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

Optimal. Leaf size=119 \[ -\frac {x}{4 \sqrt {x^4+1}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}} \]

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Rubi [C]  time = 1.05, antiderivative size = 404, normalized size of antiderivative = 3.39, number of steps used = 37, number of rules used = 16, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {6725, 220, 2073, 1222, 1179, 1196, 1211, 1699, 207, 1198, 203, 21, 1429, 409, 1217, 1707} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\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}}+\frac {i \left (\sqrt {2}+(1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{4 \sqrt {x^4+1}} \end {gather*}

Antiderivative was successfully verified.

[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 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 207

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

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 409

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 1179

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 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1198

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 1211

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 1217

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 1222

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 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 1699

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 1707

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)]*EllipticPi[Cancel[-((B*d - A*e)^2
/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2])/(4*d*e*A*q*Sqrt[a + c*x^4]), 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 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 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^{16}}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx &=\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}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(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 \tan ^{-1}(x)|\frac {1}{2}\right )}{8 \sqrt {1+x^4}}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(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} \operatorname {Subst}\left (\int \frac {1}{-1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )-\frac {1}{8} \operatorname {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 {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\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}} F\left (2 \tan ^{-1}(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 {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\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}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(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}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{16 \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 1.18, size = 227, normalized size = 1.91 \begin {gather*} \left (\frac {1}{8}+\frac {i}{8}\right ) \left (-\frac {(1-i) x}{\sqrt {x^4+1}}-3 \sqrt {2} F\left (\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (-i;\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (i;\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (-\frac {1+i}{\sqrt {2}};\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (-\frac {1-i}{\sqrt {2}};\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (\frac {1-i}{\sqrt {2}};\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (\frac {1+i}{\sqrt {2}};\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1/8 + I/8)*(((-1 + I)*x)/Sqrt[1 + x^4] - 3*Sqrt[2]*EllipticF[I*ArcSinh[((1 + I)*x)/Sqrt[2]], -1] + Sqrt[2]*El
lipticPi[-I, I*ArcSinh[((1 + I)*x)/Sqrt[2]], -1] + Sqrt[2]*EllipticPi[I, I*ArcSinh[((1 + I)*x)/Sqrt[2]], -1] +
Sqrt[2]*EllipticPi[(-1 - I)/Sqrt[2], I*ArcSinh[((1 + I)*x)/Sqrt[2]], -1] + Sqrt[2]*EllipticPi[(-1 + I)/Sqrt[2
], I*ArcSinh[((1 + I)*x)/Sqrt[2]], -1] + Sqrt[2]*EllipticPi[(1 - I)/Sqrt[2], I*ArcSinh[((1 + I)*x)/Sqrt[2]], -
1] + Sqrt[2]*EllipticPi[(1 + I)/Sqrt[2], I*ArcSinh[((1 + I)*x)/Sqrt[2]], -1])

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IntegrateAlgebraic [A]  time = 0.68, size = 119, normalized size = 1.00 \begin {gather*} -\frac {x}{4 \sqrt {1+x^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*x/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*Sq
rt[2]) - ArcTanh[(2^(1/4)*x)/Sqrt[1 + x^4]]/(4*2^(1/4)) - ArcTanh[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(8*Sqrt[2])

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fricas [B]  time = 0.62, size = 302, normalized size = 2.54 \begin {gather*} -\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )} + 2^{\frac {1}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )}\right )} + 4 \, \sqrt {x^{4} + 1} {\left (2^{\frac {3}{4}} x^{3} + 2^{\frac {1}{4}} {\left (x^{5} + x\right )}\right )}}{2 \, {\left (x^{8} + 1\right )}}\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 (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 \, \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{16} + 1}{{\left (x^{16} - 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 2.41, size = 150, normalized size = 1.26

method	result	size
elliptic	\(\frac {\left (\frac {\ln \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{16}-\frac {2^{\frac {1}{4}} \ln \left (\frac {\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}+\frac {2^{\frac {3}{4}}}{2}}{\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}-\frac {2^{\frac {3}{4}}}{2}}\right )}{8}+\frac {2^{\frac {1}{4}} \arctan \left (\frac {2^{\frac {3}{4}} \sqrt {x^{4}+1}}{2 x}\right )}{4}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{8}-\frac {\ln \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{16}-\frac {\sqrt {2}\, x}{4 \sqrt {x^{4}+1}}\right ) \sqrt {2}}{2}\)	\(150\)
default	\(\frac {3 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{4 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{8}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {\arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{6}+x^{2}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}\, \sqrt {x^{4}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}}-\frac {2 \left (-1\right )^{\frac {3}{4}} \underline {\hspace {1.25 ex}}\alpha ^{7} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \underline {\hspace {1.25 ex}}\alpha ^{6}, i\right )}{\sqrt {x^{4}+1}}\right )\right )}{16}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{4 \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{4 \sqrt {x^{4}+1}}-\frac {x}{4 \sqrt {x^{4}+1}}\)	\(272\)
risch	\(\frac {3 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{4 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{8}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {\arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{6}+x^{2}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}\, \sqrt {x^{4}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}}-\frac {2 \left (-1\right )^{\frac {3}{4}} \underline {\hspace {1.25 ex}}\alpha ^{7} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \underline {\hspace {1.25 ex}}\alpha ^{6}, i\right )}{\sqrt {x^{4}+1}}\right )\right )}{16}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{4 \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{4 \sqrt {x^{4}+1}}-\frac {x}{4 \sqrt {x^{4}+1}}\)	\(272\)
trager	\(-\frac {x}{4 \sqrt {x^{4}+1}}-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) x +\sqrt {x^{4}+1}}{\left (-1+x \right ) \left (1+x \right )}\right )}{16}+\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{16}-\frac {\RootOf \left (\textit {\_Z}^{2}-2 \RootOf \left (\textit {\_Z}^{2}-2\right )\right ) \ln \left (-\frac {x^{4} \RootOf \left (\textit {\_Z}^{2}-2 \RootOf \left (\textit {\_Z}^{2}-2\right )\right )+x^{2} \RootOf \left (\textit {\_Z}^{2}-2 \RootOf \left (\textit {\_Z}^{2}-2\right )\right ) \RootOf \left (\textit {\_Z}^{2}-2\right )+4 \sqrt {x^{4}+1}\, x +\RootOf \left (\textit {\_Z}^{2}-2 \RootOf \left (\textit {\_Z}^{2}-2\right )\right )}{-x^{4}+\RootOf \left (\textit {\_Z}^{2}-2\right ) x^{2}-1}\right )}{16}+\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \RootOf \left (\textit {\_Z}^{2}+2\right ) \RootOf \left (\textit {\_Z}^{2}-2 \RootOf \left (\textit {\_Z}^{2}-2\right )\right ) \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{2}-2\right ) \RootOf \left (\textit {\_Z}^{2}-2 \RootOf \left (\textit {\_Z}^{2}-2\right )\right ) \RootOf \left (\textit {\_Z}^{2}+2\right ) x^{4}+\RootOf \left (\textit {\_Z}^{2}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}-2 \RootOf \left (\textit {\_Z}^{2}-2\right )\right ) \RootOf \left (\textit {\_Z}^{2}+2\right ) x^{2}-\RootOf \left (\textit {\_Z}^{2}-2\right ) \RootOf \left (\textit {\_Z}^{2}+2\right ) \RootOf \left (\textit {\_Z}^{2}-2 \RootOf \left (\textit {\_Z}^{2}-2\right )\right )+8 \sqrt {x^{4}+1}\, x}{x^{4}+\RootOf \left (\textit {\_Z}^{2}-2\right ) x^{2}+1}\right )}{32}\)	\(329\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{16} + 1}{{\left (x^{16} - 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{16}+1}{\sqrt {x^4+1}\,\left (x^{16}-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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