3.290 \(\int \frac{1}{\sqrt{2+b x^2} \sqrt{3+d x^2}} \, dx\)

Optimal. Leaf size=78 \[ \frac{\sqrt{b x^2+2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right ),1-\frac{3 b}{2 d}\right )}{\sqrt{2} \sqrt{d} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}} \]

[Out]

(Sqrt[2 + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[3]], 1 - (3*b)/(2*d)])/(Sqrt[2]*Sqrt[d]*Sqrt[(2 + b*x^2)/(3
+ d*x^2)]*Sqrt[3 + d*x^2])

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Rubi [A]  time = 0.0137813, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {418} \[ \frac{\sqrt{b x^2+2} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right )|1-\frac{3 b}{2 d}\right )}{\sqrt{2} \sqrt{d} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[2 + b*x^2]*Sqrt[3 + d*x^2]),x]

[Out]

(Sqrt[2 + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[3]], 1 - (3*b)/(2*d)])/(Sqrt[2]*Sqrt[d]*Sqrt[(2 + b*x^2)/(3
+ d*x^2)]*Sqrt[3 + d*x^2])

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]

Rubi steps

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

Mathematica [A]  time = 0.0119633, size = 37, normalized size = 0.47 \[ \frac{\text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{-b} x}{\sqrt{2}}\right ),\frac{2 d}{3 b}\right )}{\sqrt{3} \sqrt{-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[2 + b*x^2]*Sqrt[3 + d*x^2]),x]

[Out]

EllipticF[ArcSin[(Sqrt[-b]*x)/Sqrt[2]], (2*d)/(3*b)]/(Sqrt[3]*Sqrt[-b])

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Maple [A]  time = 0.023, size = 38, normalized size = 0.5 \begin{align*}{\frac{\sqrt{2}}{2}{\it EllipticF} \left ({\frac{x\sqrt{3}}{3}\sqrt{-d}},{\frac{\sqrt{2}\sqrt{3}}{2}\sqrt{{\frac{b}{d}}}} \right ){\frac{1}{\sqrt{-d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*2^(1/2)*EllipticF(1/3*x*3^(1/2)*(-d)^(1/2),1/2*2^(1/2)*3^(1/2)*(1/d*b)^(1/2))/(-d)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + 2} \sqrt{d x^{2} + 3}}{b d x^{4} +{\left (3 \, b + 2 \, d\right )} x^{2} + 6}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^2 + 2)*sqrt(d*x^2 + 3)/(b*d*x^4 + (3*b + 2*d)*x^2 + 6), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+2)**(1/2)/(d*x**2+3)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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