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

Optimal. Leaf size=69 \[ \frac{\left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{4}\right )}{4 \sqrt{x^4+x^2+1}}+\frac{1}{2} \tan ^{-1}\left (\frac{x}{\sqrt{x^4+x^2+1}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.058729, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1210, 1103, 1698, 203} \[ \frac{1}{2} \tan ^{-1}\left (\frac{x}{\sqrt{x^4+x^2+1}}\right )+\frac{\left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{4 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

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]

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])

Rubi steps

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

Mathematica [C]  time = 0.0686553, size = 73, normalized size = 1.06 \[ -\frac{(-1)^{2/3} \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} \Pi \left (\sqrt [3]{-1};-i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )}{\sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((-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)])/Sqrt[1 + x^2 + x^4])

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Maple [C]  time = 0.016, size = 104, normalized size = 1.5 \begin{align*}{\frac{1}{\sqrt{-{\frac{1}{2}}+{\frac{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(1/(x^2+1)/(x^4+x^2+1)^(1/2),x)

[Out]

1/(-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{1}{\sqrt{x^{4} + x^{2} + 1}{\left (x^{2} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(x^4 + x^2 + 1)*(x^2 + 1)), 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} + 2 \, x^{4} + 2 \, x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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