \, x^{4} - i\right )} \log \left (\frac {2^{\frac {3}{4}} {\left (-i \, x^{8} - 4 i \, x^{4} - i\right )} - 4 \, {\left (x^{5} - \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} - 4 \cdot 2^{\frac {1}{4}} {\left (-i \, x^{6} - i \, x^{2}\right )}}{x^{8} + 1}\right ) + 2 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) - \sqrt {2} {\left (x^{4} + 1\right )} \log \left (\frac {x^{4} - 2 \, \sqrt {2} \sqrt {x^{4} + 1} x + 2 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) + 8 \, \sqrt {x^{4} + 1} x}{32 \, {\left (x^{4} + 1\right )}} \]

[In]

integrate((x^16+1)/(x^4+1)^(1/2)/(x^16-1),x, algorithm="fricas")

[Out]

-1/32*(2^(3/4)*(x^4 + 1)*log(-(2^(3/4)*(x^8 + 4*x^4 + 1) + 4*(x^5 + sqrt(2)*x^3 + x)*sqrt(x^4 + 1) + 4*2^(1/4)
*(x^6 + x^2))/(x^8 + 1)) - 2^(3/4)*(x^4 + 1)*log((2^(3/4)*(x^8 + 4*x^4 + 1) - 4*(x^5 + sqrt(2)*x^3 + x)*sqrt(x
^4 + 1) + 4*2^(1/4)*(x^6 + x^2))/(x^8 + 1)) - 2^(3/4)*(I*x^4 + I)*log((2^(3/4)*(I*x^8 + 4*I*x^4 + I) - 4*(x^5
- sqrt(2)*x^3 + x)*sqrt(x^4 + 1) - 4*2^(1/4)*(I*x^6 + I*x^2))/(x^8 + 1)) - 2^(3/4)*(-I*x^4 - I)*log((2^(3/4)*(
-I*x^8 - 4*I*x^4 - I) - 4*(x^5 - sqrt(2)*x^3 + x)*sqrt(x^4 + 1) - 4*2^(1/4)*(-I*x^6 - I*x^2))/(x^8 + 1)) + 2*s
qrt(2)*(x^4 + 1)*arctan(sqrt(2)*x/sqrt(x^4 + 1)) - sqrt(2)*(x^4 + 1)*log((x^4 - 2*sqrt(2)*sqrt(x^4 + 1)*x + 2*
x^2 + 1)/(x^4 - 2*x^2 + 1)) + 8*sqrt(x^4 + 1)*x)/(x^4 + 1)

Sympy [F(-1)]

Timed out. \[ \int \frac {1+x^{16}}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx=\text {Timed out} \]

[In]

integrate((x**16+1)/(x**4+1)**(1/2)/(x**16-1),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1+x^{16}}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx=\int { \frac {x^{16} + 1}{{\left (x^{16} - 1\right )} \sqrt {x^{4} + 1}} \,d x } \]

[In]

integrate((x^16+1)/(x^4+1)^(1/2)/(x^16-1),x, algorithm="maxima")

[Out]

integrate((x^16 + 1)/((x^16 - 1)*sqrt(x^4 + 1)), x)

Giac [F]

\[ \int \frac {1+x^{16}}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx=\int { \frac {x^{16} + 1}{{\left (x^{16} - 1\right )} \sqrt {x^{4} + 1}} \,d x } \]

[In]

integrate((x^16+1)/(x^4+1)^(1/2)/(x^16-1),x, algorithm="giac")

[Out]

integrate((x^16 + 1)/((x^16 - 1)*sqrt(x^4 + 1)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1+x^{16}}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx=\int \frac {x^{16}+1}{\sqrt {x^4+1}\,\left (x^{16}-1\right )} \,d x \]

[In]

int((x^16 + 1)/((x^4 + 1)^(1/2)*(x^16 - 1)),x)

[Out]

int((x^16 + 1)/((x^4 + 1)^(1/2)*(x^16 - 1)), x)