3.993 \(\int \frac{x^2}{\sqrt{4-x^2} \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=87 \[ \frac{\sqrt{c+d x^2} E\left (\sin ^{-1}\left (\frac{x}{2}\right )|-\frac{4 d}{c}\right )}{d \sqrt{\frac{d x^2}{c}+1}}-\frac{c \sqrt{\frac{d x^2}{c}+1} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x}{2}\right ),-\frac{4 d}{c}\right )}{d \sqrt{c+d x^2}} \]

[Out]

(Sqrt[c + d*x^2]*EllipticE[ArcSin[x/2], (-4*d)/c])/(d*Sqrt[1 + (d*x^2)/c]) - (c*Sqrt[1 + (d*x^2)/c]*EllipticF[
ArcSin[x/2], (-4*d)/c])/(d*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.0697967, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {493, 426, 424, 421, 419} \[ \frac{\sqrt{c+d x^2} E\left (\sin ^{-1}\left (\frac{x}{2}\right )|-\frac{4 d}{c}\right )}{d \sqrt{\frac{d x^2}{c}+1}}-\frac{c \sqrt{\frac{d x^2}{c}+1} F\left (\sin ^{-1}\left (\frac{x}{2}\right )|-\frac{4 d}{c}\right )}{d \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(Sqrt[4 - x^2]*Sqrt[c + d*x^2]),x]

[Out]

(Sqrt[c + d*x^2]*EllipticE[ArcSin[x/2], (-4*d)/c])/(d*Sqrt[1 + (d*x^2)/c]) - (c*Sqrt[1 + (d*x^2)/c]*EllipticF[
ArcSin[x/2], (-4*d)/c])/(d*Sqrt[c + d*x^2])

Rule 493

Int[(x_)^(n_)/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[1/b, Int[Sqrt[a +
b*x^n]/Sqrt[c + d*x^n], x], x] - Dist[a/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b, c,
d}, x] && NeQ[b*c - a*d, 0] && (EqQ[n, 2] || EqQ[n, 4]) &&  !(EqQ[n, 2] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rubi steps

\begin{align*} \int \frac{x^2}{\sqrt{4-x^2} \sqrt{c+d x^2}} \, dx &=\frac{\int \frac{\sqrt{c+d x^2}}{\sqrt{4-x^2}} \, dx}{d}-\frac{c \int \frac{1}{\sqrt{4-x^2} \sqrt{c+d x^2}} \, dx}{d}\\ &=\frac{\sqrt{c+d x^2} \int \frac{\sqrt{1+\frac{d x^2}{c}}}{\sqrt{4-x^2}} \, dx}{d \sqrt{1+\frac{d x^2}{c}}}-\frac{\left (c \sqrt{1+\frac{d x^2}{c}}\right ) \int \frac{1}{\sqrt{4-x^2} \sqrt{1+\frac{d x^2}{c}}} \, dx}{d \sqrt{c+d x^2}}\\ &=\frac{\sqrt{c+d x^2} E\left (\sin ^{-1}\left (\frac{x}{2}\right )|-\frac{4 d}{c}\right )}{d \sqrt{1+\frac{d x^2}{c}}}-\frac{c \sqrt{1+\frac{d x^2}{c}} F\left (\sin ^{-1}\left (\frac{x}{2}\right )|-\frac{4 d}{c}\right )}{d \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0524682, size = 59, normalized size = 0.68 \[ \frac{c \sqrt{\frac{d x^2}{c}+1} \left (E\left (\sin ^{-1}\left (\frac{x}{2}\right )|-\frac{4 d}{c}\right )-\text{EllipticF}\left (\sin ^{-1}\left (\frac{x}{2}\right ),-\frac{4 d}{c}\right )\right )}{d \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(Sqrt[4 - x^2]*Sqrt[c + d*x^2]),x]

[Out]

(c*Sqrt[1 + (d*x^2)/c]*(EllipticE[ArcSin[x/2], (-4*d)/c] - EllipticF[ArcSin[x/2], (-4*d)/c]))/(d*Sqrt[c + d*x^
2])

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Maple [A]  time = 0.02, size = 59, normalized size = 0.7 \begin{align*}{\frac{c}{d} \left ( -{\it EllipticF} \left ({\frac{x}{2}},2\,\sqrt{-{\frac{d}{c}}} \right ) +{\it EllipticE} \left ({\frac{x}{2}},2\,\sqrt{-{\frac{d}{c}}} \right ) \right ) \sqrt{{\frac{d{x}^{2}+c}{c}}}{\frac{1}{\sqrt{d{x}^{2}+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-x^2+4)^(1/2)/(d*x^2+c)^(1/2),x)

[Out]

(-EllipticF(1/2*x,2*(-d/c)^(1/2))+EllipticE(1/2*x,2*(-d/c)^(1/2)))/(d*x^2+c)^(1/2)*c*((d*x^2+c)/c)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(sqrt(d*x^2 + c)*sqrt(-x^2 + 4)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(d*x^2 + c)*sqrt(-x^2 + 4)*x^2/(d*x^4 + (c - 4*d)*x^2 - 4*c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{- \left (x - 2\right ) \left (x + 2\right )} \sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(-x**2+4)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**2/(sqrt(-(x - 2)*(x + 2))*sqrt(c + d*x**2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(sqrt(d*x^2 + c)*sqrt(-x^2 + 4)), x)