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

Optimal. Leaf size=89 \[ \frac{2 \sqrt{1-b^2 x^4} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{b} x\right ),-1\right )}{\sqrt{b} \sqrt{b^2 x^4-1}}-\frac{\sqrt{1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt{b} x\right )\right |-1\right )}{\sqrt{b} \sqrt{b^2 x^4-1}} \]

[Out]

-((Sqrt[1 - b^2*x^4]*EllipticE[ArcSin[Sqrt[b]*x], -1])/(Sqrt[b]*Sqrt[-1 + b^2*x^4])) + (2*Sqrt[1 - b^2*x^4]*El
lipticF[ArcSin[Sqrt[b]*x], -1])/(Sqrt[b]*Sqrt[-1 + b^2*x^4])

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Rubi [A]  time = 0.0455572, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {1200, 1199, 423, 424, 248, 221} \[ \frac{2 \sqrt{1-b^2 x^4} F\left (\left .\sin ^{-1}\left (\sqrt{b} x\right )\right |-1\right )}{\sqrt{b} \sqrt{b^2 x^4-1}}-\frac{\sqrt{1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt{b} x\right )\right |-1\right )}{\sqrt{b} \sqrt{b^2 x^4-1}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - b*x^2)/Sqrt[-1 + b^2*x^4],x]

[Out]

-((Sqrt[1 - b^2*x^4]*EllipticE[ArcSin[Sqrt[b]*x], -1])/(Sqrt[b]*Sqrt[-1 + b^2*x^4])) + (2*Sqrt[1 - b^2*x^4]*El
lipticF[ArcSin[Sqrt[b]*x], -1])/(Sqrt[b]*Sqrt[-1 + b^2*x^4])

Rule 1200

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

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

Mathematica [C]  time = 0.0209353, size = 74, normalized size = 0.83 \[ -\frac{\sqrt{1-b^2 x^4} \left (b x^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};b^2 x^4\right )-3 x \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};b^2 x^4\right )\right )}{3 \sqrt{b^2 x^4-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 108, normalized size = 1.2 \begin{align*} -{\sqrt{b{x}^{2}+1}\sqrt{-b{x}^{2}+1} \left ({\it EllipticF} \left ( x\sqrt{-b},i \right ) -{\it EllipticE} \left ( x\sqrt{-b},i \right ) \right ){\frac{1}{\sqrt{-b}}}{\frac{1}{\sqrt{{b}^{2}{x}^{4}-1}}}}+{\sqrt{b{x}^{2}+1}\sqrt{-b{x}^{2}+1}{\it EllipticF} \left ( x\sqrt{-b},i \right ){\frac{1}{\sqrt{-b}}}{\frac{1}{\sqrt{{b}^{2}{x}^{4}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.86707, size = 60, normalized size = 0.67 \begin{align*} \frac{i b 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 |{b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} - \frac{i 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 |{b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*b*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b**2*x**4)/(4*gamma(7/4)) - I*x*gamma(1/4)*hyper((1/4, 1/2), (5/
4,), b**2*x**4)/(4*gamma(5/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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