3.106 \(\int \frac{\sqrt{c+d x^2}}{(a+b x^2)^{3/2} \sqrt{e+f x^2}} \, dx\)

Optimal. Leaf size=148 \[ \frac{\sqrt{e} \sqrt{c+d x^2} \sqrt{\frac{a \left (e+f x^2\right )}{e \left (a+b x^2\right )}} E\left (\sin ^{-1}\left (\frac{\sqrt{b e-a f} x}{\sqrt{e} \sqrt{b x^2+a}}\right )|\frac{(b c-a d) e}{c (b e-a f)}\right )}{a \sqrt{e+f x^2} \sqrt{b e-a f} \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \]

[Out]

(Sqrt[e]*Sqrt[c + d*x^2]*Sqrt[(a*(e + f*x^2))/(e*(a + b*x^2))]*EllipticE[ArcSin[(Sqrt[b*e - a*f]*x)/(Sqrt[e]*S
qrt[a + b*x^2])], ((b*c - a*d)*e)/(c*(b*e - a*f))])/(a*Sqrt[b*e - a*f]*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*S
qrt[e + f*x^2])

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Rubi [A]  time = 0.0846888, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {554, 424} \[ \frac{\sqrt{e} \sqrt{c+d x^2} \sqrt{\frac{a \left (e+f x^2\right )}{e \left (a+b x^2\right )}} E\left (\sin ^{-1}\left (\frac{\sqrt{b e-a f} x}{\sqrt{e} \sqrt{b x^2+a}}\right )|\frac{(b c-a d) e}{c (b e-a f)}\right )}{a \sqrt{e+f x^2} \sqrt{b e-a f} \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e]*Sqrt[c + d*x^2]*Sqrt[(a*(e + f*x^2))/(e*(a + b*x^2))]*EllipticE[ArcSin[(Sqrt[b*e - a*f]*x)/(Sqrt[e]*S
qrt[a + b*x^2])], ((b*c - a*d)*e)/(c*(b*e - a*f))])/(a*Sqrt[b*e - a*f]*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*S
qrt[e + f*x^2])

Rule 554

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)^(3/2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[(Sqrt
[c + d*x^2]*Sqrt[(a*(e + f*x^2))/(e*(a + b*x^2))])/(a*Sqrt[e + f*x^2]*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]),
Subst[Int[Sqrt[1 - ((b*c - a*d)*x^2)/c]/Sqrt[1 - ((b*e - a*f)*x^2)/e], x], x, x/Sqrt[a + b*x^2]], x] /; FreeQ[
{a, b, c, d, e, f}, x]

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{c+d x^2}}{\left (a+b x^2\right )^{3/2} \sqrt{e+f x^2}} \, dx &=\frac{\left (\sqrt{c+d x^2} \sqrt{\frac{a \left (e+f x^2\right )}{e \left (a+b x^2\right )}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{(b c-a d) x^2}{c}}}{\sqrt{1-\frac{(b e-a f) x^2}{e}}} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{a \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} \sqrt{e+f x^2}}\\ &=\frac{\sqrt{e} \sqrt{c+d x^2} \sqrt{\frac{a \left (e+f x^2\right )}{e \left (a+b x^2\right )}} E\left (\sin ^{-1}\left (\frac{\sqrt{b e-a f} x}{\sqrt{e} \sqrt{a+b x^2}}\right )|\frac{(b c-a d) e}{c (b e-a f)}\right )}{a \sqrt{b e-a f} \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} \sqrt{e+f x^2}}\\ \end{align*}

Mathematica [A]  time = 0.081812, size = 148, normalized size = 1. \[ \frac{\sqrt{e} \sqrt{c+d x^2} \sqrt{\frac{a \left (e+f x^2\right )}{e \left (a+b x^2\right )}} E\left (\sin ^{-1}\left (\frac{\sqrt{b e-a f} x}{\sqrt{e} \sqrt{b x^2+a}}\right )|\frac{(b c-a d) e}{c (b e-a f)}\right )}{a \sqrt{e+f x^2} \sqrt{b e-a f} \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e]*Sqrt[c + d*x^2]*Sqrt[(a*(e + f*x^2))/(e*(a + b*x^2))]*EllipticE[ArcSin[(Sqrt[b*e - a*f]*x)/(Sqrt[e]*S
qrt[a + b*x^2])], ((b*c - a*d)*e)/(c*(b*e - a*f))])/(a*Sqrt[b*e - a*f]*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*S
qrt[e + f*x^2])

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Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{d{x}^{2}+c} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{f{x}^{2}+e}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^(1/2)/(b*x^2+a)^(3/2)/(f*x^2+e)^(1/2),x)

[Out]

int((d*x^2+c)^(1/2)/(b*x^2+a)^(3/2)/(f*x^2+e)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x^{2} + c}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}} \sqrt{f x^{2} + e}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*x^2 + c)/((b*x^2 + a)^(3/2)*sqrt(f*x^2 + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(f*x^2 + e)/(b^2*f*x^6 + (b^2*e + 2*a*b*f)*x^4 + a^2*e + (2*a*b*e
+ a^2*f)*x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c + d x^{2}}}{\left (a + b x^{2}\right )^{\frac{3}{2}} \sqrt{e + f x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(1/2)/(b*x**2+a)**(3/2)/(f*x**2+e)**(1/2),x)

[Out]

Integral(sqrt(c + d*x**2)/((a + b*x**2)**(3/2)*sqrt(e + f*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x^{2} + c}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}} \sqrt{f x^{2} + e}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*x^2 + c)/((b*x^2 + a)^(3/2)*sqrt(f*x^2 + e)), x)