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

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

[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])

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Rubi [A]  time = 0.0135721, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {1196} \[ \frac{\left (b x^2+1\right ) \sqrt{\frac{b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt{b} x\right )|\frac{1}{2}\right )}{\sqrt{b} \sqrt{b^2 x^4+1}}-\frac{x \sqrt{b^2 x^4+1}}{b x^2+1} \]

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])

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 &=-\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}}\\ \end{align*}

Mathematica [C]  time = 0.0116996, size = 47, normalized size = 0.53 \[ x \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-b^2 x^4\right )-\frac{1}{3} b x^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-b^2 x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.153, size = 120, normalized size = 1.4 \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*}{\rm integral}\left (-\frac{b x^{2} - 1}{\sqrt{b^{2} x^{4} + 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(-(b*x^2 - 1)/sqrt(b^2*x^4 + 1), x)

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Sympy [C]  time = 1.83919, size = 66, normalized size = 0.74 \begin{align*} - \frac{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{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]

-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)) + x*gamma(1/4)*hyper((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)