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

Optimal. Leaf size=166 \[ \frac {\sqrt {x^4+x^2+1} x}{6 \left (x^2+1\right )^2}+\frac {\sqrt {x^4+x^2+1} x}{6 \left (x^2+1\right )^3}+\frac {1}{4} \tan ^{-1}\left (\frac {x}{\sqrt {x^4+x^2+1}}\right )-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{8 \sqrt {x^4+x^2+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{3 \sqrt {x^4+x^2+1}} \]

[Out]

1/4*arctan(x/(x^4+x^2+1)^(1/2))+1/6*x*(x^4+x^2+1)^(1/2)/(x^2+1)^3+1/6*x*(x^4+x^2+1)^(1/2)/(x^2+1)^2+1/3*(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)-1/8*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticF(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]  time = 0.62, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 14, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {1228, 1223, 1696, 1586, 1197, 1103, 1195, 1593, 1712, 1700, 1698, 203, 12, 1317} \[ \frac {\sqrt {x^4+x^2+1} x}{6 \left (x^2+1\right )^2}+\frac {\sqrt {x^4+x^2+1} x}{6 \left (x^2+1\right )^3}+\frac {1}{4} \tan ^{-1}\left (\frac {x}{\sqrt {x^4+x^2+1}}\right )-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{8 \sqrt {x^4+x^2+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{3 \sqrt {x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[1 + x^2 + x^4])/(6*(1 + x^2)^3) + (x*Sqrt[1 + x^2 + x^4])/(6*(1 + x^2)^2) + ArcTan[x/Sqrt[1 + x^2 + x^
4]]/4 + ((1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/4])/(3*Sqrt[1 + x^2 + x^4]) - ((
1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/4])/(8*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 203

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

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 1223

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)*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])/Sqrt[a + b*x^2 + c*x^4], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b
^2 - 4*a*c, 0] && ILtQ[q, -1]

Rule 1228

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 1317

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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1696

Int[((P4x_)*((d_) + (e_.)*(x_)^2)^(q_))/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]}, -Simp[((C*d^2 - B*d*e + A*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)*Simp[a*d*(C*d - B*e) + A*(a*e^2*(2*q + 3) + 2*d*(c*d - b*e)*(q + 1)) - 2*((B*d
- A*e)*(b*e*(q + 2) - c*d*(q + 1)) - C*d*(b*d + a*e*(q + 1)))*x^2 + c*(C*d^2 - B*d*e + A*e^2)*(2*q + 5)*x^4, x
])/Sqrt[a + b*x^2 + c*x^4], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[P4x, x^2] && LeQ[Expon[P4x, x], 4] &
& NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[q, -1]

Rule 1698

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 1700

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

Rule 1712

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*} \int \frac {\sqrt {1+x^2+x^4}}{\left (1+x^2\right )^4} \, dx &=\int \left (\frac {1}{\left (1+x^2\right )^4 \sqrt {1+x^2+x^4}}-\frac {1}{\left (1+x^2\right )^3 \sqrt {1+x^2+x^4}}+\frac {1}{\left (1+x^2\right )^2 \sqrt {1+x^2+x^4}}\right ) \, dx\\ &=\int \frac {1}{\left (1+x^2\right )^4 \sqrt {1+x^2+x^4}} \, dx-\int \frac {1}{\left (1+x^2\right )^3 \sqrt {1+x^2+x^4}} \, dx+\int \frac {1}{\left (1+x^2\right )^2 \sqrt {1+x^2+x^4}} \, dx\\ &=\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^3}-\frac {x \sqrt {1+x^2+x^4}}{4 \left (1+x^2\right )^2}+\frac {x \sqrt {1+x^2+x^4}}{2 \left (1+x^2\right )}-\frac {1}{6} \int \frac {-5+2 x^2-3 x^4}{\left (1+x^2\right )^3 \sqrt {1+x^2+x^4}} \, dx+\frac {1}{4} \int \frac {-3+2 x^2-x^4}{\left (1+x^2\right )^2 \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 \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^3}+\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^2}-\frac {x \sqrt {1+x^2+x^4}}{4 \left (1+x^2\right )}+\frac {1}{24} \int \frac {10-8 x^2+10 x^4}{\left (1+x^2\right )^2 \sqrt {1+x^2+x^4}} \, dx-\frac {1}{8} \int \frac {-10 x^2-6 x^4}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx+\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 {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^3}+\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^2}-\frac {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 )}{2 \sqrt {1+x^2+x^4}}-\frac {1}{48} \int \frac {8+36 x^2+28 x^4}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx-\frac {1}{8} \int \frac {x^2 \left (-10-6 x^2\right )}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx-\int \frac {x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx\\ &=\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^3}+\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^2}-\frac {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 )}{2 \sqrt {1+x^2+x^4}}-\frac {1}{48} \int \frac {8+28 x^2}{\sqrt {1+x^2+x^4}} \, dx-\frac {1}{8} \int \frac {-6-10 x^2}{\left (1+x^2\right ) \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 {3}{4} \int \frac {1-x^2}{\sqrt {1+x^2+x^4}} \, dx\\ &=\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^3}+\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^2}+\frac {7 x \sqrt {1+x^2+x^4}}{12 \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 )}{4 \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}{4} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {7}{12} \int \frac {1-x^2}{\sqrt {1+x^2+x^4}} \, dx-\frac {3}{4} \int \frac {1}{\sqrt {1+x^2+x^4}} \, dx+\int \frac {1}{\sqrt {1+x^2+x^4}} \, dx\\ &=\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^3}+\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^2}+\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 )}{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 )}{8 \sqrt {1+x^2+x^4}}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {1+x^2+x^4}}\right )\\ &=\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^3}+\frac {x \sqrt {1+x^2+x^4}}{6 \left (1+x^2\right )^2}+\frac {1}{4} \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 )}{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 )}{8 \sqrt {1+x^2+x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.41, size = 240, normalized size = 1.45 \[ \frac {-(-1)^{2/3} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} F\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+3 (-1)^{2/3} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} \Pi \left (\sqrt [3]{-1};i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-2 \sqrt [3]{-1} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} \left (E\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-F\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )\right )+\frac {x \left (x^4+x^2+1\right ) \left (2 x^4+5 x^2+4\right )}{\left (x^2+1\right )^3}}{6 \sqrt {x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + x^{2} + 1}}{x^{8} + 4 \, x^{6} + 6 \, x^{4} + 4 \, x^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + x^2 + 1)/(x^8 + 4*x^6 + 6*x^4 + 4*x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.03, size = 438, normalized size = 2.64 \[ \frac {\sqrt {x^{4}+x^{2}+1}\, x}{6 \left (x^{2}+1\right )^{3}}+\frac {\sqrt {x^{4}+x^{2}+1}\, x}{6 \left (x^{2}+1\right )^{2}}+\frac {\sqrt {x^{4}+x^{2}+1}\, x}{3 x^{2}+3}-\frac {4 \sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticE \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}-\frac {\sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticF \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {4 \sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticF \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {\sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticPi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{2 \sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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