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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.02, 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]

-1/3*(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]))/Sqrt[-1 + b^2*x^4]

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b^{2} x^{4} - 1}}{b x^{2} + 1}, x\right ) \]

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

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)

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maple [A]  time = 0.01, size = 108, normalized size = 1.21 \[ \frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \EllipticF \left (\sqrt {-b}\, x , i\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (-\EllipticE \left (\sqrt {-b}\, x , i\right )+\EllipticF \left (\sqrt {-b}\, x , i\right )\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}} \]

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

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {b\,x^2-1}{\sqrt {b^2\,x^4-1}} \,d x \]

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]

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

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sympy [A]  time = 2.19, size = 60, normalized size = 0.67 \[ \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

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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