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

Optimal. Leaf size=148 \[ \frac{\sqrt{3 x^2+2} \text{EllipticF}\left (\tan ^{-1}(2 x),\frac{5}{8}\right )}{2 \sqrt{2} \sqrt{\frac{3 x^2+2}{4 x^2+1}} \sqrt{4 x^2+1}}+\frac{4 \sqrt{3 x^2+2} x}{3 \sqrt{4 x^2+1}}-\frac{2 \sqrt{2} \sqrt{3 x^2+2} E\left (\tan ^{-1}(2 x)|\frac{5}{8}\right )}{3 \sqrt{\frac{3 x^2+2}{4 x^2+1}} \sqrt{4 x^2+1}} \]

[Out]

(4*x*Sqrt[2 + 3*x^2])/(3*Sqrt[1 + 4*x^2]) - (2*Sqrt[2]*Sqrt[2 + 3*x^2]*EllipticE[ArcTan[2*x], 5/8])/(3*Sqrt[(2
+ 3*x^2)/(1 + 4*x^2)]*Sqrt[1 + 4*x^2]) + (Sqrt[2 + 3*x^2]*EllipticF[ArcTan[2*x], 5/8])/(2*Sqrt[2]*Sqrt[(2 + 3
*x^2)/(1 + 4*x^2)]*Sqrt[1 + 4*x^2])

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Rubi [A]  time = 0.0497495, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {422, 418, 492, 411} \[ \frac{4 \sqrt{3 x^2+2} x}{3 \sqrt{4 x^2+1}}+\frac{\sqrt{3 x^2+2} F\left (\tan ^{-1}(2 x)|\frac{5}{8}\right )}{2 \sqrt{2} \sqrt{\frac{3 x^2+2}{4 x^2+1}} \sqrt{4 x^2+1}}-\frac{2 \sqrt{2} \sqrt{3 x^2+2} E\left (\tan ^{-1}(2 x)|\frac{5}{8}\right )}{3 \sqrt{\frac{3 x^2+2}{4 x^2+1}} \sqrt{4 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + 4*x^2]/Sqrt[2 + 3*x^2],x]

[Out]

(4*x*Sqrt[2 + 3*x^2])/(3*Sqrt[1 + 4*x^2]) - (2*Sqrt[2]*Sqrt[2 + 3*x^2]*EllipticE[ArcTan[2*x], 5/8])/(3*Sqrt[(2
+ 3*x^2)/(1 + 4*x^2)]*Sqrt[1 + 4*x^2]) + (Sqrt[2 + 3*x^2]*EllipticF[ArcTan[2*x], 5/8])/(2*Sqrt[2]*Sqrt[(2 + 3
*x^2)/(1 + 4*x^2)]*Sqrt[1 + 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{1+4 x^2}}{\sqrt{2+3 x^2}} \, dx &=4 \int \frac{x^2}{\sqrt{2+3 x^2} \sqrt{1+4 x^2}} \, dx+\int \frac{1}{\sqrt{2+3 x^2} \sqrt{1+4 x^2}} \, dx\\ &=\frac{4 x \sqrt{2+3 x^2}}{3 \sqrt{1+4 x^2}}+\frac{\sqrt{2+3 x^2} F\left (\tan ^{-1}(2 x)|\frac{5}{8}\right )}{2 \sqrt{2} \sqrt{\frac{2+3 x^2}{1+4 x^2}} \sqrt{1+4 x^2}}-\frac{4}{3} \int \frac{\sqrt{2+3 x^2}}{\left (1+4 x^2\right )^{3/2}} \, dx\\ &=\frac{4 x \sqrt{2+3 x^2}}{3 \sqrt{1+4 x^2}}-\frac{2 \sqrt{2} \sqrt{2+3 x^2} E\left (\tan ^{-1}(2 x)|\frac{5}{8}\right )}{3 \sqrt{\frac{2+3 x^2}{1+4 x^2}} \sqrt{1+4 x^2}}+\frac{\sqrt{2+3 x^2} F\left (\tan ^{-1}(2 x)|\frac{5}{8}\right )}{2 \sqrt{2} \sqrt{\frac{2+3 x^2}{1+4 x^2}} \sqrt{1+4 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0045624, size = 27, normalized size = 0.18 \[ -\frac{i E\left (i \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )|\frac{8}{3}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + 4*x^2]/Sqrt[2 + 3*x^2],x]

[Out]

((-I)*EllipticE[I*ArcSinh[Sqrt[3/2]*x], 8/3])/Sqrt[3]

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Maple [C]  time = 0.023, size = 20, normalized size = 0.1 \begin{align*} -{\frac{i}{3}}{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{6},{\frac{2\,\sqrt{6}}{3}} \right ) \sqrt{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(4*x**2 + 1)/sqrt(3*x**2 + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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