3.6.57 \(\int \frac {1+x^6}{\sqrt {1+x^4} (-1+x^6)} \, dx\)

Optimal. Leaf size=43 \[ -\frac {2}{3} \tan ^{-1}\left (\frac {x}{\sqrt {x^4+1}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{3 \sqrt {2}} \]

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Rubi [C]  time = 1.32, antiderivative size = 354, normalized size of antiderivative = 8.23, number of steps used = 41, number of rules used = 13, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {6725, 220, 2073, 1211, 1699, 207, 6728, 1725, 1217, 1707, 1248, 725, 206} \begin {gather*} -\frac {2}{3} \tan ^{-1}\left (\frac {x}{\sqrt {x^4+1}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{3 \sqrt {2}}-\frac {\left (1+i \sqrt {3}\right ) \tanh ^{-1}\left (\frac {2-\left (1-i \sqrt {3}\right ) x^2}{\sqrt {2 \left (1-i \sqrt {3}\right )} \sqrt {x^4+1}}\right )}{6 \sqrt {2 \left (1-i \sqrt {3}\right )}}+\frac {\left (1+i \sqrt {3}\right ) \tanh ^{-1}\left (\frac {4+\left (1+i \sqrt {3}\right )^2 x^2}{2 \sqrt {2 \left (1-i \sqrt {3}\right )} \sqrt {x^4+1}}\right )}{6 \sqrt {2 \left (1-i \sqrt {3}\right )}}-\frac {\left (1+i \sqrt {3}\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 )}{6 \sqrt {x^4+1}}-\frac {\left (1-i \sqrt {3}\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 )}{6 \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 )}{3 \sqrt {x^4+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[x/Sqrt[1 + x^4]])/3 - ArcTanh[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(3*Sqrt[2]) - ((1 + I*Sqrt[3])*ArcTanh[(2
- (1 - I*Sqrt[3])*x^2)/(Sqrt[2*(1 - I*Sqrt[3])]*Sqrt[1 + x^4])])/(6*Sqrt[2*(1 - I*Sqrt[3])]) + ((1 + I*Sqrt[3]
)*ArcTanh[(4 + (1 + I*Sqrt[3])^2*x^2)/(2*Sqrt[2*(1 - I*Sqrt[3])]*Sqrt[1 + x^4])])/(6*Sqrt[2*(1 - I*Sqrt[3])])
+ ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(3*Sqrt[1 + x^4]) - ((1 - I*Sqrt[3])*(1
+ x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(6*Sqrt[1 + x^4]) - ((1 + I*Sqrt[3])*(1 + x^2)
*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(6*Sqrt[1 + x^4])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 725

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

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 1248

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

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 1725

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

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]

Rule 6728

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*} \int \frac {1+x^6}{\sqrt {1+x^4} \left (-1+x^6\right )} \, dx &=\int \left (\frac {1}{\sqrt {1+x^4}}+\frac {2}{\sqrt {1+x^4} \left (-1+x^6\right )}\right ) \, dx\\ &=2 \int \frac {1}{\sqrt {1+x^4} \left (-1+x^6\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}{3 \left (-1+x^2\right ) \sqrt {1+x^4}}+\frac {-2+x}{6 \left (1-x+x^2\right ) \sqrt {1+x^4}}+\frac {-2-x}{6 \left (1+x+x^2\right ) \sqrt {1+x^4}}\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}{3} \int \frac {-2+x}{\left (1-x+x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \int \frac {-2-x}{\left (1+x+x^2\right ) \sqrt {1+x^4}} \, dx+\frac {2}{3} \int \frac {1}{\left (-1+x^2\right ) \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}}-\frac {1}{3} \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{3} \int \frac {-1-x^2}{\left (-1+x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \int \left (\frac {1+i \sqrt {3}}{\left (-1-i \sqrt {3}+2 x\right ) \sqrt {1+x^4}}+\frac {1-i \sqrt {3}}{\left (-1+i \sqrt {3}+2 x\right ) \sqrt {1+x^4}}\right ) \, dx+\frac {1}{3} \int \left (\frac {-1+i \sqrt {3}}{\left (1-i \sqrt {3}+2 x\right ) \sqrt {1+x^4}}+\frac {-1-i \sqrt {3}}{\left (1+i \sqrt {3}+2 x\right ) \sqrt {1+x^4}}\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 )}{3 \sqrt {1+x^4}}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )+\frac {1}{3} \left (-1-i \sqrt {3}\right ) \int \frac {1}{\left (1+i \sqrt {3}+2 x\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (1-i \sqrt {3}\right ) \int \frac {1}{\left (-1+i \sqrt {3}+2 x\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (-1+i \sqrt {3}\right ) \int \frac {1}{\left (1-i \sqrt {3}+2 x\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (1+i \sqrt {3}\right ) \int \frac {1}{\left (-1-i \sqrt {3}+2 x\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \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 )}{3 \sqrt {1+x^4}}+\frac {1}{3} \left (2 \left (1-i \sqrt {3}\right )\right ) \int \frac {1}{\left (\left (-1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (2 \left (1-i \sqrt {3}\right )\right ) \int \frac {x}{\left (\left (1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{3} \left (2 \left (1-i \sqrt {3}\right )\right ) \int \frac {x}{\left (\left (-1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (2 \left (1-i \sqrt {3}\right )\right ) \int \frac {1}{\left (\left (1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{3} \left (2 \left (1+i \sqrt {3}\right )\right ) \int \frac {x}{\left (\left (-1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (2 \left (1+i \sqrt {3}\right )\right ) \int \frac {1}{\left (\left (1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (2 \left (1+i \sqrt {3}\right )\right ) \int \frac {1}{\left (\left (-1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (2 \left (1+i \sqrt {3}\right )\right ) \int \frac {x}{\left (\left (1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \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 )}{3 \sqrt {1+x^4}}+\frac {1}{3} \left (1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (1-i \sqrt {3}\right )^2-4 x\right ) \sqrt {1+x^2}} \, dx,x,x^2\right )-\frac {1}{3} \left (1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (-1+i \sqrt {3}\right )^2-4 x\right ) \sqrt {1+x^2}} \, dx,x,x^2\right )-\frac {1}{3} \left (1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (-1-i \sqrt {3}\right )^2-4 x\right ) \sqrt {1+x^2}} \, dx,x,x^2\right )+\frac {1}{3} \left (1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (1+i \sqrt {3}\right )^2-4 x\right ) \sqrt {1+x^2}} \, dx,x,x^2\right )+2 \frac {\left (i-\sqrt {3}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{3 \left (i+\sqrt {3}\right )}+\frac {\left (4 \left (i-\sqrt {3}\right )\right ) \int \frac {1+x^2}{\left (\left (1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx}{3 \left (i+\sqrt {3}\right )}+\frac {\left (4 \left (i-\sqrt {3}\right )\right ) \int \frac {1+x^2}{\left (\left (-1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx}{3 \left (i+\sqrt {3}\right )}+2 \frac {\left (i+\sqrt {3}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{3 \left (i-\sqrt {3}\right )}+\frac {\left (4 \left (i+\sqrt {3}\right )\right ) \int \frac {1+x^2}{\left (\left (-1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx}{3 \left (i-\sqrt {3}\right )}+\frac {\left (4 \left (i+\sqrt {3}\right )\right ) \int \frac {1+x^2}{\left (\left (1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx}{3 \left (i-\sqrt {3}\right )}\\ &=-\frac {2}{3} \tan ^{-1}\left (\frac {x}{\sqrt {1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \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 )}{3 \sqrt {1+x^4}}+\frac {\left (i-\sqrt {3}\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 )}{3 \left (i+\sqrt {3}\right ) \sqrt {1+x^4}}+\frac {\left (i+\sqrt {3}\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 )}{3 \left (i-\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {1}{3} \left (-1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{16+\left (-1-i \sqrt {3}\right )^4-x^2} \, dx,x,\frac {-4-\left (-1-i \sqrt {3}\right )^2 x^2}{\sqrt {1+x^4}}\right )+\frac {1}{3} \left (-1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{16+\left (1+i \sqrt {3}\right )^4-x^2} \, dx,x,\frac {-4-\left (1+i \sqrt {3}\right )^2 x^2}{\sqrt {1+x^4}}\right )+\frac {1}{3} \left (-1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{16+\left (1-i \sqrt {3}\right )^4-x^2} \, dx,x,\frac {-4-\left (1-i \sqrt {3}\right )^2 x^2}{\sqrt {1+x^4}}\right )-\frac {1}{3} \left (-1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{16+\left (-1+i \sqrt {3}\right )^4-x^2} \, dx,x,\frac {-4-\left (-1+i \sqrt {3}\right )^2 x^2}{\sqrt {1+x^4}}\right )\\ &=-\frac {2}{3} \tan ^{-1}\left (\frac {x}{\sqrt {1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \sqrt {2}}-\frac {\left (1+i \sqrt {3}\right ) \tanh ^{-1}\left (\frac {2-\left (1-i \sqrt {3}\right ) x^2}{\sqrt {2 \left (1-i \sqrt {3}\right )} \sqrt {1+x^4}}\right )}{6 \sqrt {2 \left (1-i \sqrt {3}\right )}}+\frac {\left (1+i \sqrt {3}\right ) \tanh ^{-1}\left (\frac {4+\left (1+i \sqrt {3}\right )^2 x^2}{2 \sqrt {2 \left (1-i \sqrt {3}\right )} \sqrt {1+x^4}}\right )}{6 \sqrt {2 \left (1-i \sqrt {3}\right )}}+\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 )}{3 \sqrt {1+x^4}}+\frac {\left (i-\sqrt {3}\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 )}{3 \left (i+\sqrt {3}\right ) \sqrt {1+x^4}}+\frac {\left (i+\sqrt {3}\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 )}{3 \left (i-\sqrt {3}\right ) \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.18, size = 114, normalized size = 2.65 \begin {gather*} \frac {1}{3} \sqrt [4]{-1} \left (2 \left (\Pi \left (i;\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\Pi \left (-\frac {i}{2}-\frac {\sqrt {3}}{2};\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\Pi \left (\frac {1}{2} \left (-i+\sqrt {3}\right );\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )\right )-3 F\left (\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^6)/(Sqrt[1 + x^4]*(-1 + x^6)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.70, size = 43, normalized size = 1.00 \begin {gather*} -\frac {2}{3} \tan ^{-1}\left (\frac {x}{\sqrt {1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^6)/(Sqrt[1 + x^4]*(-1 + x^6)),x]

[Out]

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

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fricas [B]  time = 0.52, size = 68, normalized size = 1.58 \begin {gather*} \frac {1}{12} \, \sqrt {2} \log \left (\frac {x^{4} - 2 \, \sqrt {2} \sqrt {x^{4} + 1} x + 2 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) - \frac {1}{3} \, \arctan \left (\frac {2 \, \sqrt {x^{4} + 1} x}{x^{4} - x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(2)*log((x^4 - 2*sqrt(2)*sqrt(x^4 + 1)*x + 2*x^2 + 1)/(x^4 - 2*x^2 + 1)) - 1/3*arctan(2*sqrt(x^4 + 1)
*x/(x^4 - x^2 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.35, size = 64, normalized size = 1.49

method	result	size
elliptic	\(\frac {\left (-\frac {\ln \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{6}+\frac {2 \sqrt {2}\, \arctan \left (\frac {\sqrt {x^{4}+1}}{x}\right )}{3}+\frac {\ln \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{6}\right ) \sqrt {2}}{2}\)	\(64\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) x +\sqrt {x^{4}+1}}{\left (-1+x \right ) \left (1+x \right )}\right )}{6}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \sqrt {x^{4}+1}\, x +\RootOf \left (\textit {\_Z}^{2}+1\right )}{\left (x^{2}+x +1\right ) \left (x^{2}-x +1\right )}\right )}{3}\)	\(106\)
default	\(\frac {\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 )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {\arctanh \left (\frac {\sqrt {\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \left (x^{2}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{\sqrt {x^{4}+1}}\right )}{2 \sqrt {\frac {1}{2}+\frac {i \sqrt {3}}{2}}}+\frac {\left (-1\right )^{\frac {3}{4}} \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ), i\right )}{\sqrt {x^{4}+1}}\right )}{3}+\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (\frac {\arctanh \left (\frac {\sqrt {\frac {1}{2}-\frac {i \sqrt {3}}{2}}\, \left (x^{2}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\sqrt {x^{4}+1}}\right )}{2 \sqrt {\frac {1}{2}-\frac {i \sqrt {3}}{2}}}+\frac {\left (-1\right )^{\frac {3}{4}} \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ), i\right )}{\sqrt {x^{4}+1}}\right )}{3}+\frac {2 \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 )}{3 \sqrt {x^{4}+1}}+\frac {\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {\arctanh \left (\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x^{2}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\sqrt {\frac {1}{2}-\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+1}}\right )}{2 \sqrt {\frac {1}{2}-\frac {i \sqrt {3}}{2}}}+\frac {\left (-1\right )^{\frac {3}{4}} \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ), i\right )}{\sqrt {x^{4}+1}}\right )}{3}+\frac {\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (-\frac {\arctanh \left (\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (x^{2}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{\sqrt {\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+1}}\right )}{2 \sqrt {\frac {1}{2}+\frac {i \sqrt {3}}{2}}}+\frac {\left (-1\right )^{\frac {3}{4}} \left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ), i\right )}{\sqrt {x^{4}+1}}\right )}{3}\)	\(571\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6+1)/(x^4+1)^(1/2)/(x^6-1),x,method=_RETURNVERBOSE)

[Out]

1/2*(-1/6*ln(1+1/2*2^(1/2)/x*(x^4+1)^(1/2))+2/3*2^(1/2)*arctan((x^4+1)^(1/2)/x)+1/6*ln(-1+1/2*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^{6} + 1}{{\left (x^{6} - 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^6+1}{\sqrt {x^4+1}\,\left (x^6-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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