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

Optimal. Leaf size=152 \[ \frac{x \sqrt{b^2 x^4+1}}{b x^2+1}+\frac{\left (b x^2+1\right ) \sqrt{\frac{b^2 x^4+1}{\left (b x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt{b} x\right )|\frac{1}{2}\right )}{\sqrt{b} \sqrt{b^2 x^4+1}}-\frac{\left (b x^2+1\right ) \sqrt{\frac{b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt{b} x\right )|\frac{1}{2}\right )}{\sqrt{b} \sqrt{b^2 x^4+1}} \]

[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 = 0.0995589, antiderivative size = 152, 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 \[ \frac{x \sqrt{b^2 x^4+1}}{b x^2+1}+\frac{\left (b x^2+1\right ) \sqrt{\frac{b^2 x^4+1}{\left (b x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt{b} x\right )|\frac{1}{2}\right )}{\sqrt{b} \sqrt{b^2 x^4+1}}-\frac{\left (b x^2+1\right ) \sqrt{\frac{b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt{b} x\right )|\frac{1}{2}\right )}{\sqrt{b} \sqrt{b^2 x^4+1}} \]

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])

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Rubi in Sympy [A]  time = 13.2283, size = 134, normalized size = 0.88 \[ \frac{x \sqrt{b^{2} x^{4} + 1}}{b x^{2} + 1} - \frac{\sqrt{\frac{b^{2} x^{4} + 1}{\left (b x^{2} + 1\right )^{2}}} \left (b x^{2} + 1\right ) E\left (2 \operatorname{atan}{\left (\sqrt{b} x \right )}\middle | \frac{1}{2}\right )}{\sqrt{b} \sqrt{b^{2} x^{4} + 1}} + \frac{\sqrt{\frac{b^{2} x^{4} + 1}{\left (b x^{2} + 1\right )^{2}}} \left (b x^{2} + 1\right ) F\left (2 \operatorname{atan}{\left (\sqrt{b} x \right )}\middle | \frac{1}{2}\right )}{\sqrt{b} \sqrt{b^{2} x^{4} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x**2+1)/(b**2*x**4+1)**(1/2),x)

[Out]

x*sqrt(b**2*x**4 + 1)/(b*x**2 + 1) - sqrt((b**2*x**4 + 1)/(b*x**2 + 1)**2)*(b*x*
*2 + 1)*elliptic_e(2*atan(sqrt(b)*x), 1/2)/(sqrt(b)*sqrt(b**2*x**4 + 1)) + sqrt(
(b**2*x**4 + 1)/(b*x**2 + 1)**2)*(b*x**2 + 1)*elliptic_f(2*atan(sqrt(b)*x), 1/2)
/(sqrt(b)*sqrt(b**2*x**4 + 1))

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Mathematica [C]  time = 0.0512056, size = 51, normalized size = 0.34 \[ \frac{E\left (\left .i \sinh ^{-1}\left (\sqrt{i b} x\right )\right |-1\right )-(1+i) F\left (\left .i \sinh ^{-1}\left (\sqrt{i b} x\right )\right |-1\right )}{\sqrt{i b}} \]

Antiderivative was successfully verified.

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

[Out]

(EllipticE[I*ArcSinh[Sqrt[I*b]*x], -1] - (1 + I)*EllipticF[I*ArcSinh[Sqrt[I*b]*x
], -1])/Sqrt[I*b]

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Maple [C]  time = 0.006, size = 120, normalized size = 0.8 \[{1\sqrt{1-ib{x}^{2}}\sqrt{1+ib{x}^{2}}{\it EllipticF} \left ( x\sqrt{ib},i \right ){\frac{1}{\sqrt{ib}}}{\frac{1}{\sqrt{{b}^{2}{x}^{4}+1}}}}+{i\sqrt{1-ib{x}^{2}}\sqrt{1+ib{x}^{2}} \left ({\it EllipticF} \left ( x\sqrt{ib},i \right ) -{\it EllipticE} \left ( x\sqrt{ib},i \right ) \right ){\frac{1}{\sqrt{ib}}}{\frac{1}{\sqrt{{b}^{2}{x}^{4}+1}}}} \]

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/(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)+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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + 1)/sqrt(b^2*x^4 + 1),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. \[{\rm integral}\left (\frac{b x^{2} + 1}{\sqrt{b^{2} x^{4} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + 1)/sqrt(b^2*x^4 + 1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.39688, size = 66, normalized size = 0.43 \[ \frac{b x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{1}{2} \\ \frac{5}{4} \end{matrix}\middle |{b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} \]

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*gam
ma(5/4))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{b x^{2} + 1}{\sqrt{b^{2} x^{4} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + 1)/sqrt(b^2*x^4 + 1),x, algorithm="giac")

[Out]

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