3.9.91 \(\int \frac {-1+x^{12}}{\sqrt {1+x^4} (1+x^{12})} \, dx\)

Optimal. Leaf size=67 \[ -\frac {x}{3 \sqrt {x^4+1}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {x^4+1}}\right )}{3 \sqrt [4]{3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {x^4+1}}\right )}{3 \sqrt [4]{3}} \]

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Rubi [C]  time = 3.22, antiderivative size = 1382, normalized size of antiderivative = 20.63, number of steps used = 25, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6725, 220, 2073, 414, 523, 409, 1217, 1707}

result too large to display

Warning: Unable to verify antiderivative.

[In]

Int[(-1 + x^12)/(Sqrt[1 + x^4]*(1 + x^12)),x]

[Out]

(2*x)/(Sqrt[3]*(3*I - Sqrt[3])*Sqrt[1 + x^4]) - (2*x)/(Sqrt[3]*(3*I + Sqrt[3])*Sqrt[1 + x^4]) + ((2*I)*2^(1/4)
*ArcTan[(Sqrt[3 - I*Sqrt[3]]*x)/((2*(1 - I*Sqrt[3]))^(1/4)*Sqrt[1 + x^4])])/(Sqrt[3]*(1 - I*Sqrt[3])^(3/4)*(3
- I*Sqrt[3])^(3/2)) - ((2*I)*2^(1/4)*ArcTan[(Sqrt[-3 + I*Sqrt[3]]*x)/((2*(1 - I*Sqrt[3]))^(1/4)*Sqrt[1 + x^4])
])/(Sqrt[3]*(1 - I*Sqrt[3])^(3/4)*(-3 + I*Sqrt[3])^(3/2)) + ((2*I)*2^(1/4)*ArcTan[(Sqrt[-3 - I*Sqrt[3]]*x)/((2
*(1 + I*Sqrt[3]))^(1/4)*Sqrt[1 + x^4])])/(Sqrt[3]*(-3 - I*Sqrt[3])^(3/2)*(1 + I*Sqrt[3])^(3/4)) - ((2*I)*2^(1/
4)*ArcTan[(Sqrt[3 + I*Sqrt[3]]*x)/((2*(1 + I*Sqrt[3]))^(1/4)*Sqrt[1 + x^4])])/(Sqrt[3]*(1 + I*Sqrt[3])^(3/4)*(
3 + I*Sqrt[3])^(3/2)) + ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(2*Sqrt[1 + x^4])
+ ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(Sqrt[3]*(3*I - Sqrt[3])*Sqrt[1 + x^4])
- ((1/6 - I/6)*(1 - Sqrt[3])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/((I - Sqrt[3])
*Sqrt[1 + x^4]) + ((1/6 + I/6)*(1 - Sqrt[3])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2]
)/((I + Sqrt[3])*Sqrt[1 + x^4]) - ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(Sqrt[3]
*(3*I + Sqrt[3])*Sqrt[1 + x^4]) + ((1 + 1/Sqrt[(1 - I*Sqrt[3])/2])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*Ellip
ticF[2*ArcTan[x], 1/2])/(Sqrt[3]*(3*I - Sqrt[3])*Sqrt[1 + x^4]) - ((1 + 1/Sqrt[(1 + I*Sqrt[3])/2])*(1 + x^2)*S
qrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(Sqrt[3]*(3*I + Sqrt[3])*Sqrt[1 + x^4]) + ((3 + Sqrt[3
]*(3*I + (2*I)*Sqrt[2 - (2*I)*Sqrt[3]]))*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[-1/4*(Sqrt[2] - Sqrt
[1 - I*Sqrt[3]])^2/Sqrt[2*(1 - I*Sqrt[3])], 2*ArcTan[x], 1/2])/(12*(3 - I*Sqrt[3])*Sqrt[1 + x^4]) + ((3*I - Sq
rt[3]*(3 - 2*Sqrt[2 - (2*I)*Sqrt[3]]))*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[(Sqrt[2] + Sqrt[1 - I*
Sqrt[3]])^2/(4*Sqrt[2*(1 - I*Sqrt[3])]), 2*ArcTan[x], 1/2])/(12*(3*I + Sqrt[3])*Sqrt[1 + x^4]) + ((3*I + 3*Sqr
t[3] + 2*Sqrt[6 + (6*I)*Sqrt[3]])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[-1/4*(Sqrt[2] - Sqrt[1 + I*
Sqrt[3]])^2/Sqrt[2*(1 + I*Sqrt[3])], 2*ArcTan[x], 1/2])/(12*(3*I - Sqrt[3])*Sqrt[1 + x^4]) + ((3*I + Sqrt[3]*(
3 - 2*Sqrt[2 + (2*I)*Sqrt[3]]))*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[(Sqrt[2] + Sqrt[1 + I*Sqrt[3]
])^2/(4*Sqrt[2*(1 + I*Sqrt[3])]), 2*ArcTan[x], 1/2])/(12*(3*I - Sqrt[3])*Sqrt[1 + x^4])

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 414

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

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 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^{12}}{\sqrt {1+x^4} \left (1+x^{12}\right )} \, dx &=\int \left (\frac {1}{\sqrt {1+x^4}}-\frac {2}{\sqrt {1+x^4} \left (1+x^{12}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\sqrt {1+x^4} \left (1+x^{12}\right )} \, dx\right )+\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 {2 i}{\sqrt {3} \left (1+i \sqrt {3}-2 x^4\right ) \left (1+x^4\right )^{3/2}}+\frac {2 i}{\sqrt {3} \left (1+x^4\right )^{3/2} \left (-1+i \sqrt {3}+2 x^4\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 {(4 i) \int \frac {1}{\left (1+i \sqrt {3}-2 x^4\right ) \left (1+x^4\right )^{3/2}} \, dx}{\sqrt {3}}-\frac {(4 i) \int \frac {1}{\left (1+x^4\right )^{3/2} \left (-1+i \sqrt {3}+2 x^4\right )} \, dx}{\sqrt {3}}\\ &=\frac {2 x}{\sqrt {3} \left (3 i-\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {2 x}{\sqrt {3} \left (3 i+\sqrt {3}\right ) \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 {(2 i) \int \frac {5-i \sqrt {3}-2 x^4}{\sqrt {1+x^4} \left (-1+i \sqrt {3}+2 x^4\right )} \, dx}{\sqrt {3} \left (3-i \sqrt {3}\right )}+\frac {(2 i) \int \frac {-5-i \sqrt {3}+2 x^4}{\left (1+i \sqrt {3}-2 x^4\right ) \sqrt {1+x^4}} \, dx}{\sqrt {3} \left (3+i \sqrt {3}\right )}\\ &=\frac {2 x}{\sqrt {3} \left (3 i-\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {2 x}{\sqrt {3} \left (3 i+\sqrt {3}\right ) \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 {(2 i) \int \frac {1}{\sqrt {1+x^4}} \, dx}{\sqrt {3} \left (3-i \sqrt {3}\right )}-\frac {(8 i) \int \frac {1}{\sqrt {1+x^4} \left (-1+i \sqrt {3}+2 x^4\right )} \, dx}{\sqrt {3} \left (3-i \sqrt {3}\right )}-\frac {(2 i) \int \frac {1}{\sqrt {1+x^4}} \, dx}{\sqrt {3} \left (3+i \sqrt {3}\right )}-\frac {(8 i) \int \frac {1}{\left (1+i \sqrt {3}-2 x^4\right ) \sqrt {1+x^4}} \, dx}{\sqrt {3} \left (3+i \sqrt {3}\right )}\\ &=\frac {2 x}{\sqrt {3} \left (3 i-\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {2 x}{\sqrt {3} \left (3 i+\sqrt {3}\right ) \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 {\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 {3} \left (3 i-\sqrt {3}\right ) \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 )}{\sqrt {3} \left (3 i+\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {1}{3} \int \frac {1}{\left (1-\frac {x^2}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}}\right ) \sqrt {1+x^4}} \, dx-\frac {1}{3} \int \frac {1}{\left (1+\frac {x^2}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}}\right ) \sqrt {1+x^4}} \, dx-\frac {1}{3} \int \frac {1}{\left (1-\frac {x^2}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}}\right ) \sqrt {1+x^4}} \, dx-\frac {1}{3} \int \frac {1}{\left (1+\frac {x^2}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}}\right ) \sqrt {1+x^4}} \, dx\\ &=\frac {2 x}{\sqrt {3} \left (3 i-\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {2 x}{\sqrt {3} \left (3 i+\sqrt {3}\right ) \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 {\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 {3} \left (3 i-\sqrt {3}\right ) \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 )}{\sqrt {3} \left (3 i+\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {\left (\left (\frac {1}{3}-\frac {i}{3}\right ) \left (1-\sqrt {3}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{i-\sqrt {3}}--\frac {\left (\left (\frac {1}{3}+\frac {i}{3}\right ) \left (1-\sqrt {3}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{i+\sqrt {3}}-\frac {\left (1+\frac {1}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{3 \left (1-\frac {2 i}{i-\sqrt {3}}\right )}+\frac {\left (1+\frac {1}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}}\right ) \int \frac {1+x^2}{\left (1+\frac {x^2}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}}\right ) \sqrt {1+x^4}} \, dx}{3 \sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )} \left (1-\frac {2}{1+i \sqrt {3}}\right )}-\frac {\left (2-\sqrt {2-2 i \sqrt {3}}\right ) \int \frac {1+x^2}{\left (1-\frac {x^2}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}}\right ) \sqrt {1+x^4}} \, dx}{3 \left (1+i \sqrt {3}\right )}-\frac {\left (2+\sqrt {2-2 i \sqrt {3}}\right ) \int \frac {1+x^2}{\left (1+\frac {x^2}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}}\right ) \sqrt {1+x^4}} \, dx}{3 \left (1+i \sqrt {3}\right )}-\frac {\left (2-\sqrt {2+2 i \sqrt {3}}\right ) \int \frac {1+x^2}{\left (1-\frac {x^2}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}}\right ) \sqrt {1+x^4}} \, dx}{3 \left (1-i \sqrt {3}\right )}-\frac {\left (1+\frac {1}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{3 \left (1-\frac {2 i}{i+\sqrt {3}}\right )}\\ &=\frac {2 x}{\sqrt {3} \left (3 i-\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {2 x}{\sqrt {3} \left (3 i+\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {\sqrt [4]{1-i \sqrt {3}} \tan ^{-1}\left (\frac {\sqrt {3-i \sqrt {3}} x}{\sqrt [4]{2 \left (1-i \sqrt {3}\right )} \sqrt {1+x^4}}\right )}{3\ 2^{3/4} \sqrt {3-i \sqrt {3}}}-\frac {\sqrt [4]{1-i \sqrt {3}} \tan ^{-1}\left (\frac {\sqrt {-3+i \sqrt {3}} x}{\sqrt [4]{2 \left (1-i \sqrt {3}\right )} \sqrt {1+x^4}}\right )}{3\ 2^{3/4} \sqrt {-3+i \sqrt {3}}}-\frac {\sqrt [4]{1+i \sqrt {3}} \tan ^{-1}\left (\frac {\sqrt {-3-i \sqrt {3}} x}{\sqrt [4]{2 \left (1+i \sqrt {3}\right )} \sqrt {1+x^4}}\right )}{3\ 2^{3/4} \sqrt {-3-i \sqrt {3}}}-\frac {2 \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt {3+i \sqrt {3}} x}{\sqrt [4]{2 \left (1+i \sqrt {3}\right )} \sqrt {1+x^4}}\right )}{3 \left (1-i \sqrt {3}\right ) \left (1+i \sqrt {3}\right )^{3/4} \sqrt {3+i \sqrt {3}}}+\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 {\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 {3} \left (3 i-\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {\left (\frac {1}{6}-\frac {i}{6}\right ) \left (1-\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 )}{\left (i-\sqrt {3}\right ) \sqrt {1+x^4}}+\frac {\left (\frac {1}{6}+\frac {i}{6}\right ) \left (1-\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 )}{\left (i+\sqrt {3}\right ) \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 )}{\sqrt {3} \left (3 i+\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {\left (1+\frac {1}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}}\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 )}{6 \left (1-\frac {2 i}{i-\sqrt {3}}\right ) \sqrt {1+x^4}}-\frac {\left (1+\frac {1}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}}\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 )}{6 \left (1-\frac {2 i}{i+\sqrt {3}}\right ) \sqrt {1+x^4}}-\frac {\left (2+\sqrt {2-2 i \sqrt {3}}\right )^2 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \Pi \left (-\frac {\left (\sqrt {2}-\sqrt {1-i \sqrt {3}}\right )^2}{4 \sqrt {2 \left (1-i \sqrt {3}\right )}};2 \tan ^{-1}(x)|\frac {1}{2}\right )}{24 \left (1+i \sqrt {3}\right ) \sqrt {1+x^4}}-\frac {\left (2-\sqrt {2-2 i \sqrt {3}}\right )^2 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \Pi \left (\frac {\left (\sqrt {2}+\sqrt {1-i \sqrt {3}}\right )^2}{4 \sqrt {2 \left (1-i \sqrt {3}\right )}};2 \tan ^{-1}(x)|\frac {1}{2}\right )}{24 \left (1+i \sqrt {3}\right ) \sqrt {1+x^4}}-\frac {\left (3 i-\sqrt {3}+2 i \sqrt {2+2 i \sqrt {3}}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \Pi \left (-\frac {\left (\sqrt {2}-\sqrt {1+i \sqrt {3}}\right )^2}{4 \sqrt {2 \left (1+i \sqrt {3}\right )}};2 \tan ^{-1}(x)|\frac {1}{2}\right )}{12 \left (i+\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {\left (2-\sqrt {2+2 i \sqrt {3}}\right )^2 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \Pi \left (\frac {\left (\sqrt {2}+\sqrt {1+i \sqrt {3}}\right )^2}{4 \sqrt {2 \left (1+i \sqrt {3}\right )}};2 \tan ^{-1}(x)|\frac {1}{2}\right )}{24 \left (1-i \sqrt {3}\right ) \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.81, size = 175, normalized size = 2.61 \begin {gather*} \frac {1}{3} \left (-\frac {(-1)^{2/3} x}{\sqrt {x^4+1}}+\frac {\sqrt [3]{-1} x}{\sqrt {x^4+1}}-\frac {2 x}{\sqrt {x^4+1}}-2 \sqrt [4]{-1} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\sqrt [4]{-1} \Pi \left (-\sqrt [3]{-1};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\sqrt [4]{-1} \Pi \left (\sqrt [3]{-1};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\sqrt [4]{-1} \Pi \left (-(-1)^{2/3};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\sqrt [4]{-1} \Pi \left ((-1)^{2/3};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^12)/(Sqrt[1 + x^4]*(1 + x^12)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.52, size = 67, normalized size = 1.00 \begin {gather*} -\frac {x}{3 \sqrt {1+x^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {1+x^4}}\right )}{3 \sqrt [4]{3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {1+x^4}}\right )}{3 \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x^12)/(Sqrt[1 + x^4]*(1 + x^12)),x]

[Out]

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

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fricas [B]  time = 0.60, size = 247, normalized size = 3.69 \begin {gather*} -\frac {4 \cdot 3^{\frac {3}{4}} {\left (x^{4} + 1\right )} \arctan \left (\frac {3^{\frac {3}{4}} {\left (2 \cdot 3^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )} + 3^{\frac {1}{4}} {\left (x^{8} + 5 \, x^{4} + 1\right )}\right )} + 6 \, \sqrt {x^{4} + 1} {\left (3^{\frac {3}{4}} x^{3} + 3^{\frac {1}{4}} {\left (x^{5} + x\right )}\right )}}{3 \, {\left (x^{8} - x^{4} + 1\right )}}\right ) + 3^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (-\frac {3^{\frac {3}{4}} {\left (x^{8} + 5 \, x^{4} + 1\right )} + 6 \, {\left (x^{5} + \sqrt {3} x^{3} + x\right )} \sqrt {x^{4} + 1} + 6 \cdot 3^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} - x^{4} + 1}\right ) - 3^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {3^{\frac {3}{4}} {\left (x^{8} + 5 \, x^{4} + 1\right )} - 6 \, {\left (x^{5} + \sqrt {3} x^{3} + x\right )} \sqrt {x^{4} + 1} + 6 \cdot 3^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} - x^{4} + 1}\right ) + 12 \, \sqrt {x^{4} + 1} x}{36 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^12-1)/(x^4+1)^(1/2)/(x^12+1),x, algorithm="fricas")

[Out]

-1/36*(4*3^(3/4)*(x^4 + 1)*arctan(1/3*(3^(3/4)*(2*3^(3/4)*(x^6 + x^2) + 3^(1/4)*(x^8 + 5*x^4 + 1)) + 6*sqrt(x^
4 + 1)*(3^(3/4)*x^3 + 3^(1/4)*(x^5 + x)))/(x^8 - x^4 + 1)) + 3^(3/4)*(x^4 + 1)*log(-(3^(3/4)*(x^8 + 5*x^4 + 1)
+ 6*(x^5 + sqrt(3)*x^3 + x)*sqrt(x^4 + 1) + 6*3^(1/4)*(x^6 + x^2))/(x^8 - x^4 + 1)) - 3^(3/4)*(x^4 + 1)*log((
3^(3/4)*(x^8 + 5*x^4 + 1) - 6*(x^5 + sqrt(3)*x^3 + x)*sqrt(x^4 + 1) + 6*3^(1/4)*(x^6 + x^2))/(x^8 - x^4 + 1))
+ 12*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^{12} - 1}{{\left (x^{12} + 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^12-1)/(x^4+1)^(1/2)/(x^12+1),x, algorithm="giac")

[Out]

integrate((x^12 - 1)/((x^12 + 1)*sqrt(x^4 + 1)), x)

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maple [B]  time = 1.32, size = 104, normalized size = 1.55

method	result	size
elliptic	\(\frac {\left (\frac {\sqrt {2}\, 3^{\frac {3}{4}} \arctan \left (\frac {3^{\frac {3}{4}} \sqrt {x^{4}+1}}{3 x}\right )}{9}-\frac {\sqrt {2}\, 3^{\frac {3}{4}} \ln \left (\frac {\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}+\frac {\sqrt {2}\, 3^{\frac {1}{4}}}{2}}{\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}-\frac {\sqrt {2}\, 3^{\frac {1}{4}}}{2}}\right )}{18}-\frac {\sqrt {2}\, x}{3 \sqrt {x^{4}+1}}\right ) \sqrt {2}}{2}\)	\(104\)
default	\(\frac {2 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{3 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {\arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{6}+\underline {\hspace {1.25 ex}}\alpha ^{2}+x^{2}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}\, \sqrt {x^{4}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}}+\frac {2 \left (-1\right )^{\frac {3}{4}} \left (-\underline {\hspace {1.25 ex}}\alpha ^{7}+\underline {\hspace {1.25 ex}}\alpha ^{3}\right ) \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \underline {\hspace {1.25 ex}}\alpha ^{6}-i \underline {\hspace {1.25 ex}}\alpha ^{2}, i\right )}{\sqrt {x^{4}+1}}\right )\right )}{12}-\frac {x}{3 \sqrt {x^{4}+1}}\)	\(193\)
risch	\(\frac {2 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{3 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {\arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{6}+\underline {\hspace {1.25 ex}}\alpha ^{2}+x^{2}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}\, \sqrt {x^{4}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}}+\frac {2 \left (-1\right )^{\frac {3}{4}} \left (-\underline {\hspace {1.25 ex}}\alpha ^{7}+\underline {\hspace {1.25 ex}}\alpha ^{3}\right ) \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \underline {\hspace {1.25 ex}}\alpha ^{6}-i \underline {\hspace {1.25 ex}}\alpha ^{2}, i\right )}{\sqrt {x^{4}+1}}\right )\right )}{12}-\frac {x}{3 \sqrt {x^{4}+1}}\)	\(193\)
trager	\(-\frac {x}{3 \sqrt {x^{4}+1}}+\frac {\RootOf \left (\textit {\_Z}^{4}-27\right ) \ln \left (-\frac {-x^{2} \RootOf \left (\textit {\_Z}^{4}-27\right )^{3}-3 \RootOf \left (\textit {\_Z}^{4}-27\right ) x^{4}+18 \sqrt {x^{4}+1}\, x -3 \RootOf \left (\textit {\_Z}^{4}-27\right )}{\RootOf \left (\textit {\_Z}^{4}-27\right )^{2} x^{2}-3 x^{4}-3}\right )}{18}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right ) \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right ) x^{2} \RootOf \left (\textit {\_Z}^{4}-27\right )^{2}+3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right ) x^{4}+18 \sqrt {x^{4}+1}\, x +3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right )}{3 x^{4}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2} x^{2}+3}\right )}{18}\)	\(196\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^12-1)/(x^4+1)^(1/2)/(x^12+1),x,method=_RETURNVERBOSE)

[Out]

1/2*(1/9*2^(1/2)*3^(3/4)*arctan(1/3*3^(3/4)/x*(x^4+1)^(1/2))-1/18*2^(1/2)*3^(3/4)*ln((1/2*2^(1/2)/x*(x^4+1)^(1
/2)+1/2*2^(1/2)*3^(1/4))/(1/2*2^(1/2)/x*(x^4+1)^(1/2)-1/2*2^(1/2)*3^(1/4)))-1/3*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^{12} - 1}{{\left (x^{12} + 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^12-1)/(x^4+1)^(1/2)/(x^12+1),x, algorithm="maxima")

[Out]

integrate((x^12 - 1)/((x^12 + 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^{12}-1}{\sqrt {x^4+1}\,\left (x^{12}+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^12 - 1)/((x^4 + 1)^(1/2)*(x^12 + 1)),x)

[Out]

int((x^12 - 1)/((x^4 + 1)^(1/2)*(x^12 + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right ) \left (x^{4} - x^{2} + 1\right )}{\left (x^{4} + 1\right )^{\frac {3}{2}} \left (x^{8} - x^{4} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**12-1)/(x**4+1)**(1/2)/(x**12+1),x)

[Out]

Integral((x - 1)*(x + 1)*(x**2 + 1)*(x**2 - x + 1)*(x**2 + x + 1)*(x**4 - x**2 + 1)/((x**4 + 1)**(3/2)*(x**8 -
x**4 + 1)), x)

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