3.230 \(\int \frac{\sqrt{1+x^2+x^4}}{(1+x^2)^3} \, dx\)

Optimal. Leaf size=93 \[ \frac{\sqrt{x^4+x^2+1} x}{4 \left (x^2+1\right )^2}+\frac{1}{4} \tan ^{-1}\left (\frac{x}{\sqrt{x^4+x^2+1}}\right )+\frac{\left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{4 \sqrt{x^4+x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.508155, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 13, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.65, Rules used = {1228, 1223, 1696, 1593, 1712, 1195, 1700, 1103, 1698, 203, 12, 1317, 1210} \[ \frac{\sqrt{x^4+x^2+1} x}{4 \left (x^2+1\right )^2}+\frac{1}{4} \tan ^{-1}\left (\frac{x}{\sqrt{x^4+x^2+1}}\right )+\frac{\left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{4 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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]

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 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 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 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 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 12

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

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 1210

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{1+x^2+x^4}}{\left (1+x^2\right )^3} \, dx &=\int \left (\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}}+\frac{1}{\left (1+x^2\right ) \sqrt{1+x^2+x^4}}\right ) \, 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+\int \frac{1}{\left (1+x^2\right ) \sqrt{1+x^2+x^4}} \, dx\\ &=\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}{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}{\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 \sqrt{1+x^2+x^4}}{4 \left (1+x^2\right )^2}+\frac{x \sqrt{1+x^2+x^4}}{4 \left (1+x^2\right )}+\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}{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{1}{2} \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}}{4 \left (1+x^2\right )^2}+\frac{3 x \sqrt{1+x^2+x^4}}{4 \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 )}{2 \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}{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}}{4 \left (1+x^2\right )^2}+\frac{3 x \sqrt{1+x^2+x^4}}{4 \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 )}{2 \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}{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}}{4 \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 )}{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 )}{2 \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 )-\int \frac{1}{\sqrt{1+x^2+x^4}} \, dx\\ &=\frac{x \sqrt{1+x^2+x^4}}{4 \left (1+x^2\right )^2}+\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{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}}{4 \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 )}{4 \sqrt{1+x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.301769, size = 176, normalized size = 1.89 \[ \frac{\sqrt [3]{-1} \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} \left (\text{EllipticF}\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )-E\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )\right )+\frac{x \left (x^2+2\right ) \left (x^4+x^2+1\right )}{\left (x^2+1\right )^2}-2 (-1)^{2/3} \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} \Pi \left (\sqrt [3]{-1};-i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )}{4 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.022, size = 333, normalized size = 3.6 \begin{align*}{\frac{x}{4\, \left ({x}^{2}+1 \right ) ^{2}}\sqrt{{x}^{4}+{x}^{2}+1}}+{\frac{x}{4\,{x}^{2}+4}\sqrt{{x}^{4}+{x}^{2}+1}}+{\frac{1}{\sqrt{-2+2\,i\sqrt{3}} \left ( i\sqrt{3}+1 \right ) }\sqrt{1+{\frac{{x}^{2}}{2}}-{\frac{i}{2}}{x}^{2}\sqrt{3}}\sqrt{1+{\frac{{x}^{2}}{2}}+{\frac{i}{2}}{x}^{2}\sqrt{3}}{\it EllipticF} \left ({\frac{x\sqrt{-2+2\,i\sqrt{3}}}{2}},{\frac{\sqrt{-2+2\,i\sqrt{3}}}{2}} \right ){\frac{1}{\sqrt{{x}^{4}+{x}^{2}+1}}}}-{\frac{1}{\sqrt{-2+2\,i\sqrt{3}} \left ( i\sqrt{3}+1 \right ) }\sqrt{1+{\frac{{x}^{2}}{2}}-{\frac{i}{2}}{x}^{2}\sqrt{3}}\sqrt{1+{\frac{{x}^{2}}{2}}+{\frac{i}{2}}{x}^{2}\sqrt{3}}{\it EllipticE} \left ({\frac{x\sqrt{-2+2\,i\sqrt{3}}}{2}},{\frac{\sqrt{-2+2\,i\sqrt{3}}}{2}} \right ){\frac{1}{\sqrt{{x}^{4}+{x}^{2}+1}}}}+{\frac{1}{2\,\sqrt{-1/2+i/2\sqrt{3}}}\sqrt{1+{\frac{{x}^{2}}{2}}-{\frac{i}{2}}{x}^{2}\sqrt{3}}\sqrt{1+{\frac{{x}^{2}}{2}}+{\frac{i}{2}}{x}^{2}\sqrt{3}}{\it EllipticPi} \left ( \sqrt{-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3}}x,- \left ( -{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) ^{-1},{\frac{\sqrt{-{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3}}}{\sqrt{-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3}}}} \right ){\frac{1}{\sqrt{{x}^{4}+{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{4} + x^{2} + 1}}{{\left (x^{2} + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{4} + x^{2} + 1}}{x^{6} + 3 \, x^{4} + 3 \, x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + x^2 + 1)/(x^6 + 3*x^4 + 3*x^2 + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )}}{\left (x^{2} + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{4} + x^{2} + 1}}{{\left (x^{2} + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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