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

Optimal. Leaf size=96 \[ -\frac{\sqrt{x^4+x^2+1} x}{3 \left (x^2+1\right )}+\frac{\left (x^2+2\right ) x}{3 \sqrt{x^4+x^2+1}}+\frac{\left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{3 \sqrt{x^4+x^2+1}} \]

[Out]

(x*(2 + x^2))/(3*Sqrt[1 + x^2 + x^4]) - (x*Sqrt[1 + x^2 + x^4])/(3*(1 + x^2)) + ((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])

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Rubi [A]  time = 0.0206474, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1178, 1195} \[ -\frac{\sqrt{x^4+x^2+1} x}{3 \left (x^2+1\right )}+\frac{\left (x^2+2\right ) x}{3 \sqrt{x^4+x^2+1}}+\frac{\left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{3 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(2 + x^2))/(3*Sqrt[1 + x^2 + x^4]) - (x*Sqrt[1 + x^2 + x^4])/(3*(1 + x^2)) + ((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])

Rule 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), 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] && LtQ[p, -1] && IntegerQ[2*p]

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]

Rubi steps

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

Mathematica [F]  time = 0, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [C]  time = 0.007, size = 247, normalized size = 2.6 \begin{align*} -2\,{\frac{-1/3\,{x}^{3}-x/6}{\sqrt{{x}^{4}+{x}^{2}+1}}}+{\frac{2}{3\,\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}}}}+{\frac{4}{3\,\sqrt{-2+2\,i\sqrt{3}} \left ( i\sqrt{3}+1 \right ) }\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}} \left ({\it EllipticF} \left ({\frac{x\sqrt{-2+2\,i\sqrt{3}}}{2}},{\frac{\sqrt{-2+2\,i\sqrt{3}}}{2}} \right ) -{\it EllipticE} \left ({\frac{x\sqrt{-2+2\,i\sqrt{3}}}{2}},{\frac{\sqrt{-2+2\,i\sqrt{3}}}{2}} \right ) \right ){\frac{1}{\sqrt{{x}^{4}+{x}^{2}+1}}}}-2\,{\frac{-x/6+1/6\,{x}^{3}}{\sqrt{{x}^{4}+{x}^{2}+1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-1/3*x^3-1/6*x)/(x^4+x^2+1)^(1/2)+2/3/(-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))+
4/3/(-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)))-2*(-1/6*x+1/6*x^3)/(x^4+x^2+1)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^2 + 1)/(x^4 + x^2 + 1)^(3/2), 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}{\left (x^{2} + 1\right )}}{x^{8} + 2 \, x^{6} + 3 \, x^{4} + 2 \, x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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