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

Optimal. Leaf size=110 \[ \frac{x \sqrt{b x^2+2}}{b \sqrt{d x^2+3}}-\frac{\sqrt{2} \sqrt{b x^2+2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right )|1-\frac{3 b}{2 d}\right )}{b \sqrt{d} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}} \]

[Out]

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

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Rubi [A]  time = 0.0451851, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {492, 411} \[ \frac{x \sqrt{b x^2+2}}{b \sqrt{d x^2+3}}-\frac{\sqrt{2} \sqrt{b x^2+2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right )|1-\frac{3 b}{2 d}\right )}{b \sqrt{d} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.0477002, size = 72, normalized size = 0.65 \[ -\frac{i \sqrt{3} \left (E\left (i \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{2}}\right )|\frac{2 d}{3 b}\right )-\text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{2}}\right ),\frac{2 d}{3 b}\right )\right )}{\sqrt{b} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[3]*(EllipticE[I*ArcSinh[(Sqrt[b]*x)/Sqrt[2]], (2*d)/(3*b)] - EllipticF[I*ArcSinh[(Sqrt[b]*x)/Sqrt[2
]], (2*d)/(3*b)]))/(Sqrt[b]*d)

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Maple [A]  time = 0.021, size = 70, normalized size = 0.6 \begin{align*}{\frac{\sqrt{2}}{b} \left ( -{\it EllipticF} \left ({\frac{x\sqrt{3}}{3}\sqrt{-d}},{\frac{\sqrt{2}\sqrt{3}}{2}\sqrt{{\frac{b}{d}}}} \right ) +{\it EllipticE} \left ({\frac{x\sqrt{3}}{3}\sqrt{-d}},{\frac{\sqrt{2}\sqrt{3}}{2}\sqrt{{\frac{b}{d}}}} \right ) \right ){\frac{1}{\sqrt{-d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(x^2/(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^{2}}{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(x^2/(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^2/(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{x^{2}}{\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(x**2/(b*x**2+2)**(1/2)/(d*x**2+3)**(1/2),x)

[Out]

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

[Out]

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