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

Optimal. Leaf size=182 \[ \frac{\sqrt{2} \sqrt{b x^2+2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right ),1-\frac{3 b}{2 d}\right )}{\sqrt{d} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}}+\frac{x \sqrt{b x^2+2}}{\sqrt{d x^2+3}}-\frac{\sqrt{2} \sqrt{b x^2+2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right )|1-\frac{3 b}{2 d}\right )}{\sqrt{d} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}} \]

[Out]

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

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Rubi [A]  time = 0.0752468, antiderivative size = 182, 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} \[ \frac{x \sqrt{b x^2+2}}{\sqrt{d x^2+3}}+\frac{\sqrt{2} \sqrt{b x^2+2} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right )|1-\frac{3 b}{2 d}\right )}{\sqrt{d} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}}-\frac{\sqrt{2} \sqrt{b x^2+2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right )|1-\frac{3 b}{2 d}\right )}{\sqrt{d} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0089538, size = 37, normalized size = 0.2 \[ \frac{\sqrt{2} E\left (\sin ^{-1}\left (\frac{\sqrt{-d} x}{\sqrt{3}}\right )|\frac{3 b}{2 d}\right )}{\sqrt{-d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0., size = 37, normalized size = 0.2 \begin{align*}{\sqrt{2}{\it EllipticE} \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((b*x^2+2)^(1/2)/(d*x^2+3)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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((b*x^2+2)^(1/2)/(d*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

integrate(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}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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((b*x**2+2)**(1/2)/(d*x**2+3)**(1/2),x)

[Out]

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

[Out]

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