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

Optimal. Leaf size=43 \[ \frac{\sqrt{1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt{b} x\right )\right |-1\right )}{\sqrt{b} \sqrt{b^2 x^4-1}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0254657, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {1200, 1199, 424} \[ \frac{\sqrt{1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt{b} x\right )\right |-1\right )}{\sqrt{b} \sqrt{b^2 x^4-1}} \]

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])

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 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]

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} E\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 = 0.0225834, size = 74, normalized size = 1.72 \[ \frac{\sqrt{1-b^2 x^4} \left (b x^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};b^2 x^4\right )+3 x \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};b^2 x^4\right )\right )}{3 \sqrt{b^2 x^4-1}} \]

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])

________________________________________________________________________________________

Maple [B]  time = 0.054, size = 107, normalized size = 2.5 \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)

________________________________________________________________________________________

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)

________________________________________________________________________________________

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)

________________________________________________________________________________________

Sympy [A]  time = 1.50467, size = 61, normalized size = 1.42 \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))

________________________________________________________________________________________

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)