\(\int \frac {1}{(1+x^2)^2 (1+x^2+x^4)^{3/2}} \, dx\) [242]

Optimal result

Integrand size = 20, antiderivative size = 111 \[ \int \frac {1}{\left (1+x^2\right )^2 \left (1+x^2+x^4\right )^{3/2}} \, dx=-\frac {x \left (2+x^2\right )}{3 \sqrt {1+x^2+x^4}}+\frac {x \sqrt {1+x^2+x^4}}{3 \left (1+x^2\right )}+\arctan \left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{6 \sqrt {1+x^2+x^4}} \]

[Out]

arctan(x/(x^4+x^2+1)^(1/2))-1/3*x*(x^2+2)/(x^4+x^2+1)^(1/2)+1/3*x*(x^4+x^2+1)^(1/2)/(x^2+1)+1/6*(x^2+1)*(cos(2
*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticE(sin(2*arctan(x)),1/2)*((x^4+x^2+1)/(x^2+1)^2)^(1/2)/(x^4+x^2+1
)^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {1242, 1192, 1209, 1237, 1726, 12, 1331, 1117, 1712, 209, 1224} \[ \int \frac {1}{\left (1+x^2\right )^2 \left (1+x^2+x^4\right )^{3/2}} \, dx=\arctan \left (\frac {x}{\sqrt {x^4+x^2+1}}\right )+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{6 \sqrt {x^4+x^2+1}}+\frac {\sqrt {x^4+x^2+1} x}{3 \left (x^2+1\right )}-\frac {\left (x^2+2\right ) x}{3 \sqrt {x^4+x^2+1}} \]

[In]

Int[1/((1 + x^2)^2*(1 + x^2 + x^4)^(3/2)),x]

[Out]

-1/3*(x*(2 + x^2))/Sqrt[1 + x^2 + x^4] + (x*Sqrt[1 + x^2 + x^4])/(3*(1 + x^2)) + ArcTan[x/Sqrt[1 + x^2 + x^4]]
+ ((1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/4])/(6*Sqrt[1 + x^2 + x^4])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 1117

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)]/(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(
4*c))], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1192

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*(a*b*e - d*(b^2 - 2*a
*c) - c*(b*d - 2*a*e)*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), 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] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1209

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)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c))], 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 1224

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

Rule 1237

Int[((d_) + (e_.)*(x_)^2)^(q_)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> Simp[(-e^2)*x*(d + e*x^2
)^(q + 1)*(Sqrt[a + b*x^2 + c*x^4]/(2*d*(q + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/(2*d*(q + 1)*(c*d^2 - b
*d*e + a*e^2)), Int[((d + e*x^2)^(q + 1)/Sqrt[a + b*x^2 + c*x^4])*Simp[a*e^2*(2*q + 3) + 2*d*(c*d - b*e)*(q +
1) - 2*e*(c*d*(q + 1) - b*e*(q + 2))*x^2 + c*e^2*(2*q + 5)*x^4, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ
[b^2 - 4*a*c, 0] && ILtQ[q, -1]

Rule 1242

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{aa, bb, cc}, In
t[ExpandIntegrand[1/Sqrt[aa + bb*x^2 + cc*x^4], (d + e*x^2)^q*(aa + bb*x^2 + cc*x^4)^(p + 1/2), x] /. {aa -> a
, bb -> b, cc -> c}, 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]
&& ILtQ[q, 0] && IntegerQ[p + 1/2]

Rule 1331

Int[(x_)^2/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[d/(2*d*e), Int[
1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[d/(2*d*e), Int[(d - e*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] && PosQ[c/a] && EqQ[c
*d^2 - a*e^2, 0]

Rule 1712

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[
A, Subst[Int[1/(d - (b*d - 2*a*e)*x^2), x], x, x/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] && EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 1726

Int[(P4x_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{A = Coeff[P4x,
x, 0], B = Coeff[P4x, x, 2], C = Coeff[P4x, x, 4]}, Dist[-C/e^2, Int[(d - e*x^2)/Sqrt[a + b*x^2 + c*x^4], x],
x] + Dist[1/e^2, Int[(C*d^2 + A*e^2 + B*e^2*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a,
b, c, d, e}, x] && PolyQ[P4x, x^2, 2] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a
*e^2, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-1-x^2}{\left (1+x^2+x^4\right )^{3/2}}+\frac {1}{\left (1+x^2\right )^2 \sqrt {1+x^2+x^4}}+\frac {1}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}}\right ) \, dx \\ & = \int \frac {-1-x^2}{\left (1+x^2+x^4\right )^{3/2}} \, dx+\int \frac {1}{\left (1+x^2\right )^2 \sqrt {1+x^2+x^4}} \, dx+\int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx \\ & = -\frac {x \left (2+x^2\right )}{3 \sqrt {1+x^2+x^4}}+\frac {x \sqrt {1+x^2+x^4}}{2 \left (1+x^2\right )}+\frac {1}{3} \int \frac {-1+x^2}{\sqrt {1+x^2+x^4}} \, dx+\frac {1}{2} \int \frac {1}{\sqrt {1+x^2+x^4}} \, dx+\frac {1}{2} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx-\frac {1}{2} \int \frac {-1+2 x^2+x^4}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx \\ & = -\frac {x \left (2+x^2\right )}{3 \sqrt {1+x^2+x^4}}+\frac {5 x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )}-\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{3 \sqrt {1+x^2+x^4}}+\frac {\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 {1}{2} \int \frac {1-x^2}{\sqrt {1+x^2+x^4}} \, dx-\frac {1}{2} \int \frac {2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {1+x^2+x^4}}\right ) \\ & = -\frac {x \left (2+x^2\right )}{3 \sqrt {1+x^2+x^4}}+\frac {x \sqrt {1+x^2+x^4}}{3 \left (1+x^2\right )}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{6 \sqrt {1+x^2+x^4}}+\frac {\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}}-\int \frac {x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx \\ & = -\frac {x \left (2+x^2\right )}{3 \sqrt {1+x^2+x^4}}+\frac {x \sqrt {1+x^2+x^4}}{3 \left (1+x^2\right )}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{6 \sqrt {1+x^2+x^4}}+\frac {\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 {1}{2} \int \frac {1}{\sqrt {1+x^2+x^4}} \, dx+\frac {1}{2} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx \\ & = -\frac {x \left (2+x^2\right )}{3 \sqrt {1+x^2+x^4}}+\frac {x \sqrt {1+x^2+x^4}}{3 \left (1+x^2\right )}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{6 \sqrt {1+x^2+x^4}}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {1+x^2+x^4}}\right ) \\ & = -\frac {x \left (2+x^2\right )}{3 \sqrt {1+x^2+x^4}}+\frac {x \sqrt {1+x^2+x^4}}{3 \left (1+x^2\right )}+\tan ^{-1}\left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{6 \sqrt {1+x^2+x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.29 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.51 \[ \int \frac {1}{\left (1+x^2\right )^2 \left (1+x^2+x^4\right )^{3/2}} \, dx=\frac {-2 x \left (1+x^2\right ) \left (2+x^2\right )+3 x \left (1+x^2+x^4\right )-\sqrt [3]{-1} \left (1+x^2\right ) \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} \left (E\left (i \text {arcsinh}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+\left (-1+5 \sqrt [3]{-1}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )-12 \sqrt [3]{-1} \operatorname {EllipticPi}\left (\sqrt [3]{-1},i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )\right )}{6 \left (1+x^2\right ) \sqrt {1+x^2+x^4}} \]

[In]

Integrate[1/((1 + x^2)^2*(1 + x^2 + x^4)^(3/2)),x]

[Out]

(-2*x*(1 + x^2)*(2 + x^2) + 3*x*(1 + x^2 + x^4) - (-1)^(1/3)*(1 + x^2)*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)^
(2/3)*x^2]*(EllipticE[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)] + (-1 + 5*(-1)^(1/3))*EllipticF[I*ArcSinh[(-1)^(5/6
)*x], (-1)^(2/3)] - 12*(-1)^(1/3)*EllipticPi[(-1)^(1/3), I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)]))/(6*(1 + x^2)*S
qrt[1 + x^2 + x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.62 (sec) , antiderivative size = 339, normalized size of antiderivative = 3.05

method	result	size
risch	\(\frac {x \left (x^{4}-3 x^{2}-1\right )}{6 \left (x^{2}+1\right ) \sqrt {x^{4}+x^{2}+1}}-\frac {5 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {2 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {2 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, \Pi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}\)	\(339\)
default	\(\frac {x \sqrt {x^{4}+x^{2}+1}}{2 x^{2}+2}-\frac {2 \left (\frac {1}{6} x^{3}+\frac {1}{3} x \right )}{\sqrt {x^{4}+x^{2}+1}}-\frac {5 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {2 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}-\frac {2 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {2 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, \Pi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}\)	\(419\)
elliptic	\(\frac {x \sqrt {x^{4}+x^{2}+1}}{2 x^{2}+2}-\frac {2 \left (\frac {1}{6} x^{3}+\frac {1}{3} x \right )}{\sqrt {x^{4}+x^{2}+1}}-\frac {5 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {2 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}-\frac {2 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {2 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, \Pi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}\)	\(419\)

[In]

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

[Out]

1/6*x*(x^4-3*x^2-1)/(x^2+1)/(x^4+x^2+1)^(1/2)-5/3/(-2+2*I*3^(1/2))^(1/2)*(1-(-1/2+1/2*I*3^(1/2))*x^2)^(1/2)*(1
-(-1/2-1/2*I*3^(1/2))*x^2)^(1/2)/(x^4+x^2+1)^(1/2)*EllipticF(1/2*x*(-2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))
^(1/2))+2/3/(-2+2*I*3^(1/2))^(1/2)*(1-(-1/2+1/2*I*3^(1/2))*x^2)^(1/2)*(1-(-1/2-1/2*I*3^(1/2))*x^2)^(1/2)/(x^4+
x^2+1)^(1/2)/(1+I*3^(1/2))*(EllipticF(1/2*x*(-2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2))-EllipticE(1/2*x
*(-2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2)))+2/(-1/2+1/2*I*3^(1/2))^(1/2)*(1+1/2*x^2-1/2*I*x^2*3^(1/2)
)^(1/2)*(1+1/2*x^2+1/2*I*x^2*3^(1/2))^(1/2)/(x^4+x^2+1)^(1/2)*EllipticPi((-1/2+1/2*I*3^(1/2))^(1/2)*x,-1/(-1/2
+1/2*I*3^(1/2)),(-1/2-1/2*I*3^(1/2))^(1/2)/(-1/2+1/2*I*3^(1/2))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.75 \[ \int \frac {1}{\left (1+x^2\right )^2 \left (1+x^2+x^4\right )^{3/2}} \, dx=-\frac {2 \, \sqrt {2} \sqrt {-3} {\left (x^{6} + 2 \, x^{4} + 2 \, x^{2} + 1\right )} \sqrt {\sqrt {-3} - 1} F(\arcsin \left (\frac {1}{2} \, \sqrt {2} x \sqrt {\sqrt {-3} - 1}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) + \sqrt {2} {\left (x^{6} + 2 \, x^{4} + 2 \, x^{2} - \sqrt {-3} {\left (x^{6} + 2 \, x^{4} + 2 \, x^{2} + 1\right )} + 1\right )} \sqrt {\sqrt {-3} - 1} E(\arcsin \left (\frac {1}{2} \, \sqrt {2} x \sqrt {\sqrt {-3} - 1}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) - 24 \, {\left (x^{6} + 2 \, x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (\frac {x}{\sqrt {x^{4} + x^{2} + 1}}\right ) - 4 \, {\left (x^{5} - 3 \, x^{3} - x\right )} \sqrt {x^{4} + x^{2} + 1}}{24 \, {\left (x^{6} + 2 \, x^{4} + 2 \, x^{2} + 1\right )}} \]

[In]

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

[Out]

-1/24*(2*sqrt(2)*sqrt(-3)*(x^6 + 2*x^4 + 2*x^2 + 1)*sqrt(sqrt(-3) - 1)*elliptic_f(arcsin(1/2*sqrt(2)*x*sqrt(sq
rt(-3) - 1)), 1/2*sqrt(-3) - 1/2) + sqrt(2)*(x^6 + 2*x^4 + 2*x^2 - sqrt(-3)*(x^6 + 2*x^4 + 2*x^2 + 1) + 1)*sqr
t(sqrt(-3) - 1)*elliptic_e(arcsin(1/2*sqrt(2)*x*sqrt(sqrt(-3) - 1)), 1/2*sqrt(-3) - 1/2) - 24*(x^6 + 2*x^4 + 2
*x^2 + 1)*arctan(x/sqrt(x^4 + x^2 + 1)) - 4*(x^5 - 3*x^3 - x)*sqrt(x^4 + x^2 + 1))/(x^6 + 2*x^4 + 2*x^2 + 1)

Sympy [F]

\[ \int \frac {1}{\left (1+x^2\right )^2 \left (1+x^2+x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {3}{2}} \left (x^{2} + 1\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\left (1+x^2\right )^2 \left (1+x^2+x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{4} + x^{2} + 1\right )}^{\frac {3}{2}} {\left (x^{2} + 1\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\left (1+x^2\right )^2 \left (1+x^2+x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{4} + x^{2} + 1\right )}^{\frac {3}{2}} {\left (x^{2} + 1\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (1+x^2\right )^2 \left (1+x^2+x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (x^2+1\right )}^2\,{\left (x^4+x^2+1\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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