3.978 \(\int \frac{1}{\sqrt{a+b x^2-c x^4}} \, dx\)

Optimal. Leaf size=169 \[ \frac{\sqrt{\sqrt{4 a c+b^2}+b} \sqrt{1-\frac{2 c x^2}{b-\sqrt{4 a c+b^2}}} \sqrt{1-\frac{2 c x^2}{\sqrt{4 a c+b^2}+b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{4 a c+b^2}+b}}\right ),\frac{\sqrt{4 a c+b^2}+b}{b-\sqrt{4 a c+b^2}}\right )}{\sqrt{2} \sqrt{c} \sqrt{a+b x^2-c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0663954, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {1104, 419} \[ \frac{\sqrt{\sqrt{4 a c+b^2}+b} \sqrt{1-\frac{2 c x^2}{b-\sqrt{4 a c+b^2}}} \sqrt{1-\frac{2 c x^2}{\sqrt{4 a c+b^2}+b}} F\left (\sin ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2+4 a c}}}\right )|\frac{b+\sqrt{b^2+4 a c}}{b-\sqrt{b^2+4 a c}}\right )}{\sqrt{2} \sqrt{c} \sqrt{a+b x^2-c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*x^2 - c*x^4],x]

[Out]

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

Rule 1104

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(Sqrt[1 + (2*
c*x^2)/(b - q)]*Sqrt[1 + (2*c*x^2)/(b + q)])/Sqrt[a + b*x^2 + c*x^4], Int[1/(Sqrt[1 + (2*c*x^2)/(b - q)]*Sqrt[
1 + (2*c*x^2)/(b + q)]), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[c/a]

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

Mathematica [C]  time = 0.0811004, size = 177, normalized size = 1.05 \[ -\frac{i \sqrt{\frac{2 c x^2}{\sqrt{4 a c+b^2}-b}+1} \sqrt{1-\frac{2 c x^2}{\sqrt{4 a c+b^2}+b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{-\frac{c}{\sqrt{4 a c+b^2}+b}}\right ),-\frac{\sqrt{4 a c+b^2}+b}{\sqrt{4 a c+b^2}-b}\right )}{\sqrt{2} \sqrt{-\frac{c}{\sqrt{4 a c+b^2}+b}} \sqrt{a+b x^2-c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*x^2 - c*x^4],x]

[Out]

((-I)*Sqrt[1 + (2*c*x^2)/(-b + Sqrt[b^2 + 4*a*c])]*Sqrt[1 - (2*c*x^2)/(b + Sqrt[b^2 + 4*a*c])]*EllipticF[I*Arc
Sinh[Sqrt[2]*Sqrt[-(c/(b + Sqrt[b^2 + 4*a*c]))]*x], -((b + Sqrt[b^2 + 4*a*c])/(-b + Sqrt[b^2 + 4*a*c]))])/(Sqr
t[2]*Sqrt[-(c/(b + Sqrt[b^2 + 4*a*c]))]*Sqrt[a + b*x^2 - c*x^4])

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Maple [A]  time = 0.21, size = 145, normalized size = 0.9 \begin{align*}{\frac{\sqrt{2}}{4}\sqrt{4-2\,{\frac{ \left ( -b+\sqrt{4\,ac+{b}^{2}} \right ){x}^{2}}{a}}}\sqrt{4+2\,{\frac{ \left ( b+\sqrt{4\,ac+{b}^{2}} \right ){x}^{2}}{a}}}{\it EllipticF} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{4\,ac+{b}^{2}} \right ) }}},{\frac{1}{2}\sqrt{-4-2\,{\frac{b \left ( b+\sqrt{4\,ac+{b}^{2}} \right ) }{ac}}}} \right ){\frac{1}{\sqrt{{\frac{1}{a} \left ( -b+\sqrt{4\,ac+{b}^{2}} \right ) }}}}{\frac{1}{\sqrt{-c{x}^{4}+b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*2^(1/2)/((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*(4-2*(-b+(4*a*c+b^2)^(1/2))/a*x^2)^(1/2)*(4+2*(b+(4*a*c+b^2)^(1/2
))/a*x^2)^(1/2)/(-c*x^4+b*x^2+a)^(1/2)*EllipticF(1/2*x*2^(1/2)*((-b+(4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4-2*b*(b
+(4*a*c+b^2)^(1/2))/a/c)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-c x^{4} + b x^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(-c*x^4 + b*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b x^{2} - c x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a + b*x**2 - c*x**4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-c x^{4} + b x^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(-c*x^4 + b*x^2 + a), x)