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

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{4} x^{4} + 1}}{c^{2} x^{2} + 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {c^{2} x^{2} - 1}{\sqrt {-c^{4} x^{4} + 1}}\,{d x} \]

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)

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maple [B]  time = 0.01, size = 117, normalized size = 5.09 \[ \frac {\sqrt {-c^{2} x^{2}+1}\, \sqrt {c^{2} x^{2}+1}\, \EllipticF \left (\sqrt {c^{2}}\, x , i\right )}{\sqrt {c^{2}}\, \sqrt {-c^{4} x^{4}+1}}+\frac {\sqrt {-c^{2} x^{2}+1}\, \sqrt {c^{2} x^{2}+1}\, \left (-\EllipticE \left (\sqrt {c^{2}}\, x , i\right )+\EllipticF \left (\sqrt {c^{2}}\, x , i\right )\right )}{\sqrt {c^{2}}\, \sqrt {-c^{4} x^{4}+1}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {c^{2} x^{2} - 1}{\sqrt {-c^{4} x^{4} + 1}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ -\int \frac {c^2\,x^2-1}{\sqrt {1-c^4\,x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 2.03, size = 71, normalized size = 3.09 \[ - \frac {c^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {c^{4} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {c^{4} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

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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