3.196 \(\int \frac{\sqrt{4+x^2}}{\sqrt{2+3 x^2}} \, dx\)

Optimal. Leaf size=136 \[ \frac{2 \sqrt{2} \sqrt{3 x^2+2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{x}{2}\right ),-5\right )}{\sqrt{x^2+4} \sqrt{\frac{3 x^2+2}{x^2+4}}}+\frac{\sqrt{3 x^2+2} x}{3 \sqrt{x^2+4}}-\frac{\sqrt{2} \sqrt{3 x^2+2} E\left (\left .\tan ^{-1}\left (\frac{x}{2}\right )\right |-5\right )}{3 \sqrt{x^2+4} \sqrt{\frac{3 x^2+2}{x^2+4}}} \]

[Out]

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

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Rubi [A]  time = 0.0442971, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {422, 418, 492, 411} \[ \frac{\sqrt{3 x^2+2} x}{3 \sqrt{x^2+4}}+\frac{2 \sqrt{2} \sqrt{3 x^2+2} F\left (\left .\tan ^{-1}\left (\frac{x}{2}\right )\right |-5\right )}{\sqrt{x^2+4} \sqrt{\frac{3 x^2+2}{x^2+4}}}-\frac{\sqrt{2} \sqrt{3 x^2+2} E\left (\left .\tan ^{-1}\left (\frac{x}{2}\right )\right |-5\right )}{3 \sqrt{x^2+4} \sqrt{\frac{3 x^2+2}{x^2+4}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[4 + x^2]/Sqrt[2 + 3*x^2],x]

[Out]

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

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[a, Int[1/(Sqrt[a + b*x^2]*Sqrt[c +
d*x^2]), x], x] + Dist[b, Int[x^2/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[
d/c] && PosQ[b/a]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

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

Mathematica [C]  time = 0.0039403, size = 27, normalized size = 0.2 \[ -\frac{2 i E\left (i \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )|\frac{1}{6}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[4 + x^2]/Sqrt[2 + 3*x^2],x]

[Out]

((-2*I)*EllipticE[I*ArcSinh[Sqrt[3/2]*x], 1/6])/Sqrt[3]

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Maple [A]  time = 0.013, size = 26, normalized size = 0.2 \begin{align*} -{\frac{i}{3}} \left ( 5\,{\it EllipticF} \left ( i/2x,\sqrt{6} \right ) +{\it EllipticE} \left ({\frac{i}{2}}x,\sqrt{6} \right ) \right ) \sqrt{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I*(5*EllipticF(1/2*I*x,6^(1/2))+EllipticE(1/2*I*x,6^(1/2)))*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{2} + 4}}{\sqrt{3 \, x^{2} + 2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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