3.25 \(\int \frac{1-c^2 x^2}{\sqrt{1-c^4 x^4}} \, dx\)

Optimal. Leaf size=23 \[ \frac{2 \text{EllipticF}\left (\sin ^{-1}(c x),-1\right )}{c}-\frac{E\left (\left .\sin ^{-1}(c x)\right |-1\right )}{c} \]

[Out]

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

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Rubi [A]  time = 0.0334165, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1199, 423, 424, 248, 221} \[ \frac{2 F\left (\left .\sin ^{-1}(c x)\right |-1\right )}{c}-\frac{E\left (\left .\sin ^{-1}(c x)\right |-1\right )}{c} \]

Antiderivative was successfully verified.

[In]

Int[(1 - c^2*x^2)/Sqrt[1 - c^4*x^4],x]

[Out]

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

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 423

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

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 248

Int[((a1_.) + (b1_.)*(x_)^(n_))^(p_.)*((a2_.) + (b2_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[(a1*a2 + b1*b2*x^(2*
n))^p, x] /; FreeQ[{a1, b1, a2, b2, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a
2, 0]))

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rubi steps

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

Mathematica [C]  time = 0.0117478, size = 47, normalized size = 2.04 \[ x \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};c^4 x^4\right )-\frac{1}{3} c^2 x^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};c^4 x^4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - c^2*x^2)/Sqrt[1 - c^4*x^4],x]

[Out]

x*Hypergeometric2F1[1/4, 1/2, 5/4, c^4*x^4] - (c^2*x^3*Hypergeometric2F1[1/2, 3/4, 7/4, c^4*x^4])/3

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Maple [B]  time = 0.048, size = 117, normalized size = 5.1 \begin{align*}{\sqrt{-{c}^{2}{x}^{2}+1}\sqrt{{c}^{2}{x}^{2}+1}{\it EllipticF} \left ( x\sqrt{{c}^{2}},i \right ){\frac{1}{\sqrt{{c}^{2}}}}{\frac{1}{\sqrt{-{c}^{4}{x}^{4}+1}}}}+{\sqrt{-{c}^{2}{x}^{2}+1}\sqrt{{c}^{2}{x}^{2}+1} \left ({\it EllipticF} \left ( x\sqrt{{c}^{2}},i \right ) -{\it EllipticE} \left ( x\sqrt{{c}^{2}},i \right ) \right ){\frac{1}{\sqrt{{c}^{2}}}}{\frac{1}{\sqrt{-{c}^{4}{x}^{4}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(c^2)^(1/2)*(-c^2*x^2+1)^(1/2)*(c^2*x^2+1)^(1/2)/(-c^4*x^4+1)^(1/2)*EllipticF(x*(c^2)^(1/2),I)+1/(c^2)^(1/2)
*(-c^2*x^2+1)^(1/2)*(c^2*x^2+1)^(1/2)/(-c^4*x^4+1)^(1/2)*(EllipticF(x*(c^2)^(1/2),I)-EllipticE(x*(c^2)^(1/2),I
))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-c^{4} x^{4} + 1}}{c^{2} x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-c^4*x^4 + 1)/(c^2*x^2 + 1), x)

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Sympy [B]  time = 1.93362, size = 71, normalized size = 3.09 \begin{align*} - \frac{c^{2} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{c^{4} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{1}{2} \\ \frac{5}{4} \end{matrix}\middle |{c^{4} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c**2*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), c**4*x**4*exp_polar(2*I*pi))/(4*gamma(7/4)) + x*gamma(1/4)*hyp
er((1/4, 1/2), (5/4,), c**4*x**4*exp_polar(2*I*pi))/(4*gamma(5/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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