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 + 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])

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Rubi [A]  time = 0.0106672, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, 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*}

Mathematica [C]  time = 0.352647, size = 164, normalized size = 3.35 \[ \frac{(-1)^{2/3} \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} \text{EllipticF}\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )+\sqrt [3]{-1} \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} \left (\text{EllipticF}\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )-E\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )\right )+\frac{x^5+x^3+x}{x^2+1}}{2 \sqrt{x^4+x^2+1}} \]

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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Maple [C]  time = 0.017, size = 224, normalized size = 4.6 \begin{align*}{\frac{x}{2\,{x}^{2}+2}\sqrt{{x}^{4}+{x}^{2}+1}}+{\frac{1}{\sqrt{-2+2\,i\sqrt{3}}}\sqrt{1- \left ( -{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ){x}^{2}}\sqrt{1- \left ( -{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ){x}^{2}}{\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}}}}+2\,{\frac{\sqrt{1- \left ( -1/2+i/2\sqrt{3} \right ){x}^{2}}\sqrt{1- \left ( -1/2-i/2\sqrt{3} \right ){x}^{2}} \left ({\it EllipticF} \left ( 1/2\,x\sqrt{-2+2\,i\sqrt{3}},1/2\,\sqrt{-2+2\,i\sqrt{3}} \right ) -{\it EllipticE} \left ( 1/2\,x\sqrt{-2+2\,i\sqrt{3}},1/2\,\sqrt{-2+2\,i\sqrt{3}} \right ) \right ) }{\sqrt{-2+2\,i\sqrt{3}}\sqrt{{x}^{4}+{x}^{2}+1} \left ( i\sqrt{3}+1 \right ) }} \end{align*}

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-(-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/(-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)/(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))-EllipticE(1/2*x*(-2+2*I*3^(1/2))^
(1/2),1/2*(-2+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 )}^{2}}\,{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)^2,x, algorithm="maxima")

[Out]

integrate(sqrt(x^4 + x^2 + 1)/(x^2 + 1)^2, 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^{4} + 2 \, 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)^2,x, algorithm="fricas")

[Out]

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

[Out]

Integral(sqrt((x**2 - x + 1)*(x**2 + x + 1))/(x**2 + 1)**2, 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 )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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