3.24.50 \(\int \frac {(-2+x^4) \sqrt {2+x^4}}{4+3 x^4+x^8} \, dx\)

Optimal. Leaf size=186 \[ -4 \sqrt [4]{2} \text {RootSum}\left [\text {$\#$1}^8-4 \text {$\#$1}^6-122 \text {$\#$1}^4-4 \text {$\#$1}^2+1\& ,\frac {\text {$\#$1}^3 \left (-\log \left (-x^2+2^{3/4} x-\sqrt {2}\right )\right )+\text {$\#$1}^3 \log \left (\text {$\#$1} x^2-2^{3/4} \text {$\#$1} x+\sqrt {2} \text {$\#$1}+\sqrt {x^4+2}\right )-\text {$\#$1} \log \left (-x^2+2^{3/4} x-\sqrt {2}\right )+\text {$\#$1} \log \left (\text {$\#$1} x^2-2^{3/4} \text {$\#$1} x+\sqrt {2} \text {$\#$1}+\sqrt {x^4+2}\right )}{\text {$\#$1}^6-3 \text {$\#$1}^4-61 \text {$\#$1}^2-1}\& \right ] \]

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Rubi [C]  time = 3.12, antiderivative size = 1539, normalized size of antiderivative = 8.27, 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[((-2 + x^4)*Sqrt[2 + x^4])/(4 + 3*x^4 + x^8),x]

[Out]

-1/2*(Sqrt[1 - I*Sqrt[7]]*(5*I + Sqrt[7])*ArcTan[(Sqrt[1 - I*Sqrt[7]]*x)/((2*(-3 - I*Sqrt[7]))^(1/4)*Sqrt[2 +
x^4])])/((3*I - Sqrt[7])*(2*(-3 - I*Sqrt[7]))^(3/4)) - ((1 - (3*I)*Sqrt[7])*ArcTan[(Sqrt[-1 - I*Sqrt[7]]*x)/((
2*(-3 + I*Sqrt[7]))^(1/4)*Sqrt[2 + x^4])])/(2^(3/4)*Sqrt[-1 - I*Sqrt[7]]*(-3 + I*Sqrt[7])^(7/4)) - ((1 + (3*I)
*Sqrt[7])*ArcTan[(Sqrt[-1 + I*Sqrt[7]]*x)/((2*(-3 - I*Sqrt[7]))^(1/4)*Sqrt[2 + x^4])])/(2^(3/4)*(-3 - I*Sqrt[7
])^(7/4)*Sqrt[-1 + I*Sqrt[7]]) - ((5*I - Sqrt[7])*Sqrt[1 + I*Sqrt[7]]*ArcTan[(Sqrt[1 + I*Sqrt[7]]*x)/((2*(-3 +
I*Sqrt[7]))^(1/4)*Sqrt[2 + x^4])])/(2*(2*(-3 + I*Sqrt[7]))^(3/4)*(3*I + Sqrt[7])) + ((1 - I*Sqrt[7])*(Sqrt[2]
+ x^2)*Sqrt[(2 + x^4)/(Sqrt[2] + x^2)^2]*EllipticF[2*ArcTan[x/2^(1/4)], 1/2])/(4*2^(1/4)*Sqrt[2 + x^4]) + ((1
+ I*Sqrt[7])*(Sqrt[2] + x^2)*Sqrt[(2 + x^4)/(Sqrt[2] + x^2)^2]*EllipticF[2*ArcTan[x/2^(1/4)], 1/2])/(4*2^(1/4
)*Sqrt[2 + x^4]) + ((1 - I*Sqrt[7])^2*(1 - 2/Sqrt[-3 - I*Sqrt[7]])*(Sqrt[2] + x^2)*Sqrt[(2 + x^4)/(Sqrt[2] + x
^2)^2]*EllipticF[2*ArcTan[x/2^(1/4)], 1/2])/(8*2^(1/4)*(7 + I*Sqrt[7])*Sqrt[2 + x^4]) + ((1 - I*Sqrt[7])^2*(1
+ 2/Sqrt[-3 - I*Sqrt[7]])*(Sqrt[2] + x^2)*Sqrt[(2 + x^4)/(Sqrt[2] + x^2)^2]*EllipticF[2*ArcTan[x/2^(1/4)], 1/2
])/(8*2^(1/4)*(7 + I*Sqrt[7])*Sqrt[2 + x^4]) + ((1 + I*Sqrt[7])^2*(1 - 2/Sqrt[-3 + I*Sqrt[7]])*(Sqrt[2] + x^2)
*Sqrt[(2 + x^4)/(Sqrt[2] + x^2)^2]*EllipticF[2*ArcTan[x/2^(1/4)], 1/2])/(8*2^(1/4)*(7 - I*Sqrt[7])*Sqrt[2 + x^
4]) + ((1 + I*Sqrt[7])^2*(1 + 2/Sqrt[-3 + I*Sqrt[7]])*(Sqrt[2] + x^2)*Sqrt[(2 + x^4)/(Sqrt[2] + x^2)^2]*Ellipt
icF[2*ArcTan[x/2^(1/4)], 1/2])/(8*2^(1/4)*(7 - I*Sqrt[7])*Sqrt[2 + x^4]) - ((3*I - Sqrt[7])*(2 + Sqrt[-3 - I*S
qrt[7]])^2*(Sqrt[2] + x^2)*Sqrt[(2 + x^4)/(Sqrt[2] + x^2)^2]*EllipticPi[-1/8*(2 - Sqrt[-3 - I*Sqrt[7]])^2/Sqrt
[-3 - I*Sqrt[7]], 2*ArcTan[x/2^(1/4)], 1/2])/(16*2^(1/4)*(7*I - 5*Sqrt[7])*Sqrt[2 + x^4]) - ((3*I - Sqrt[7])*(
2 - Sqrt[-3 - I*Sqrt[7]])^2*(Sqrt[2] + x^2)*Sqrt[(2 + x^4)/(Sqrt[2] + x^2)^2]*EllipticPi[(2 + Sqrt[-3 - I*Sqrt
[7]])^2/(8*Sqrt[-3 - I*Sqrt[7]]), 2*ArcTan[x/2^(1/4)], 1/2])/(16*2^(1/4)*(7*I - 5*Sqrt[7])*Sqrt[2 + x^4]) - ((
3*I + Sqrt[7])*(2 + Sqrt[-3 + I*Sqrt[7]])^2*(Sqrt[2] + x^2)*Sqrt[(2 + x^4)/(Sqrt[2] + x^2)^2]*EllipticPi[-1/8*
(2 - Sqrt[-3 + I*Sqrt[7]])^2/Sqrt[-3 + I*Sqrt[7]], 2*ArcTan[x/2^(1/4)], 1/2])/(16*2^(1/4)*(7*I + 5*Sqrt[7])*Sq
rt[2 + x^4]) - ((3*I + Sqrt[7])*(2 - Sqrt[-3 + I*Sqrt[7]])^2*(Sqrt[2] + x^2)*Sqrt[(2 + x^4)/(Sqrt[2] + x^2)^2]
*EllipticPi[(2 + Sqrt[-3 + I*Sqrt[7]])^2/(8*Sqrt[-3 + I*Sqrt[7]]), 2*ArcTan[x/2^(1/4)], 1/2])/(16*2^(1/4)*(7*I
+ 5*Sqrt[7])*Sqrt[2 + 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 (-2+x^4\right ) \sqrt {2+x^4}}{4+3 x^4+x^8} \, dx &=\int \left (\frac {\left (1+i \sqrt {7}\right ) \sqrt {2+x^4}}{3-i \sqrt {7}+2 x^4}+\frac {\left (1-i \sqrt {7}\right ) \sqrt {2+x^4}}{3+i \sqrt {7}+2 x^4}\right ) \, dx\\ &=\left (1-i \sqrt {7}\right ) \int \frac {\sqrt {2+x^4}}{3+i \sqrt {7}+2 x^4} \, dx+\left (1+i \sqrt {7}\right ) \int \frac {\sqrt {2+x^4}}{3-i \sqrt {7}+2 x^4} \, dx\\ &=\left (-3-i \sqrt {7}\right ) \int \frac {1}{\sqrt {2+x^4} \left (3+i \sqrt {7}+2 x^4\right )} \, dx+\frac {1}{2} \left (1-i \sqrt {7}\right ) \int \frac {1}{\sqrt {2+x^4}} \, dx+\left (-3+i \sqrt {7}\right ) \int \frac {1}{\sqrt {2+x^4} \left (3-i \sqrt {7}+2 x^4\right )} \, dx+\frac {1}{2} \left (1+i \sqrt {7}\right ) \int \frac {1}{\sqrt {2+x^4}} \, dx\\ &=\frac {\left (1-i \sqrt {7}\right ) \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} \sqrt {2+x^4}}+\frac {\left (1+i \sqrt {7}\right ) \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} \sqrt {2+x^4}}-\frac {1}{2} \int \frac {1}{\left (1-\frac {x^2}{\sqrt {\frac {1}{2} \left (-3-i \sqrt {7}\right )}}\right ) \sqrt {2+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+\frac {x^2}{\sqrt {\frac {1}{2} \left (-3-i \sqrt {7}\right )}}\right ) \sqrt {2+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1-\frac {x^2}{\sqrt {\frac {1}{2} \left (-3+i \sqrt {7}\right )}}\right ) \sqrt {2+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+\frac {x^2}{\sqrt {\frac {1}{2} \left (-3+i \sqrt {7}\right )}}\right ) \sqrt {2+x^4}} \, dx\\ &=\frac {\left (1-i \sqrt {7}\right ) \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} \sqrt {2+x^4}}+\frac {\left (1+i \sqrt {7}\right ) \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} \sqrt {2+x^4}}-\frac {\left (1-\frac {2}{\sqrt {-3-i \sqrt {7}}}\right ) \int \frac {1}{\sqrt {2+x^4}} \, dx}{2 \left (1+\frac {4 i}{3 i-\sqrt {7}}\right )}-\frac {\left (1+\frac {2}{\sqrt {-3-i \sqrt {7}}}\right ) \int \frac {1}{\sqrt {2+x^4}} \, dx}{2 \left (1+\frac {4 i}{3 i-\sqrt {7}}\right )}-\frac {\left (2-\sqrt {-3-i \sqrt {7}}\right ) \int \frac {1+\frac {x^2}{\sqrt {2}}}{\left (1-\frac {x^2}{\sqrt {\frac {1}{2} \left (-3-i \sqrt {7}\right )}}\right ) \sqrt {2+x^4}} \, dx}{7+i \sqrt {7}}-\frac {\left (2+\sqrt {-3-i \sqrt {7}}\right ) \int \frac {1+\frac {x^2}{\sqrt {2}}}{\left (1+\frac {x^2}{\sqrt {\frac {1}{2} \left (-3-i \sqrt {7}\right )}}\right ) \sqrt {2+x^4}} \, dx}{7+i \sqrt {7}}-\frac {\left (2-\sqrt {-3+i \sqrt {7}}\right ) \int \frac {1+\frac {x^2}{\sqrt {2}}}{\left (1-\frac {x^2}{\sqrt {\frac {1}{2} \left (-3+i \sqrt {7}\right )}}\right ) \sqrt {2+x^4}} \, dx}{7-i \sqrt {7}}-\frac {\left (2+\sqrt {-3+i \sqrt {7}}\right ) \int \frac {1+\frac {x^2}{\sqrt {2}}}{\left (1+\frac {x^2}{\sqrt {\frac {1}{2} \left (-3+i \sqrt {7}\right )}}\right ) \sqrt {2+x^4}} \, dx}{7-i \sqrt {7}}-\frac {\left (1-\frac {2}{\sqrt {-3+i \sqrt {7}}}\right ) \int \frac {1}{\sqrt {2+x^4}} \, dx}{2 \left (1+\frac {4 i}{3 i+\sqrt {7}}\right )}-\frac {\left (1+\frac {2}{\sqrt {-3+i \sqrt {7}}}\right ) \int \frac {1}{\sqrt {2+x^4}} \, dx}{2 \left (1+\frac {4 i}{3 i+\sqrt {7}}\right )}\\ &=-\frac {\sqrt [4]{-3-i \sqrt {7}} \tan ^{-1}\left (\frac {\sqrt {1-i \sqrt {7}} x}{\sqrt [4]{2 \left (-3-i \sqrt {7}\right )} \sqrt {2+x^4}}\right )}{2\ 2^{3/4} \sqrt {1-i \sqrt {7}}}-\frac {\sqrt [4]{-3+i \sqrt {7}} \tan ^{-1}\left (\frac {\sqrt {-1-i \sqrt {7}} x}{\sqrt [4]{2 \left (-3+i \sqrt {7}\right )} \sqrt {2+x^4}}\right )}{2\ 2^{3/4} \sqrt {-1-i \sqrt {7}}}-\frac {\sqrt [4]{-3-i \sqrt {7}} \tan ^{-1}\left (\frac {\sqrt {-1+i \sqrt {7}} x}{\sqrt [4]{2 \left (-3-i \sqrt {7}\right )} \sqrt {2+x^4}}\right )}{2\ 2^{3/4} \sqrt {-1+i \sqrt {7}}}-\frac {\sqrt [4]{-3+i \sqrt {7}} \tan ^{-1}\left (\frac {\sqrt {1+i \sqrt {7}} x}{\sqrt [4]{2 \left (-3+i \sqrt {7}\right )} \sqrt {2+x^4}}\right )}{2\ 2^{3/4} \sqrt {1+i \sqrt {7}}}+\frac {\left (1-i \sqrt {7}\right ) \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} \sqrt {2+x^4}}+\frac {\left (1+i \sqrt {7}\right ) \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} \sqrt {2+x^4}}-\frac {\left (1-\frac {2}{\sqrt {-3-i \sqrt {7}}}\right ) \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} \left (1+\frac {4 i}{3 i-\sqrt {7}}\right ) \sqrt {2+x^4}}-\frac {\left (1+\frac {2}{\sqrt {-3-i \sqrt {7}}}\right ) \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} \left (1+\frac {4 i}{3 i-\sqrt {7}}\right ) \sqrt {2+x^4}}-\frac {\left (1-\frac {2}{\sqrt {-3+i \sqrt {7}}}\right ) \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} \left (1+\frac {4 i}{3 i+\sqrt {7}}\right ) \sqrt {2+x^4}}-\frac {\left (1+\frac {2}{\sqrt {-3+i \sqrt {7}}}\right ) \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} \left (1+\frac {4 i}{3 i+\sqrt {7}}\right ) \sqrt {2+x^4}}-\frac {\left (2+\sqrt {-3-i \sqrt {7}}\right )^2 \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} \Pi \left (-\frac {\left (2-\sqrt {-3-i \sqrt {7}}\right )^2}{8 \sqrt {-3-i \sqrt {7}}};2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{2} \left (7+i \sqrt {7}\right ) \sqrt {2+x^4}}-\frac {\left (2-\sqrt {-3-i \sqrt {7}}\right )^2 \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} \Pi \left (\frac {\left (2+\sqrt {-3-i \sqrt {7}}\right )^2}{8 \sqrt {-3-i \sqrt {7}}};2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{2} \left (7+i \sqrt {7}\right ) \sqrt {2+x^4}}-\frac {\left (2+\sqrt {-3+i \sqrt {7}}\right )^2 \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} \Pi \left (-\frac {\left (2-\sqrt {-3+i \sqrt {7}}\right )^2}{8 \sqrt {-3+i \sqrt {7}}};2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{2} \left (7-i \sqrt {7}\right ) \sqrt {2+x^4}}-\frac {\left (2-\sqrt {-3+i \sqrt {7}}\right )^2 \left (\sqrt {2}+x^2\right ) \sqrt {\frac {2+x^4}{\left (\sqrt {2}+x^2\right )^2}} \Pi \left (\frac {\left (2+\sqrt {-3+i \sqrt {7}}\right )^2}{8 \sqrt {-3+i \sqrt {7}}};2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{2} \left (7-i \sqrt {7}\right ) \sqrt {2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.36, size = 166, normalized size = 0.89 \begin {gather*} \frac {1}{2} \sqrt [4]{-\frac {1}{2}} \left (-2 F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac {1}{2}} x\right )\right |-1\right )+\Pi \left (-\frac {2 \sqrt {2}}{-i+\sqrt {7}};\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac {1}{2}} x\right )\right |-1\right )+\Pi \left (\frac {2 \sqrt {2}}{-i+\sqrt {7}};\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac {1}{2}} x\right )\right |-1\right )+\Pi \left (-\frac {2 \sqrt {2}}{i+\sqrt {7}};\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac {1}{2}} x\right )\right |-1\right )+\Pi \left (\frac {2 \sqrt {2}}{i+\sqrt {7}};\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac {1}{2}} x\right )\right |-1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2 + x^4)*Sqrt[2 + x^4])/(4 + 3*x^4 + x^8),x]

[Out]

((-1/2)^(1/4)*(-2*EllipticF[I*ArcSinh[(-1/2)^(1/4)*x], -1] + EllipticPi[(-2*Sqrt[2])/(-I + Sqrt[7]), I*ArcSinh
[(-1/2)^(1/4)*x], -1] + EllipticPi[(2*Sqrt[2])/(-I + Sqrt[7]), I*ArcSinh[(-1/2)^(1/4)*x], -1] + EllipticPi[(-2
*Sqrt[2])/(I + Sqrt[7]), I*ArcSinh[(-1/2)^(1/4)*x], -1] + EllipticPi[(2*Sqrt[2])/(I + Sqrt[7]), I*ArcSinh[(-1/
2)^(1/4)*x], -1]))/2

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IntegrateAlgebraic [A]  time = 0.50, size = 33, normalized size = 0.18 \begin {gather*} -\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt {2+x^4}}\right )-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {2+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-2 + x^4)*Sqrt[2 + x^4])/(4 + 3*x^4 + x^8),x]

[Out]

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

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fricas [B]  time = 0.67, size = 60, normalized size = 0.32 \begin {gather*} -\frac {1}{4} \, \arctan \left (\frac {2 \, \sqrt {x^{4} + 2} x}{x^{4} - x^{2} + 2}\right ) + \frac {1}{4} \, \log \left (\frac {x^{4} + x^{2} - 2 \, \sqrt {x^{4} + 2} x + 2}{x^{4} - x^{2} + 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.62, size = 41, normalized size = 0.22

method	result	size
elliptic	\(\frac {\left (\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {x^{4}+2}}{x}\right )}{2}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {x^{4}+2}}{x}\right )}{2}\right ) \sqrt {2}}{2}\)	\(41\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \sqrt {x^{4}+2}\, x +2 \RootOf \left (\textit {\_Z}^{2}+1\right )}{x^{4}+x^{2}+2}\right )}{4}-\frac {\ln \left (-\frac {x^{4}+2 \sqrt {x^{4}+2}\, x +x^{2}+2}{x^{4}-x^{2}+2}\right )}{4}\)	\(98\)
default	\(\frac {\sqrt {2}\, \sqrt {4-2 i \sqrt {2}\, x^{2}}\, \sqrt {4+2 i \sqrt {2}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {2}\, \sqrt {i \sqrt {2}}}{2}, i\right )}{4 \sqrt {i \sqrt {2}}\, \sqrt {x^{4}+2}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{4}-\textit {\_Z}^{2}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {4 \arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{2}+x^{2}+1\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {x^{4}+2}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}}-\frac {2 \,2^{\frac {1}{4}} \left (-\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha \right ) \sqrt {2-i \sqrt {2}\, x^{2}}\, \sqrt {2+i \sqrt {2}\, x^{2}}\, \EllipticPi \left (\frac {x \sqrt {2}\, \sqrt {i \sqrt {2}}}{2}, \frac {i \sqrt {2}\, \underline {\hspace {1.25 ex}}\alpha ^{2}}{2}-\frac {i \sqrt {2}}{2}, \frac {\sqrt {-\frac {i \sqrt {2}}{2}}\, \sqrt {2}}{\sqrt {i \sqrt {2}}}\right )}{\sqrt {i}\, \sqrt {x^{4}+2}}\right )\right )}{32}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{4}+\textit {\_Z}^{2}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {4 \arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{2}+x^{2}-1\right )}{\sqrt {-\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {x^{4}+2}}\right )}{\sqrt {-\underline {\hspace {1.25 ex}}\alpha ^{2}}}-\frac {2 \,2^{\frac {1}{4}} \left (-\underline {\hspace {1.25 ex}}\alpha ^{3}-\underline {\hspace {1.25 ex}}\alpha \right ) \sqrt {2-i \sqrt {2}\, x^{2}}\, \sqrt {2+i \sqrt {2}\, x^{2}}\, \EllipticPi \left (\frac {x \sqrt {2}\, \sqrt {i \sqrt {2}}}{2}, \frac {i \sqrt {2}\, \underline {\hspace {1.25 ex}}\alpha ^{2}}{2}+\frac {i \sqrt {2}}{2}, \frac {\sqrt {-\frac {i \sqrt {2}}{2}}\, \sqrt {2}}{\sqrt {i \sqrt {2}}}\right )}{\sqrt {i}\, \sqrt {x^{4}+2}}\right )\right )}{32}\)	\(361\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (x^4-2\right )\,\sqrt {x^4+2}}{x^8+3\,x^4+4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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