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

Optimal. Leaf size=49 \[ \frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{2 \sqrt {x^4+x^2+1}} \]

[Out]

1/2*(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)

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Rubi [A]  time = 0.01, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {1225} \[ \frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{2 \sqrt {x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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])

Rule 1225

Int[Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> With[{q = Rt[e/d, 2]}, Simp
[(c*(d + e*x^2)*Sqrt[(e^2*(a + b*x^2 + c*x^4))/(c*(d + e*x^2)^2)]*EllipticE[2*ArcTan[q*x], (2*c*d - b*e)/(4*c*
d)])/(2*d*e^2*q*Sqrt[a + b*x^2 + c*x^4]), 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] && EqQ[c*d^2 - a*e^2, 0] && PosQ[e/d]

Rubi steps

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

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Mathematica [C]  time = 0.35, size = 164, normalized size = 3.35 \[ \frac {(-1)^{2/3} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} F\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+\sqrt [3]{-1} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} \left (F\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-E\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )\right )+\frac {x^5+x^3+x}{x^2+1}}{2 \sqrt {x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + x^{2} + 1}}{x^{4} + 2 \, x^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 224, normalized size = 4.57 \[ \frac {\sqrt {x^{4}+x^{2}+1}\, x}{2 x^{2}+2}+\frac {\sqrt {-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {2 \sqrt {-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )+\EllipticF \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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