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

Optimal. Leaf size=118 \[ \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt {x^4+x^2+1}}\right )-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{4 \sqrt {x^4+x^2+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{2 \sqrt {x^4+x^2+1}} \]

[Out]

1/2*arctan(x/(x^4+x^2+1)^(1/2))+1/2*(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)-1/4*(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 = 0.14, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1223, 1712, 1195, 12, 1317, 1103, 1698, 203} \[ \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt {x^4+x^2+1}}\right )-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{4 \sqrt {x^4+x^2+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{2 \sqrt {x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[x/Sqrt[1 + x^2 + x^4]]/2 + ((1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/4])/(2
*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])/(4*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 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 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 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 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 {1}{\left (1+x^2\right )^2 \sqrt {1+x^2+x^4}} \, dx &=\frac {x \sqrt {1+x^2+x^4}}{2 \left (1+x^2\right )}-\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}}{2 \left (1+x^2\right )}+\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 {\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}}-\int \frac {x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx\\ &=\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}{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 {\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}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {1+x^2+x^4}}\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}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.40, size = 226, normalized size = 1.92 \[ \frac {-(-1)^{2/3} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} F\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+2 (-1)^{2/3} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} \Pi \left (\sqrt [3]{-1};i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+\sqrt [3]{-1} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} \left (F\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-E\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )\right )+\frac {x^5+x^3+x}{x^2+1}}{2 \sqrt {x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x + x^3 + x^5)/(1 + x^2) - (-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)] + (-1)^(1/3)*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)^(2/3)*x^2]*(-EllipticE[I*ArcSin
h[(-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)])/(2*Sqrt[1 + x^2
+ x^4])

________________________________________________________________________________________

fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + x^{2} + 1}}{x^{8} + 3 \, x^{6} + 4 \, x^{4} + 3 \, x^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + x^2 + 1)/(x^8 + 3*x^6 + 4*x^4 + 3*x^2 + 1), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.02, size = 397, normalized size = 3.36 \[ \frac {\sqrt {x^{4}+x^{2}+1}\, x}{2 x^{2}+2}-\frac {2 \sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticE \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}-\frac {\sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticF \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {2 \sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticF \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {\sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticPi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________