3.10.96 \(\int \frac {(-1+x^4) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx\)

Optimal. Leaf size=75 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x \sqrt {x^4+1}}{x^4-x^2+1}\right )}{2 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt {x^4+1}}{x^4+x^2+1}\right )}{2 \sqrt {2}} \]

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Rubi [C]  time = 1.92, antiderivative size = 1128, normalized size of antiderivative = 15.04, number of steps used = 20, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6728, 406, 220, 409, 1217, 1707}

result too large to display

Warning: Unable to verify antiderivative.

[In]

Int[((-1 + x^4)*Sqrt[1 + x^4])/(1 + 3*x^4 + x^8),x]

[Out]

((-1/4 + I/4)*ArcTan[((-1)^(1/4)*x)/Sqrt[1 + x^4]])/Sqrt[2] - ((-1)^(1/4)*(2*I - Sqrt[6 - 2*Sqrt[5]])*(2*I - S
qrt[2*(3 + Sqrt[5])])*ArcTan[((-1)^(1/4)*x)/Sqrt[1 + x^4]])/(16*Sqrt[5]) + ((1/4 + I/4)*ArcTan[((-1)^(3/4)*x)/
Sqrt[1 + x^4]])/Sqrt[2] + ((1/16 + I/16)*(2*I + Sqrt[6 - 2*Sqrt[5]])*(2 - I*Sqrt[2*(3 + Sqrt[5])])*ArcTan[((-1
)^(3/4)*x)/Sqrt[1 + x^4]])/Sqrt[10] + ((1 - Sqrt[5])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[
x], 1/2])/(4*Sqrt[1 + x^4]) + ((1 + Sqrt[5])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2]
)/(4*Sqrt[1 + x^4]) - ((3 + Sqrt[5])*(1 - I*Sqrt[2/(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*Ellip
ticF[2*ArcTan[x], 1/2])/(4*(5 + Sqrt[5])*Sqrt[1 + x^4]) - ((3 + Sqrt[5])*(1 + I*Sqrt[2/(3 + Sqrt[5])])*(1 + x^
2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(4*(5 + Sqrt[5])*Sqrt[1 + x^4]) - ((3 - Sqrt[5])*(
2 - I*Sqrt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(8*(5 - Sqrt[5
])*Sqrt[1 + x^4]) - ((3 - Sqrt[5])*(2 + I*Sqrt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*Ellipti
cF[2*ArcTan[x], 1/2])/(8*(5 - Sqrt[5])*Sqrt[1 + x^4]) + ((5 - 2*Sqrt[5])*(3 + Sqrt[5])*(1 + Sqrt[5] - (2*I)*Sq
rt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[(I/4)*Sqrt[(3 + Sqrt[5])/2]*(I - Sqrt[2/
(3 + Sqrt[5])])^2, 2*ArcTan[x], 1/2])/(160*Sqrt[1 + x^4]) + ((5 - 2*Sqrt[5])*(3 + Sqrt[5])*(1 + Sqrt[5] + (2*I
)*Sqrt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[(-1/4*I)*Sqrt[(3 + Sqrt[5])/2]*(I +
Sqrt[2/(3 + Sqrt[5])])^2, 2*ArcTan[x], 1/2])/(160*Sqrt[1 + x^4]) - ((2 + I*Sqrt[6 - 2*Sqrt[5]])*(2*I + Sqrt[2*
(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[(I/16)*Sqrt[(3 - Sqrt[5])/2]*(2*I - Sqrt[2*(3
+ Sqrt[5])])^2, 2*ArcTan[x], 1/2])/(32*Sqrt[5]*Sqrt[1 + x^4]) + ((2*I + Sqrt[6 - 2*Sqrt[5]])*(2 + I*Sqrt[2*(3
+ Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[(-1/16*I)*Sqrt[(3 - Sqrt[5])/2]*(2*I + Sqrt[2*(
3 + Sqrt[5])])^2, 2*ArcTan[x], 1/2])/(32*Sqrt[5]*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 406

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

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 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 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 ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx &=\int \left (\frac {\left (1-\sqrt {5}\right ) \sqrt {1+x^4}}{3-\sqrt {5}+2 x^4}+\frac {\left (1+\sqrt {5}\right ) \sqrt {1+x^4}}{3+\sqrt {5}+2 x^4}\right ) \, dx\\ &=\left (1-\sqrt {5}\right ) \int \frac {\sqrt {1+x^4}}{3-\sqrt {5}+2 x^4} \, dx+\left (1+\sqrt {5}\right ) \int \frac {\sqrt {1+x^4}}{3+\sqrt {5}+2 x^4} \, dx\\ &=\left (-3-\sqrt {5}\right ) \int \frac {1}{\sqrt {1+x^4} \left (3+\sqrt {5}+2 x^4\right )} \, dx+\frac {1}{2} \left (1-\sqrt {5}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx+\left (-3+\sqrt {5}\right ) \int \frac {1}{\sqrt {1+x^4} \left (3-\sqrt {5}+2 x^4\right )} \, dx+\frac {1}{2} \left (1+\sqrt {5}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=\frac {\left (1-\sqrt {5}\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 )}{4 \sqrt {1+x^4}}+\frac {\left (1+\sqrt {5}\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 )}{4 \sqrt {1+x^4}}-\frac {1}{2} \int \frac {1}{\left (1-i \sqrt {\frac {2}{3-\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+i \sqrt {\frac {2}{3-\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1-i \sqrt {\frac {2}{3+\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+i \sqrt {\frac {2}{3+\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx\\ &=\frac {\left (1-\sqrt {5}\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 )}{4 \sqrt {1+x^4}}+\frac {\left (1+\sqrt {5}\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 )}{4 \sqrt {1+x^4}}-\frac {\left (1-i \sqrt {\frac {2}{3-\sqrt {5}}}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{2 \left (1+\frac {2}{3-\sqrt {5}}\right )}-\frac {\left (1+i \sqrt {\frac {2}{3-\sqrt {5}}}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{2 \left (1+\frac {2}{3-\sqrt {5}}\right )}-\frac {\left (\left (3+\sqrt {5}\right ) \left (1-i \sqrt {\frac {2}{3+\sqrt {5}}}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{2 \left (5+\sqrt {5}\right )}-\frac {\left (\left (3+\sqrt {5}\right ) \left (1+i \sqrt {\frac {2}{3+\sqrt {5}}}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{2 \left (5+\sqrt {5}\right )}-\frac {\left (i+\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )}\right ) \int \frac {1+x^2}{\left (1-i \sqrt {\frac {2}{3-\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx}{2 \sqrt {5}}+\frac {\left (2 i-\sqrt {2 \left (3+\sqrt {5}\right )}\right ) \int \frac {1+x^2}{\left (1+i \sqrt {\frac {2}{3-\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx}{4 \sqrt {5}}-\frac {\left (2-i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \int \frac {1+x^2}{\left (1+i \sqrt {\frac {2}{3+\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx}{2 \left (5+\sqrt {5}\right )}-\frac {\left (2+i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \int \frac {1+x^2}{\left (1-i \sqrt {\frac {2}{3+\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx}{2 \left (5+\sqrt {5}\right )}\\ &=\frac {1}{4} (-1)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{-1} x}{\sqrt {1+x^4}}\right )-\frac {\sqrt [4]{-1} \left (2 i-\sqrt {6-2 \sqrt {5}}\right ) \left (2 i-\sqrt {2 \left (3+\sqrt {5}\right )}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{-1} x}{\sqrt {1+x^4}}\right )}{16 \sqrt {5}}+\frac {1}{4} \sqrt [4]{-1} \tan ^{-1}\left (\frac {(-1)^{3/4} x}{\sqrt {1+x^4}}\right )+\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) \left (2 i+\sqrt {6-2 \sqrt {5}}\right ) \left (2 i+\sqrt {2 \left (3+\sqrt {5}\right )}\right ) \tan ^{-1}\left (\frac {(-1)^{3/4} x}{\sqrt {1+x^4}}\right )}{\sqrt {10}}+\frac {\left (1-\sqrt {5}\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 )}{4 \sqrt {1+x^4}}+\frac {\left (1+\sqrt {5}\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 )}{4 \sqrt {1+x^4}}-\frac {\left (3+\sqrt {5}\right ) \left (1-i \sqrt {\frac {2}{3+\sqrt {5}}}\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 )}{4 \left (5+\sqrt {5}\right ) \sqrt {1+x^4}}-\frac {\left (3+\sqrt {5}\right ) \left (1+i \sqrt {\frac {2}{3+\sqrt {5}}}\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 )}{4 \left (5+\sqrt {5}\right ) \sqrt {1+x^4}}-\frac {\left (3-\sqrt {5}\right ) \left (1-i \sqrt {\frac {1}{2} \left (3+\sqrt {5}\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 )}{4 \left (5-\sqrt {5}\right ) \sqrt {1+x^4}}-\frac {\left (3-\sqrt {5}\right ) \left (1+i \sqrt {\frac {1}{2} \left (3+\sqrt {5}\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 )}{4 \left (5-\sqrt {5}\right ) \sqrt {1+x^4}}+\frac {\left (2 i-\sqrt {2 \left (3+\sqrt {5}\right )}\right )^2 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \Pi \left (\frac {1}{4} i \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \left (i-\sqrt {\frac {2}{3+\sqrt {5}}}\right )^2;2 \tan ^{-1}(x)|\frac {1}{2}\right )}{16 \left (5+\sqrt {5}\right ) \sqrt {1+x^4}}+\frac {\left (2 i+\sqrt {2 \left (3+\sqrt {5}\right )}\right )^2 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \Pi \left (-\frac {1}{4} i \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \left (i+\sqrt {\frac {2}{3+\sqrt {5}}}\right )^2;2 \tan ^{-1}(x)|\frac {1}{2}\right )}{16 \left (5+\sqrt {5}\right ) \sqrt {1+x^4}}-\frac {\left (2+i \sqrt {6-2 \sqrt {5}}\right ) \left (i+\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \Pi \left (\frac {1}{16} i \sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} \left (2 i-\sqrt {2 \left (3+\sqrt {5}\right )}\right )^2;2 \tan ^{-1}(x)|\frac {1}{2}\right )}{16 \sqrt {5} \sqrt {1+x^4}}+\frac {\left (2 i+\sqrt {6-2 \sqrt {5}}\right ) \left (2+i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \Pi \left (-\frac {1}{16} i \sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} \left (2 i+\sqrt {2 \left (3+\sqrt {5}\right )}\right )^2;2 \tan ^{-1}(x)|\frac {1}{2}\right )}{32 \sqrt {5} \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.55, size = 146, normalized size = 1.95 \begin {gather*} \frac {1}{2} \sqrt [4]{-1} \left (-2 F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left (-\sqrt {\frac {2}{3+\sqrt {5}}};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left (\sqrt {\frac {2}{3+\sqrt {5}}};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left (-\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^4)*Sqrt[1 + x^4])/(1 + 3*x^4 + x^8),x]

[Out]

((-1)^(1/4)*(-2*EllipticF[I*ArcSinh[(-1)^(1/4)*x], -1] + EllipticPi[-Sqrt[2/(3 + Sqrt[5])], I*ArcSinh[(-1)^(1/
4)*x], -1] + EllipticPi[Sqrt[2/(3 + Sqrt[5])], I*ArcSinh[(-1)^(1/4)*x], -1] + EllipticPi[-Sqrt[(3 + Sqrt[5])/2
], I*ArcSinh[(-1)^(1/4)*x], -1] + EllipticPi[Sqrt[(3 + Sqrt[5])/2], I*ArcSinh[(-1)^(1/4)*x], -1]))/2

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IntegrateAlgebraic [A]  time = 0.36, size = 75, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x \sqrt {1+x^4}}{1-x^2+x^4}\right )}{2 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt {1+x^4}}{1+x^2+x^4}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x^4)*Sqrt[1 + x^4])/(1 + 3*x^4 + x^8),x]

[Out]

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

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fricas [B]  time = 0.74, size = 430, normalized size = 5.73 \begin {gather*} \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {x^{8} + 3 \, x^{4} + 2 \, \sqrt {2} {\left (x^{5} - x^{3} + x\right )} \sqrt {x^{4} + 1} - {\left (4 \, \sqrt {x^{4} + 1} x^{3} + \sqrt {2} {\left (x^{8} - 2 \, x^{6} + x^{4} - 2 \, x^{2} + 1\right )}\right )} \sqrt {\frac {x^{8} + 4 \, x^{6} + 3 \, x^{4} + 2 \, \sqrt {2} {\left (x^{5} + x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \, x^{2} + 1}{x^{8} + 3 \, x^{4} + 1}} + 1}{x^{8} - 4 \, x^{6} + 3 \, x^{4} - 4 \, x^{2} + 1}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {x^{8} + 3 \, x^{4} - 2 \, \sqrt {2} {\left (x^{5} - x^{3} + x\right )} \sqrt {x^{4} + 1} - {\left (4 \, \sqrt {x^{4} + 1} x^{3} - \sqrt {2} {\left (x^{8} - 2 \, x^{6} + x^{4} - 2 \, x^{2} + 1\right )}\right )} \sqrt {\frac {x^{8} + 4 \, x^{6} + 3 \, x^{4} - 2 \, \sqrt {2} {\left (x^{5} + x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \, x^{2} + 1}{x^{8} + 3 \, x^{4} + 1}} + 1}{x^{8} - 4 \, x^{6} + 3 \, x^{4} - 4 \, x^{2} + 1}\right ) - \frac {1}{16} \, \sqrt {2} \log \left (\frac {4 \, {\left (x^{8} + 4 \, x^{6} + 3 \, x^{4} + 2 \, \sqrt {2} {\left (x^{5} + x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \, x^{2} + 1\right )}}{x^{8} + 3 \, x^{4} + 1}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (\frac {4 \, {\left (x^{8} + 4 \, x^{6} + 3 \, x^{4} - 2 \, \sqrt {2} {\left (x^{5} + x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \, x^{2} + 1\right )}}{x^{8} + 3 \, x^{4} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)*(x^4+1)^(1/2)/(x^8+3*x^4+1),x, algorithm="fricas")

[Out]

1/4*sqrt(2)*arctan(-(x^8 + 3*x^4 + 2*sqrt(2)*(x^5 - x^3 + x)*sqrt(x^4 + 1) - (4*sqrt(x^4 + 1)*x^3 + sqrt(2)*(x
^8 - 2*x^6 + x^4 - 2*x^2 + 1))*sqrt((x^8 + 4*x^6 + 3*x^4 + 2*sqrt(2)*(x^5 + x^3 + x)*sqrt(x^4 + 1) + 4*x^2 + 1
)/(x^8 + 3*x^4 + 1)) + 1)/(x^8 - 4*x^6 + 3*x^4 - 4*x^2 + 1)) - 1/4*sqrt(2)*arctan(-(x^8 + 3*x^4 - 2*sqrt(2)*(x
^5 - x^3 + x)*sqrt(x^4 + 1) - (4*sqrt(x^4 + 1)*x^3 - sqrt(2)*(x^8 - 2*x^6 + x^4 - 2*x^2 + 1))*sqrt((x^8 + 4*x^
6 + 3*x^4 - 2*sqrt(2)*(x^5 + x^3 + x)*sqrt(x^4 + 1) + 4*x^2 + 1)/(x^8 + 3*x^4 + 1)) + 1)/(x^8 - 4*x^6 + 3*x^4
- 4*x^2 + 1)) - 1/16*sqrt(2)*log(4*(x^8 + 4*x^6 + 3*x^4 + 2*sqrt(2)*(x^5 + x^3 + x)*sqrt(x^4 + 1) + 4*x^2 + 1)
/(x^8 + 3*x^4 + 1)) + 1/16*sqrt(2)*log(4*(x^8 + 4*x^6 + 3*x^4 - 2*sqrt(2)*(x^5 + x^3 + x)*sqrt(x^4 + 1) + 4*x^
2 + 1)/(x^8 + 3*x^4 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)*(x^4+1)^(1/2)/(x^8+3*x^4+1),x, algorithm="giac")

[Out]

integrate(sqrt(x^4 + 1)*(x^4 - 1)/(x^8 + 3*x^4 + 1), x)

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maple [A]  time = 0.96, size = 102, normalized size = 1.36

method	result	size
default	\(\frac {\left (\frac {\ln \left (\frac {x^{4}+1}{x^{2}}-\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}\right )}{2}-\frac {\ln \left (\frac {x^{4}+1}{x^{2}}+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}\right )}{2}\right ) \sqrt {2}}{2}\)	\(102\)
elliptic	\(\frac {\left (\frac {\ln \left (\frac {x^{4}+1}{x^{2}}-\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}\right )}{2}-\frac {\ln \left (\frac {x^{4}+1}{x^{2}}+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}\right )}{2}\right ) \sqrt {2}}{2}\)	\(102\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{5} x^{2}+\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x^{4}+\RootOf \left (\textit {\_Z}^{4}+1\right )^{3}+2 \sqrt {x^{4}+1}\, x}{\RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{2}-x^{4}-1}\right )}{4}-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x^{2}+\RootOf \left (\textit {\_Z}^{4}+1\right ) x^{4}+2 \sqrt {x^{4}+1}\, x +\RootOf \left (\textit {\_Z}^{4}+1\right )}{\RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{2}+x^{4}+1}\right )}{4}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-1)*(x^4+1)^(1/2)/(x^8+3*x^4+1),x,method=_RETURNVERBOSE)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)*(x^4+1)^(1/2)/(x^8+3*x^4+1),x, algorithm="maxima")

[Out]

integrate(sqrt(x^4 + 1)*(x^4 - 1)/(x^8 + 3*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 )\,\sqrt {x^4+1}}{x^8+3\,x^4+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((x^4 - 1)*(x^4 + 1)^(1/2))/(3*x^4 + x^8 + 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 ) \sqrt {x^{4} + 1}}{x^{8} + 3 x^{4} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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