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

Optimal. Leaf size=156 \[ \frac{\left (b x^2+1\right ) \sqrt{\frac{b^2 x^4+1}{\left (b x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt{b} x\right ),\frac{1}{2}\right )}{\sqrt{b} \sqrt{-b^2 x^4-1}}-\frac{x \sqrt{-b^2 x^4-1}}{b x^2+1}-\frac{\left (b x^2+1\right ) \sqrt{\frac{b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt{b} x\right )|\frac{1}{2}\right )}{\sqrt{b} \sqrt{-b^2 x^4-1}} \]

[Out]

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

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Rubi [A]  time = 0.03229, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1198, 220, 1196} \[ -\frac{x \sqrt{-b^2 x^4-1}}{b x^2+1}+\frac{\left (b x^2+1\right ) \sqrt{\frac{b^2 x^4+1}{\left (b x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt{b} x\right )|\frac{1}{2}\right )}{\sqrt{b} \sqrt{-b^2 x^4-1}}-\frac{\left (b x^2+1\right ) \sqrt{\frac{b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt{b} x\right )|\frac{1}{2}\right )}{\sqrt{b} \sqrt{-b^2 x^4-1}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + b*x^2)/Sqrt[-1 - b^2*x^4],x]

[Out]

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

Rule 1198

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

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

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

Mathematica [C]  time = 0.0186291, size = 76, normalized size = 0.49 \[ \frac{\sqrt{b^2 x^4+1} \left (b x^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-b^2 x^4\right )+3 x \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-b^2 x^4\right )\right )}{3 \sqrt{-b^2 x^4-1}} \]

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 [C]  time = 0.046, size = 122, normalized size = 0.8 \begin{align*}{-i\sqrt{1+ib{x}^{2}}\sqrt{1-ib{x}^{2}} \left ({\it EllipticF} \left ( x\sqrt{-ib},i \right ) -{\it EllipticE} \left ( x\sqrt{-ib},i \right ) \right ){\frac{1}{\sqrt{-ib}}}{\frac{1}{\sqrt{-{b}^{2}{x}^{4}-1}}}}+{\sqrt{1+ib{x}^{2}}\sqrt{1-ib{x}^{2}}{\it EllipticF} \left ( x\sqrt{-ib},i \right ){\frac{1}{\sqrt{-ib}}}{\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]

-I/(-I*b)^(1/2)*(1+I*b*x^2)^(1/2)*(1-I*b*x^2)^(1/2)/(-b^2*x^4-1)^(1/2)*(EllipticF(x*(-I*b)^(1/2),I)-EllipticE(
x*(-I*b)^(1/2),I))+1/(-I*b)^(1/2)*(1+I*b*x^2)^(1/2)*(1-I*b*x^2)^(1/2)/(-b^2*x^4-1)^(1/2)*EllipticF(x*(-I*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*} \frac{b x{\rm integral}\left (-\frac{\sqrt{-b^{2} x^{4} - 1}{\left (b x^{2} + 1\right )}}{b^{3} x^{6} + b x^{2}}, x\right ) - \sqrt{-b^{2} x^{4} - 1}}{b 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="fricas")

[Out]

(b*x*integral(-sqrt(-b^2*x^4 - 1)*(b*x^2 + 1)/(b^3*x^6 + b*x^2), x) - sqrt(-b^2*x^4 - 1))/(b*x)

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Sympy [C]  time = 1.51971, size = 71, normalized size = 0.46 \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} e^{i \pi }} \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} e^{i \pi }} \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*exp_polar(I*pi))/(4*gamma(7/4)) - I*x*gamma(1/4)*hype
r((1/4, 1/2), (5/4,), b**2*x**4*exp_polar(I*pi))/(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)