3.6.57 \(\int \frac {1+x^6}{\sqrt {1+x^4} (-1+x^6)} \, dx\) [557]

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

[Out]

-2/3*arctan(x/(x^4+1)^(1/2))-1/6*arctanh(2^(1/2)*x/(x^4+1)^(1/2))*2^(1/2)

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Rubi [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 1.08, 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 = {6857, 226, 2098, 1225, 1713, 213, 6860, 1739, 1231, 1721, 1262, 739, 212} \begin {gather*} -\frac {2}{3} \text {ArcTan}\left (\frac {x}{\sqrt {x^4+1}}\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 \text {ArcTan}(x)\left |\frac {1}{2}\right .\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 \text {ArcTan}(x)\left |\frac {1}{2}\right .\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 \text {ArcTan}(x)\left |\frac {1}{2}\right .\right )}{3 \sqrt {x^4+1}}-\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 )}} \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 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 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 739

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 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 1262

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 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 1739

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 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]

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*} \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} \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 [A]
time = 0.49, size = 43, normalized size = 1.00 \begin {gather*} -\frac {2}{3} \text {ArcTan}\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]

Integrate[(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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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.85, size = 571, normalized size = 13.28

method	result	size
elliptic	\(\frac {\left (\frac {2 \sqrt {2}\, \arctan \left (\frac {\sqrt {x^{4}+1}}{x}\right )}{3}+\frac {\ln \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{6}-\frac {\ln \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{6}\right ) \sqrt {2}}{2}\)	\(64\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) x +\sqrt {x^{4}+1}}{\left (1+x \right ) \left (-1+x \right )}\right )}{6}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}+\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \sqrt {x^{4}+1}\, x -\RootOf \left (\textit {\_Z}^{2}+1\right )}{\left (x^{2}+x +1\right ) \left (x^{2}-x +1\right )}\right )}{3}\)	\(108\)
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/(1/2*2^(1/2)+1/2*I*2^(1/2))*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticF(x*(1/2*2^(1/2)+1/2*I*2^(
1/2)),I)+1/3*(-1/2+1/2*I*3^(1/2))*(1/2/(1/2+1/2*I*3^(1/2))^(1/2)*arctanh((1/2+1/2*I*3^(1/2))^(1/2)*(x^2-1/2+1/
2*I*3^(1/2))/(x^4+1)^(1/2))+(-1)^(3/4)*(-1/2-1/2*I*3^(1/2))*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*Elli
pticPi((-1)^(1/4)*x,-I*(-1/2+1/2*I*3^(1/2)),I))+1/3*(-1/2-1/2*I*3^(1/2))*(1/2/(1/2-1/2*I*3^(1/2))^(1/2)*arctan
h((1/2-1/2*I*3^(1/2))^(1/2)*(x^2-1/2-1/2*I*3^(1/2))/(x^4+1)^(1/2))+(-1)^(3/4)*(-1/2+1/2*I*3^(1/2))*(1-I*x^2)^(
1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticPi((-1)^(1/4)*x,-I*(-1/2-1/2*I*3^(1/2)),I))+2/3*(-1)^(3/4)*(1-I*x^2
)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticPi((-1)^(1/4)*x,-I,(-I)^(1/2)/(-1)^(1/4))+1/3*(1/2+1/2*I*3^(1/2)
)*(-1/2/(1/2-1/2*I*3^(1/2))^(1/2)*arctanh((-1/2+1/2*I*3^(1/2))*(x^2-1/2-1/2*I*3^(1/2))/(1/2-1/2*I*3^(1/2))^(1/
2)/(x^4+1)^(1/2))+(-1)^(3/4)*(1/2-1/2*I*3^(1/2))*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticPi((-1)
^(1/4)*x,I*(1/2+1/2*I*3^(1/2)),I))+1/3*(1/2-1/2*I*3^(1/2))*(-1/2/(1/2+1/2*I*3^(1/2))^(1/2)*arctanh((-1/2-1/2*I
*3^(1/2))*(x^2-1/2+1/2*I*3^(1/2))/(1/2+1/2*I*3^(1/2))^(1/2)/(x^4+1)^(1/2))+(-1)^(3/4)*(1/2+1/2*I*3^(1/2))*(1-I
*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticPi((-1)^(1/4)*x,I*(1/2-1/2*I*3^(1/2)),I))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (31) = 62\).
time = 0.39, 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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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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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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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