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

Optimal. Leaf size=149 \[ -\frac{\sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{\sqrt{x^4+3 x^2+2}}-\frac{3 x \left (x^2+2\right )}{2 \sqrt{x^4+3 x^2+2}}+\frac{x \left (3 x^2+5\right )}{2 \sqrt{x^4+3 x^2+2}}+\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4+3 x^2+2}} \]

[Out]

(-3*x*(2 + x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (x*(5 + 3*x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (3*(1 + x^2)*Sqrt[(2
+ x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) - (Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^
2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/Sqrt[2 + 3*x^2 + x^4]

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Rubi [A]  time = 0.0417569, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {1092, 1189, 1099, 1135} \[ -\frac{3 x \left (x^2+2\right )}{2 \sqrt{x^4+3 x^2+2}}+\frac{x \left (3 x^2+5\right )}{2 \sqrt{x^4+3 x^2+2}}-\frac{\sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{x^4+3 x^2+2}}+\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x^2 + x^4)^(-3/2),x]

[Out]

(-3*x*(2 + x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (x*(5 + 3*x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (3*(1 + x^2)*Sqrt[(2
+ x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) - (Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^
2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/Sqrt[2 + 3*x^2 + x^4]

Rule 1092

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(b^2 - 2*a*c + b*c*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[(b^2 - 2*a*c + 2*(p + 1)
*(b^2 - 4*a*c) + b*c*(4*p + 7)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1189

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[d, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[e, Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b +
q)/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1099

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[((2*a + (b +
q)*x^2)*Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)]
)/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1135

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b +
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)])/(2*c*Sqrt[a + b*x^2
+ c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx &=\frac{x \left (5+3 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}-\frac{1}{2} \int \frac{4+3 x^2}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{x \left (5+3 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}-\frac{3}{2} \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx-2 \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=-\frac{3 x \left (2+x^2\right )}{2 \sqrt{2+3 x^2+x^4}}+\frac{x \left (5+3 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}+\frac{3 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{\sqrt{2} \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0399735, size = 99, normalized size = 0.66 \[ \frac{i \sqrt{x^2+1} \sqrt{x^2+2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right ),2\right )+3 x^3+3 i \sqrt{x^2+1} \sqrt{x^2+2} E\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+5 x}{2 \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x^2 + x^4)^(-3/2),x]

[Out]

(5*x + 3*x^3 + (3*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] + I*Sqrt[1 + x^2]*Sqrt[2 +
x^2]*EllipticF[I*ArcSinh[x/Sqrt[2]], 2])/(2*Sqrt[2 + 3*x^2 + x^4])

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Maple [C]  time = 0.003, size = 129, normalized size = 0.9 \begin{align*} -2\,{\frac{-3/4\,{x}^{3}-5/4\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}+{i\sqrt{2}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}}-{{\frac{3\,i}{4}}\sqrt{2} \left ({\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) -{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-3/4*x^3-5/4*x)/(x^4+3*x^2+2)^(1/2)+I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticF(
1/2*I*x*2^(1/2),2^(1/2))-3/4*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*(EllipticF(1/2*I*x*2^
(1/2),2^(1/2))-EllipticE(1/2*I*x*2^(1/2),2^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + 3*x^2 + 2)^(-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} + 3 \, x^{2} + 2}}{x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 3*x^2 + 2)/(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**4 + 3*x**2 + 2)**(-3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + 3*x^2 + 2)^(-3/2), x)