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

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

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Rubi [C]  time = 1.43, antiderivative size = 350, normalized size of antiderivative = 8.14, number of steps used = 34, number of rules used = 8, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {6725, 1208, 1197, 1103, 1195, 1216, 1706, 6728} \begin {gather*} \tanh ^{-1}\left (\frac {x}{\sqrt {x^4+x^2+1}}\right )-\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+x^2+1}}\right )-\frac {\left (\sqrt {3}+i\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{\left (\sqrt {3}+3 i\right ) \sqrt {x^4+x^2+1}}+\frac {\left (5+i \sqrt {3}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{4 \sqrt {x^4+x^2+1}}+\frac {\left (5-i \sqrt {3}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{4 \sqrt {x^4+x^2+1}}-\frac {\left (-\sqrt {3}+i\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{\left (-\sqrt {3}+3 i\right ) \sqrt {x^4+x^2+1}}-\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{2 \sqrt {x^4+x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.79, size = 284, normalized size = 6.60 \begin {gather*} \frac {\sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} \left (\left ((-1)^{2/3}-1\right ) F\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-2 \left (\sqrt [3]{-1}-2\right ) \Pi \left (-1;i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-4 (-1)^{2/3} \Pi \left (-(-1)^{2/3};i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+2 \sqrt [3]{-1} \Pi \left (-(-1)^{2/3};i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+(-1)^{2/3} \Pi \left (-(-1)^{5/6};i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-\Pi \left (-(-1)^{5/6};i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+(-1)^{2/3} \Pi \left ((-1)^{5/6};i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-\Pi \left ((-1)^{5/6};i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )\right )}{\left (1+\sqrt [3]{-1}\right ) \sqrt {x^4+x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)^(2/3)*x^2]*((-1 + (-1)^(2/3))*EllipticF[I*ArcSinh[(-1)^(5/6)*x], (-1)^
(2/3)] - 2*(-2 + (-1)^(1/3))*EllipticPi[-1, I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)] + 2*(-1)^(1/3)*EllipticPi[-(-
1)^(2/3), I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)] - 4*(-1)^(2/3)*EllipticPi[-(-1)^(2/3), I*ArcSinh[(-1)^(5/6)*x],
(-1)^(2/3)] - EllipticPi[-(-1)^(5/6), I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)] + (-1)^(2/3)*EllipticPi[-(-1)^(5/6
), I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)] - EllipticPi[(-1)^(5/6), I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)] + (-1)^(
2/3)*EllipticPi[(-1)^(5/6), I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)]))/((1 + (-1)^(1/3))*Sqrt[1 + x^2 + x^4])

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IntegrateAlgebraic [A]  time = 0.36, size = 43, normalized size = 1.00 \begin {gather*} \tanh ^{-1}\left (\frac {x}{\sqrt {1+x^2+x^4}}\right )-\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^2+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.57, size = 111, normalized size = 2.58 \begin {gather*} \frac {1}{4} \, \sqrt {2} \log \left (-\frac {x^{8} + 14 \, x^{6} + 19 \, x^{4} - 4 \, \sqrt {2} {\left (x^{5} + 3 \, x^{3} + x\right )} \sqrt {x^{4} + x^{2} + 1} + 14 \, x^{2} + 1}{x^{8} - 2 \, x^{6} + 3 \, x^{4} - 2 \, x^{2} + 1}\right ) + \frac {1}{2} \, \log \left (-\frac {x^{4} + 2 \, x^{2} + 2 \, \sqrt {x^{4} + x^{2} + 1} x + 1}{x^{4} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*log(-(x^8 + 14*x^6 + 19*x^4 - 4*sqrt(2)*(x^5 + 3*x^3 + x)*sqrt(x^4 + x^2 + 1) + 14*x^2 + 1)/(x^8 -
2*x^6 + 3*x^4 - 2*x^2 + 1)) + 1/2*log(-(x^4 + 2*x^2 + 2*sqrt(x^4 + x^2 + 1)*x + 1)/(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} + x^{2} + 1} {\left (x^{4} - 1\right )}}{{\left (x^{4} - x^{2} + 1\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)^(1/2)/(x^4+1)/(x^4-x^2+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.50, size = 70, normalized size = 1.63

method	result	size
elliptic	\(\frac {\left (-\ln \left (1+\frac {\sqrt {x^{4}+x^{2}+1}\, \sqrt {2}}{2 x}\right )+\sqrt {2}\, \arctanh \left (\frac {\sqrt {x^{4}+x^{2}+1}}{x}\right )+\ln \left (-1+\frac {\sqrt {x^{4}+x^{2}+1}\, \sqrt {2}}{2 x}\right )\right ) \sqrt {2}}{2}\)	\(70\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) x^{4}+3 \RootOf \left (\textit {\_Z}^{2}-2\right ) x^{2}+4 \sqrt {x^{4}+x^{2}+1}\, x +\RootOf \left (\textit {\_Z}^{2}-2\right )}{x^{4}-x^{2}+1}\right )}{2}-\frac {\ln \left (-\frac {-x^{4}+2 \sqrt {x^{4}+x^{2}+1}\, x -2 x^{2}-1}{x^{4}+1}\right )}{2}\)	\(103\)
default	\(\frac {2 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \EllipticF \left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {\sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{4}-\textit {\_Z}^{2}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {\arctanh \left (\frac {\left (2 \underline {\hspace {1.25 ex}}\alpha ^{2}+1\right ) \left (-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+7 x^{2}+8\right ) \sqrt {2}}{28 \sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {x^{4}+x^{2}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}}-\frac {2 \left (-\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha \right ) \sqrt {x^{2}+2-i \sqrt {3}\, x^{2}}\, \sqrt {x^{2}+2+i \sqrt {3}\, x^{2}}\, \EllipticPi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , \frac {i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}}{2}+\frac {\underline {\hspace {1.25 ex}}\alpha ^{2}}{2}-\frac {1}{2}-\frac {i \sqrt {3}}{2}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {i \sqrt {3}-1}\, \sqrt {x^{4}+x^{2}+1}}\right )\right )}{4}-\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 {\arctanh \left (\frac {\left (2 \underline {\hspace {1.25 ex}}\alpha ^{2}+1\right ) \left (-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+5 x^{2}+4\right )}{10 \sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {x^{4}+x^{2}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}}+\frac {\sqrt {2}\, \underline {\hspace {1.25 ex}}\alpha ^{3} \sqrt {x^{2}+2-i \sqrt {3}\, x^{2}}\, \sqrt {x^{2}+2+i \sqrt {3}\, x^{2}}\, \EllipticPi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , \frac {i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}}{2}+\frac {\underline {\hspace {1.25 ex}}\alpha ^{2}}{2}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {i \sqrt {3}-1}\, \sqrt {x^{4}+x^{2}+1}}\right )\right )}{4}\)	\(436\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-ln(1+1/2*(x^4+x^2+1)^(1/2)*2^(1/2)/x)+2^(1/2)*arctanh(1/x*(x^4+x^2+1)^(1/2))+ln(-1+1/2*(x^4+x^2+1)^(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 {x^{4} + x^{2} + 1} {\left (x^{4} - 1\right )}}{{\left (x^{4} - x^{2} + 1\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)^(1/2)/(x^4+1)/(x^4-x^2+1),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\left (x^4-1\right )\,\sqrt {x^4+x^2+1}}{\left (x^4+1\right )\,\left (x^4-x^2+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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