3.114 \(\int \frac{1}{\sqrt{a+b x^2} \sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx\)

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

[Out]

(Sqrt[e]*Sqrt[c + d*x^2]*Sqrt[(a*(e + f*x^2))/(e*(a + b*x^2))]*EllipticF[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))])/(c*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 = 0.0799099, 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 = {552, 419} \[ \frac{\sqrt{e} \sqrt{c+d x^2} \sqrt{\frac{a \left (e+f x^2\right )}{e \left (a+b x^2\right )}} F\left (\sin ^{-1}\left (\frac{\sqrt{b e-a f} x}{\sqrt{e} \sqrt{b x^2+a}}\right )|\frac{(b c-a d) e}{c (b e-a f)}\right )}{c \sqrt{e+f x^2} \sqrt{b e-a f} \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^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))]*EllipticF[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))])/(c*Sqrt[b*e - a*f]*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*S
qrt[e + f*x^2])

Rule 552

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^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))])/(c*Sqrt[e + f*x^2]*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]),
Subst[Int[1/(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] /; Fr
eeQ[{a, b, c, d, e, f}, x]

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{1}{\sqrt{a+b x^2} \sqrt{c+d x^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{1}{\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 )}{c \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 )}} F\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 )}{c \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 = 0.0830293, size = 148, normalized size = 1. \[ \frac{\sqrt{e} \sqrt{c+d x^2} \sqrt{\frac{a \left (e+f x^2\right )}{e \left (a+b x^2\right )}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x \sqrt{b e-a f}}{\sqrt{e} \sqrt{a+b x^2}}\right ),\frac{e (b c-a d)}{c (b e-a f)}\right )}{c \sqrt{e+f x^2} \sqrt{b e-a f} \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^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))]*EllipticF[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))])/(c*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 = 0.067, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{d{x}^{2}+c}}}{\frac{1}{\sqrt{f{x}^{2}+e}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*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 d f x^{6} +{\left (b d e +{\left (b c + a d\right )} f\right )} x^{4} + a c e +{\left (a c f +{\left (b c + a d\right )} e\right )} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)/(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*d*f*x^6 + (b*d*e + (b*c + a*d)*f)*x^4 + a*c*e + (a
*c*f + (b*c + a*d)*e)*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(a + b*x**2)*sqrt(c + d*x**2)*sqrt(e + f*x**2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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