3.11.7 \(\int \frac {(-1+x^4) (1-x^2+x^4) \sqrt {4-x^2+4 x^4}}{(1+x^4) (4+7 x^4+4 x^8)} \, dx\)

Optimal. Leaf size=76 \[ \tan ^{-1}\left (\frac {x}{\sqrt {4 x^4-x^2+4}}\right )-\frac {3}{4} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {4 x^4-x^2+4}}\right )+\frac {1}{4} \tanh ^{-1}\left (\frac {x}{\sqrt {4 x^4-x^2+4}}\right ) \]

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Rubi [C]  time = 3.67, antiderivative size = 1134, normalized size of antiderivative = 14.92, number of steps used = 50, number of rules used = 8, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.151, Rules used = {6725, 1208, 1197, 1103, 1195, 1216, 1706, 6728} \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt {4 x^4-x^2+4}}\right )-\frac {9 \sqrt {5} \left (i+\sqrt {15}\right )^2 \tan ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {4 x^4-x^2+4}}\right )}{16 \left (15 i+7 \sqrt {15}\right )}-\frac {9 \sqrt {5} \left (7 i+\sqrt {15}\right ) \tan ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {4 x^4-x^2+4}}\right )}{8 \left (15+7 i \sqrt {15}\right )}+\frac {1}{4} \tanh ^{-1}\left (\frac {x}{\sqrt {4 x^4-x^2+4}}\right )-\frac {9 \left (5 i+3 \sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \left (15 i+\sqrt {15}\right ) \sqrt {4 x^4-x^2+4}}+\frac {\left (9 i-\sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \left (15 i+\sqrt {15}\right ) \sqrt {4 x^4-x^2+4}}+\frac {\left (9 i+\sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \left (15 i-\sqrt {15}\right ) \sqrt {4 x^4-x^2+4}}-\frac {\left (4+i \sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \sqrt {4 x^4-x^2+4}}-\frac {3 \left (2+i \sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \sqrt {4 x^4-x^2+4}}-\frac {\left (4-i \sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \sqrt {4 x^4-x^2+4}}-\frac {3 \left (2-i \sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \sqrt {4 x^4-x^2+4}}-\frac {9 \left (5 i-3 \sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \left (15 i-\sqrt {15}\right ) \sqrt {4 x^4-x^2+4}}+\frac {7 \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{4 \sqrt {4 x^4-x^2+4}}+\frac {45 \left (7 i-\sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} \Pi \left (\frac {3}{8};2 \tan ^{-1}(x)|\frac {9}{16}\right )}{32 \left (15 i+7 \sqrt {15}\right ) \sqrt {4 x^4-x^2+4}}+\frac {45 \left (7 i+\sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} \Pi \left (\frac {3}{8};2 \tan ^{-1}(x)|\frac {9}{16}\right )}{32 \left (15 i-7 \sqrt {15}\right ) \sqrt {4 x^4-x^2+4}}+\frac {3 \left (7 i-\sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} \Pi \left (\frac {5}{8};2 \tan ^{-1}(x)|\frac {9}{16}\right )}{32 \left (15 i+7 \sqrt {15}\right ) \sqrt {4 x^4-x^2+4}}+\frac {3 \left (7 i+\sqrt {15}\right ) \left (x^2+1\right ) \sqrt {\frac {4 x^4-x^2+4}{\left (x^2+1\right )^2}} \Pi \left (\frac {5}{8};2 \tan ^{-1}(x)|\frac {9}{16}\right )}{32 \left (15 i-7 \sqrt {15}\right ) \sqrt {4 x^4-x^2+4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^4)*(1 - x^2 + x^4)*Sqrt[4 - x^2 + 4*x^4])/((1 + x^4)*(4 + 7*x^4 + 4*x^8)),x]

[Out]

ArcTan[x/Sqrt[4 - x^2 + 4*x^4]] - (9*Sqrt[5]*(7*I + Sqrt[15])*ArcTan[(Sqrt[3]*x)/Sqrt[4 - x^2 + 4*x^4]])/(8*(1
5 + (7*I)*Sqrt[15])) - (9*Sqrt[5]*(I + Sqrt[15])^2*ArcTan[(Sqrt[3]*x)/Sqrt[4 - x^2 + 4*x^4]])/(16*(15*I + 7*Sq
rt[15])) + ArcTanh[x/Sqrt[4 - x^2 + 4*x^4]]/4 + (7*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^2]*EllipticF[2*A
rcTan[x], 9/16])/(4*Sqrt[4 - x^2 + 4*x^4]) - (9*(5*I - 3*Sqrt[15])*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^
2]*EllipticF[2*ArcTan[x], 9/16])/(16*(15*I - Sqrt[15])*Sqrt[4 - x^2 + 4*x^4]) - (3*(2 - I*Sqrt[15])*(1 + x^2)*
Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 9/16])/(16*Sqrt[4 - x^2 + 4*x^4]) - ((4 - I*Sqrt[15
])*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 9/16])/(16*Sqrt[4 - x^2 + 4*x^4]) - (3
*(2 + I*Sqrt[15])*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 9/16])/(16*Sqrt[4 - x^2
+ 4*x^4]) - ((4 + I*Sqrt[15])*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 9/16])/(16
*Sqrt[4 - x^2 + 4*x^4]) + ((9*I + Sqrt[15])*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x
], 9/16])/(16*(15*I - Sqrt[15])*Sqrt[4 - x^2 + 4*x^4]) + ((9*I - Sqrt[15])*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1
+ x^2)^2]*EllipticF[2*ArcTan[x], 9/16])/(16*(15*I + Sqrt[15])*Sqrt[4 - x^2 + 4*x^4]) - (9*(5*I + 3*Sqrt[15])*
(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 9/16])/(16*(15*I + Sqrt[15])*Sqrt[4 - x^2
+ 4*x^4]) + (45*(7*I + Sqrt[15])*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^2]*EllipticPi[3/8, 2*ArcTan[x], 9
/16])/(32*(15*I - 7*Sqrt[15])*Sqrt[4 - x^2 + 4*x^4]) + (45*(7*I - Sqrt[15])*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(
1 + x^2)^2]*EllipticPi[3/8, 2*ArcTan[x], 9/16])/(32*(15*I + 7*Sqrt[15])*Sqrt[4 - x^2 + 4*x^4]) + (3*(7*I + Sqr
t[15])*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^2]*EllipticPi[5/8, 2*ArcTan[x], 9/16])/(32*(15*I - 7*Sqrt[15
])*Sqrt[4 - x^2 + 4*x^4]) + (3*(7*I - Sqrt[15])*(1 + x^2)*Sqrt[(4 - x^2 + 4*x^4)/(1 + x^2)^2]*EllipticPi[5/8,
2*ArcTan[x], 9/16])/(32*(15*I + 7*Sqrt[15])*Sqrt[4 - x^2 + 4*x^4])

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

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

Rule 1197

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

Rule 1208

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

Rule 1216

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

Rule 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + 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 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^
2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 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 {\left (-1+x^4\right ) \left (1-x^2+x^4\right ) \sqrt {4-x^2+4 x^4}}{\left (1+x^4\right ) \left (4+7 x^4+4 x^8\right )} \, dx &=\int \left (\frac {2 x^2 \sqrt {4-x^2+4 x^4}}{1+x^4}+\frac {\left (1-4 x^2\right ) \sqrt {4-x^2+4 x^4}}{4 \left (2-x^2+2 x^4\right )}-\frac {3 \left (1+4 x^2\right ) \sqrt {4-x^2+4 x^4}}{4 \left (2+x^2+2 x^4\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {\left (1-4 x^2\right ) \sqrt {4-x^2+4 x^4}}{2-x^2+2 x^4} \, dx-\frac {3}{4} \int \frac {\left (1+4 x^2\right ) \sqrt {4-x^2+4 x^4}}{2+x^2+2 x^4} \, dx+2 \int \frac {x^2 \sqrt {4-x^2+4 x^4}}{1+x^4} \, dx\\ &=\frac {1}{4} \int \left (-\frac {4 \sqrt {4-x^2+4 x^4}}{-1-i \sqrt {15}+4 x^2}-\frac {4 \sqrt {4-x^2+4 x^4}}{-1+i \sqrt {15}+4 x^2}\right ) \, dx-\frac {3}{4} \int \left (\frac {4 \sqrt {4-x^2+4 x^4}}{1-i \sqrt {15}+4 x^2}+\frac {4 \sqrt {4-x^2+4 x^4}}{1+i \sqrt {15}+4 x^2}\right ) \, dx+2 \int \left (-\frac {\sqrt {4-x^2+4 x^4}}{2 \left (i-x^2\right )}+\frac {\sqrt {4-x^2+4 x^4}}{2 \left (i+x^2\right )}\right ) \, dx\\ &=-\left (3 \int \frac {\sqrt {4-x^2+4 x^4}}{1-i \sqrt {15}+4 x^2} \, dx\right )-3 \int \frac {\sqrt {4-x^2+4 x^4}}{1+i \sqrt {15}+4 x^2} \, dx-\int \frac {\sqrt {4-x^2+4 x^4}}{i-x^2} \, dx+\int \frac {\sqrt {4-x^2+4 x^4}}{i+x^2} \, dx-\int \frac {\sqrt {4-x^2+4 x^4}}{-1-i \sqrt {15}+4 x^2} \, dx-\int \frac {\sqrt {4-x^2+4 x^4}}{-1+i \sqrt {15}+4 x^2} \, dx\\ &=i \int \frac {1}{\left (i-x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx+i \int \frac {1}{\left (i+x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx+\frac {1}{16} \int \frac {4+4 \left (-1-i \sqrt {15}\right )-16 x^2}{\sqrt {4-x^2+4 x^4}} \, dx+\frac {1}{16} \int \frac {4+4 \left (-1+i \sqrt {15}\right )-16 x^2}{\sqrt {4-x^2+4 x^4}} \, dx+\frac {3}{16} \int \frac {4+4 \left (1-i \sqrt {15}\right )-16 x^2}{\sqrt {4-x^2+4 x^4}} \, dx+\frac {3}{16} \int \frac {4+4 \left (1+i \sqrt {15}\right )-16 x^2}{\sqrt {4-x^2+4 x^4}} \, dx-\frac {1}{4} \left (1-i \sqrt {15}\right ) \int \frac {1}{\left (-1+i \sqrt {15}+4 x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx-\frac {1}{4} \left (9 \left (1-i \sqrt {15}\right )\right ) \int \frac {1}{\left (1-i \sqrt {15}+4 x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx-\frac {1}{4} \left (1+i \sqrt {15}\right ) \int \frac {1}{\left (-1-i \sqrt {15}+4 x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx-\frac {1}{4} \left (9 \left (1+i \sqrt {15}\right )\right ) \int \frac {1}{\left (1+i \sqrt {15}+4 x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx-\int \frac {(1+4 i)-4 x^2}{\sqrt {4-x^2+4 x^4}} \, dx+\int \frac {(-1+4 i)+4 x^2}{\sqrt {4-x^2+4 x^4}} \, dx\\ &=-\left ((-3+4 i) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx\right )+\left (-\frac {1}{2}+\frac {i}{2}\right ) \int \frac {1+x^2}{\left (i+x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx+\left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx+\left (\frac {1}{2}+\frac {i}{2}\right ) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx+\left (\frac {1}{2}+\frac {i}{2}\right ) \int \frac {1+x^2}{\left (i-x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx+2 \left (3 \int \frac {1-x^2}{\sqrt {4-x^2+4 x^4}} \, dx\right )+(3+4 i) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx-2 \left (4 \int \frac {1-x^2}{\sqrt {4-x^2+4 x^4}} \, dx\right )-\frac {\left (9 \left (5 i-3 \sqrt {15}\right )\right ) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx}{4 \left (15 i-\sqrt {15}\right )}+\frac {\left (9 \left (5 i-3 \sqrt {15}\right )\right ) \int \frac {1+x^2}{\left (1-i \sqrt {15}+4 x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx}{15 i-\sqrt {15}}+\frac {1}{4} \left (-4-i \sqrt {15}\right ) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx-\frac {1}{4} \left (3 \left (2-i \sqrt {15}\right )\right ) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx+\frac {1}{4} \left (-4+i \sqrt {15}\right ) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx-\frac {1}{4} \left (3 \left (2+i \sqrt {15}\right )\right ) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx+\frac {\left (9 i+\sqrt {15}\right ) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx}{4 \left (15 i-\sqrt {15}\right )}-\frac {\left (9 i+\sqrt {15}\right ) \int \frac {1+x^2}{\left (-1+i \sqrt {15}+4 x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx}{15 i-\sqrt {15}}+\frac {\left (9 i-\sqrt {15}\right ) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx}{4 \left (15 i+\sqrt {15}\right )}-\frac {\left (9 i-\sqrt {15}\right ) \int \frac {1+x^2}{\left (-1-i \sqrt {15}+4 x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx}{15 i+\sqrt {15}}-\frac {\left (9 \left (5 i+3 \sqrt {15}\right )\right ) \int \frac {1}{\sqrt {4-x^2+4 x^4}} \, dx}{4 \left (15 i+\sqrt {15}\right )}+\frac {\left (9 \left (5 i+3 \sqrt {15}\right )\right ) \int \frac {1+x^2}{\left (1+i \sqrt {15}+4 x^2\right ) \sqrt {4-x^2+4 x^4}} \, dx}{15 i+\sqrt {15}}+2 \int \frac {1-x^2}{\sqrt {4-x^2+4 x^4}} \, dx\\ &=\tan ^{-1}\left (\frac {x}{\sqrt {4-x^2+4 x^4}}\right )-\frac {9 \left (5 \sqrt {3}+7 i \sqrt {5}\right ) \tan ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {4-x^2+4 x^4}}\right )}{8 \left (15+7 i \sqrt {15}\right )}-\frac {9 \left (5 i \sqrt {3}+7 \sqrt {5}\right ) \tan ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {4-x^2+4 x^4}}\right )}{8 \left (15 i+7 \sqrt {15}\right )}+\frac {1}{4} \tanh ^{-1}\left (\frac {x}{\sqrt {4-x^2+4 x^4}}\right )+2 \left (-\frac {x \sqrt {4-x^2+4 x^4}}{4 \left (1+x^2\right )}+\frac {\left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{2 \sqrt {4-x^2+4 x^4}}\right )+2 \left (-\frac {3 x \sqrt {4-x^2+4 x^4}}{4 \left (1+x^2\right )}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{2 \sqrt {4-x^2+4 x^4}}\right )-2 \left (-\frac {x \sqrt {4-x^2+4 x^4}}{1+x^2}+\frac {2 \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{\sqrt {4-x^2+4 x^4}}\right )+\frac {7 \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{4 \sqrt {4-x^2+4 x^4}}-\frac {9 \left (5 i-3 \sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \left (15 i-\sqrt {15}\right ) \sqrt {4-x^2+4 x^4}}-\frac {3 \left (2-i \sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \sqrt {4-x^2+4 x^4}}-\frac {\left (4-i \sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \sqrt {4-x^2+4 x^4}}-\frac {3 \left (2+i \sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \sqrt {4-x^2+4 x^4}}-\frac {\left (4+i \sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \sqrt {4-x^2+4 x^4}}+\frac {\left (9 i+\sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \left (15 i-\sqrt {15}\right ) \sqrt {4-x^2+4 x^4}}+\frac {\left (9 i-\sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \left (15 i+\sqrt {15}\right ) \sqrt {4-x^2+4 x^4}}-\frac {9 \left (5 i+3 \sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {9}{16}\right )}{16 \left (15 i+\sqrt {15}\right ) \sqrt {4-x^2+4 x^4}}+\frac {45 \left (7 i+\sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} \Pi \left (\frac {3}{8};2 \tan ^{-1}(x)|\frac {9}{16}\right )}{32 \left (15 i-7 \sqrt {15}\right ) \sqrt {4-x^2+4 x^4}}+\frac {45 \left (7 i-\sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} \Pi \left (\frac {3}{8};2 \tan ^{-1}(x)|\frac {9}{16}\right )}{32 \left (15 i+7 \sqrt {15}\right ) \sqrt {4-x^2+4 x^4}}+\frac {3 \left (7 i+\sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} \Pi \left (\frac {5}{8};2 \tan ^{-1}(x)|\frac {9}{16}\right )}{32 \left (15 i-7 \sqrt {15}\right ) \sqrt {4-x^2+4 x^4}}+\frac {3 \left (7 i-\sqrt {15}\right ) \left (1+x^2\right ) \sqrt {\frac {4-x^2+4 x^4}{\left (1+x^2\right )^2}} \Pi \left (\frac {5}{8};2 \tan ^{-1}(x)|\frac {9}{16}\right )}{32 \left (15 i+7 \sqrt {15}\right ) \sqrt {4-x^2+4 x^4}}\\ \end {align*}

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Mathematica [C]  time = 2.54, size = 605, normalized size = 7.96 \begin {gather*} -\frac {i \sqrt {2-\frac {16 x^2}{1-3 i \sqrt {7}}} \sqrt {1-\frac {8 x^2}{1+3 i \sqrt {7}}} \left (4 F\left (i \sinh ^{-1}\left (2 \sqrt {-\frac {2}{1-3 i \sqrt {7}}} x\right )|\frac {i+3 \sqrt {7}}{i-3 \sqrt {7}}\right )+4 \Pi \left (\frac {1}{8} \left (-i-3 \sqrt {7}\right );i \sinh ^{-1}\left (2 \sqrt {-\frac {2}{1-3 i \sqrt {7}}} x\right )|\frac {i+3 \sqrt {7}}{i-3 \sqrt {7}}\right )+4 \Pi \left (\frac {1}{8} \left (i+3 \sqrt {7}\right );i \sinh ^{-1}\left (2 \sqrt {-\frac {2}{1-3 i \sqrt {7}}} x\right )|\frac {i+3 \sqrt {7}}{i-3 \sqrt {7}}\right )+\Pi \left (\frac {i+3 \sqrt {7}}{2 i-2 \sqrt {15}};i \sinh ^{-1}\left (2 \sqrt {-\frac {2}{1-3 i \sqrt {7}}} x\right )|\frac {i+3 \sqrt {7}}{i-3 \sqrt {7}}\right )-9 \Pi \left (\frac {i+3 \sqrt {7}}{2 \left (-i+\sqrt {15}\right )};i \sinh ^{-1}\left (2 \sqrt {-\frac {2}{1-3 i \sqrt {7}}} x\right )|\frac {i+3 \sqrt {7}}{i-3 \sqrt {7}}\right )-9 \Pi \left (-\frac {i+3 \sqrt {7}}{2 i+2 \sqrt {15}};i \sinh ^{-1}\left (2 \sqrt {-\frac {2}{1-3 i \sqrt {7}}} x\right )|\frac {i+3 \sqrt {7}}{i-3 \sqrt {7}}\right )+\Pi \left (\frac {i+3 \sqrt {7}}{2 i+2 \sqrt {15}};i \sinh ^{-1}\left (2 \sqrt {-\frac {2}{1-3 i \sqrt {7}}} x\right )|\frac {i+3 \sqrt {7}}{i-3 \sqrt {7}}\right )\right )}{16 \sqrt {-\frac {i}{3 \sqrt {7}+i}} \sqrt {4 x^4-x^2+4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^4)*(1 - x^2 + x^4)*Sqrt[4 - x^2 + 4*x^4])/((1 + x^4)*(4 + 7*x^4 + 4*x^8)),x]

[Out]

((-1/16*I)*Sqrt[2 - (16*x^2)/(1 - (3*I)*Sqrt[7])]*Sqrt[1 - (8*x^2)/(1 + (3*I)*Sqrt[7])]*(4*EllipticF[I*ArcSinh
[2*Sqrt[-2/(1 - (3*I)*Sqrt[7])]*x], (I + 3*Sqrt[7])/(I - 3*Sqrt[7])] + 4*EllipticPi[(-I - 3*Sqrt[7])/8, I*ArcS
inh[2*Sqrt[-2/(1 - (3*I)*Sqrt[7])]*x], (I + 3*Sqrt[7])/(I - 3*Sqrt[7])] + 4*EllipticPi[(I + 3*Sqrt[7])/8, I*Ar
cSinh[2*Sqrt[-2/(1 - (3*I)*Sqrt[7])]*x], (I + 3*Sqrt[7])/(I - 3*Sqrt[7])] + EllipticPi[(I + 3*Sqrt[7])/(2*I -
2*Sqrt[15]), I*ArcSinh[2*Sqrt[-2/(1 - (3*I)*Sqrt[7])]*x], (I + 3*Sqrt[7])/(I - 3*Sqrt[7])] - 9*EllipticPi[(I +
3*Sqrt[7])/(2*(-I + Sqrt[15])), I*ArcSinh[2*Sqrt[-2/(1 - (3*I)*Sqrt[7])]*x], (I + 3*Sqrt[7])/(I - 3*Sqrt[7])]
- 9*EllipticPi[-((I + 3*Sqrt[7])/(2*I + 2*Sqrt[15])), I*ArcSinh[2*Sqrt[-2/(1 - (3*I)*Sqrt[7])]*x], (I + 3*Sqr
t[7])/(I - 3*Sqrt[7])] + EllipticPi[(I + 3*Sqrt[7])/(2*I + 2*Sqrt[15]), I*ArcSinh[2*Sqrt[-2/(1 - (3*I)*Sqrt[7]
)]*x], (I + 3*Sqrt[7])/(I - 3*Sqrt[7])]))/(Sqrt[(-I)/(I + 3*Sqrt[7])]*Sqrt[4 - x^2 + 4*x^4])

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IntegrateAlgebraic [A]  time = 0.91, size = 76, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt {4-x^2+4 x^4}}\right )-\frac {3}{4} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {4-x^2+4 x^4}}\right )+\frac {1}{4} \tanh ^{-1}\left (\frac {x}{\sqrt {4-x^2+4 x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x^4)*(1 - x^2 + x^4)*Sqrt[4 - x^2 + 4*x^4])/((1 + x^4)*(4 + 7*x^4 + 4*x^8)),x]

[Out]

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

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fricas [A]  time = 0.60, size = 114, normalized size = 1.50 \begin {gather*} -\frac {3}{8} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {4 \, x^{4} - x^{2} + 4} x}{2 \, {\left (x^{4} - x^{2} + 1\right )}}\right ) + \frac {1}{2} \, \arctan \left (\frac {\sqrt {4 \, x^{4} - x^{2} + 4} x}{2 \, x^{4} - x^{2} + 2}\right ) + \frac {1}{8} \, \log \left (-\frac {2 \, x^{4} + \sqrt {4 \, x^{4} - x^{2} + 4} x + 2}{2 \, x^{4} - x^{2} + 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)*(x^4-x^2+1)*(4*x^4-x^2+4)^(1/2)/(x^4+1)/(4*x^8+7*x^4+4),x, algorithm="fricas")

[Out]

-3/8*sqrt(3)*arctan(1/2*sqrt(3)*sqrt(4*x^4 - x^2 + 4)*x/(x^4 - x^2 + 1)) + 1/2*arctan(sqrt(4*x^4 - x^2 + 4)*x/
(2*x^4 - x^2 + 2)) + 1/8*log(-(2*x^4 + sqrt(4*x^4 - x^2 + 4)*x + 2)/(2*x^4 - x^2 + 2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {4 \, x^{4} - x^{2} + 4} {\left (x^{4} - x^{2} + 1\right )} {\left (x^{4} - 1\right )}}{{\left (4 \, x^{8} + 7 \, x^{4} + 4\right )} {\left (x^{4} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)*(x^4-x^2+1)*(4*x^4-x^2+4)^(1/2)/(x^4+1)/(4*x^8+7*x^4+4),x, algorithm="giac")

[Out]

integrate(sqrt(4*x^4 - x^2 + 4)*(x^4 - x^2 + 1)*(x^4 - 1)/((4*x^8 + 7*x^4 + 4)*(x^4 + 1)), x)

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maple [A]  time = 0.99, size = 86, normalized size = 1.13

method	result	size
elliptic	\(\frac {\left (\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {4 x^{4}-x^{2}+4}}{x}\right )}{4}-\sqrt {2}\, \arctan \left (\frac {\sqrt {4 x^{4}-x^{2}+4}}{x}\right )+\frac {3 \sqrt {6}\, \arctan \left (\frac {\sqrt {6}\, \sqrt {4 x^{4}-x^{2}+4}\, \sqrt {2}}{6 x}\right )}{4}\right ) \sqrt {2}}{2}\)	\(86\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}-\sqrt {4 x^{4}-x^{2}+4}\, x +2 \RootOf \left (\textit {\_Z}^{2}+1\right )}{x^{4}+1}\right )}{2}-\frac {3 \RootOf \left (\textit {\_Z}^{2}+3\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}+3\right ) x^{4}-2 \RootOf \left (\textit {\_Z}^{2}+3\right ) x^{2}+3 \sqrt {4 x^{4}-x^{2}+4}\, x +2 \RootOf \left (\textit {\_Z}^{2}+3\right )}{2 x^{4}+x^{2}+2}\right )}{8}-\frac {\ln \left (-\frac {-2 x^{4}+\sqrt {4 x^{4}-x^{2}+4}\, x -2}{2 x^{4}-x^{2}+2}\right )}{8}\)	\(180\)
default	\(-\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (2 \textit {\_Z}^{4}-\textit {\_Z}^{2}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {8 \arctanh \left (\frac {\left (8 \underline {\hspace {1.25 ex}}\alpha ^{2}-1\right ) \left (-63 \underline {\hspace {1.25 ex}}\alpha ^{2}+61 x^{2}+16\right )}{122 \sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {4 x^{4}-x^{2}+4}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}}-\frac {\sqrt {8}\, \left (-2 \underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha \right ) \sqrt {-x^{2}+8-3 i x^{2} \sqrt {7}}\, \sqrt {-x^{2}+8+3 i x^{2} \sqrt {7}}\, \EllipticPi \left (\sqrt {\frac {1}{8}+\frac {3 i \sqrt {7}}{8}}\, x , \frac {3 i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {7}}{8}-\frac {\underline {\hspace {1.25 ex}}\alpha ^{2}}{8}-\frac {3 i \sqrt {7}}{16}+\frac {1}{16}, \frac {\sqrt {\frac {1}{8}-\frac {3 i \sqrt {7}}{8}}}{\sqrt {\frac {1}{8}+\frac {3 i \sqrt {7}}{8}}}\right )}{\sqrt {1+3 i \sqrt {7}}\, \sqrt {4 x^{4}-x^{2}+4}}\right )\right )}{128}+\frac {4 \sqrt {1-\left (\frac {1}{8}+\frac {3 i \sqrt {7}}{8}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{8}-\frac {3 i \sqrt {7}}{8}\right ) x^{2}}\, \EllipticF \left (\frac {x \sqrt {2+6 i \sqrt {7}}}{4}, \frac {\sqrt {-62-6 i \sqrt {7}}}{8}\right )}{\sqrt {2+6 i \sqrt {7}}\, \sqrt {4 x^{4}-x^{2}+4}}+\frac {3 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (2 \textit {\_Z}^{4}+\textit {\_Z}^{2}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {8 \sqrt {3}\, \arctanh \left (\frac {\left (8 \underline {\hspace {1.25 ex}}\alpha ^{2}-1\right ) \left (-21 \underline {\hspace {1.25 ex}}\alpha ^{2}+23 x^{2}-16\right )}{46 \sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {4 x^{4}-x^{2}+4}}\right )}{\sqrt {-\underline {\hspace {1.25 ex}}\alpha ^{2}}}-\frac {3 \sqrt {8}\, \left (-2 \underline {\hspace {1.25 ex}}\alpha ^{3}-\underline {\hspace {1.25 ex}}\alpha \right ) \sqrt {-x^{2}+8-3 i x^{2} \sqrt {7}}\, \sqrt {-x^{2}+8+3 i x^{2} \sqrt {7}}\, \EllipticPi \left (\sqrt {\frac {1}{8}+\frac {3 i \sqrt {7}}{8}}\, x , \frac {3 i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {7}}{8}-\frac {\underline {\hspace {1.25 ex}}\alpha ^{2}}{8}+\frac {3 i \sqrt {7}}{16}-\frac {1}{16}, \frac {\sqrt {\frac {1}{8}-\frac {3 i \sqrt {7}}{8}}}{\sqrt {\frac {1}{8}+\frac {3 i \sqrt {7}}{8}}}\right )}{\sqrt {1+3 i \sqrt {7}}\, \sqrt {4 x^{4}-x^{2}+4}}\right )\right )}{128}-\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{4}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {4 \arctanh \left (\frac {\left (8 \underline {\hspace {1.25 ex}}\alpha ^{2}-1\right ) \left (-63 \underline {\hspace {1.25 ex}}\alpha ^{2}+65 x^{2}-16\right )}{130 \sqrt {-\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {4 x^{4}-x^{2}+4}}\right )}{\sqrt {-\underline {\hspace {1.25 ex}}\alpha ^{2}}}+\frac {\sqrt {8}\, \underline {\hspace {1.25 ex}}\alpha ^{3} \sqrt {-x^{2}+8-3 i x^{2} \sqrt {7}}\, \sqrt {-x^{2}+8+3 i x^{2} \sqrt {7}}\, \EllipticPi \left (\sqrt {\frac {1}{8}+\frac {3 i \sqrt {7}}{8}}\, x , \frac {3 i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {7}}{8}-\frac {\underline {\hspace {1.25 ex}}\alpha ^{2}}{8}, \frac {\sqrt {\frac {1}{8}-\frac {3 i \sqrt {7}}{8}}}{\sqrt {\frac {1}{8}+\frac {3 i \sqrt {7}}{8}}}\right )}{\sqrt {1+3 i \sqrt {7}}\, \sqrt {4 x^{4}-x^{2}+4}}\right )\right )}{16}\)	\(671\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-1)*(x^4-x^2+1)*(4*x^4-x^2+4)^(1/2)/(x^4+1)/(4*x^8+7*x^4+4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {4 \, x^{4} - x^{2} + 4} {\left (x^{4} - x^{2} + 1\right )} {\left (x^{4} - 1\right )}}{{\left (4 \, x^{8} + 7 \, x^{4} + 4\right )} {\left (x^{4} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)*(x^4-x^2+1)*(4*x^4-x^2+4)^(1/2)/(x^4+1)/(4*x^8+7*x^4+4),x, algorithm="maxima")

[Out]

integrate(sqrt(4*x^4 - x^2 + 4)*(x^4 - x^2 + 1)*(x^4 - 1)/((4*x^8 + 7*x^4 + 4)*(x^4 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (x^4-1\right )\,\left (x^4-x^2+1\right )\,\sqrt {4\,x^4-x^2+4}}{\left (x^4+1\right )\,\left (4\,x^8+7\,x^4+4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 - 1)*(x^4 - x^2 + 1)*(4*x^4 - x^2 + 4)^(1/2))/((x^4 + 1)*(7*x^4 + 4*x^8 + 4)),x)

[Out]

int(((x^4 - 1)*(x^4 - x^2 + 1)*(4*x^4 - x^2 + 4)^(1/2))/((x^4 + 1)*(7*x^4 + 4*x^8 + 4)), 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**4-1)*(x**4-x**2+1)*(4*x**4-x**2+4)**(1/2)/(x**4+1)/(4*x**8+7*x**4+4),x)

[Out]

Timed out

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