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

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*(b^2*x^4+1)^(1/2)/(b*x^2+1)-(b*x^2+1)*(cos(2*arctan(x*b^(1/2)))^2)^(1/2)/cos(2*arctan(x*b^(1/2)))*EllipticE(
sin(2*arctan(x*b^(1/2))),1/2*2^(1/2))*((b^2*x^4+1)/(b*x^2+1)^2)^(1/2)/b^(1/2)/(b^2*x^4+1)^(1/2)+(b*x^2+1)*(cos
(2*arctan(x*b^(1/2)))^2)^(1/2)/cos(2*arctan(x*b^(1/2)))*EllipticF(sin(2*arctan(x*b^(1/2))),1/2*2^(1/2))*((b^2*
x^4+1)/(b*x^2+1)^2)^(1/2)/b^(1/2)/(b^2*x^4+1)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 152, normalized size of antiderivative = 1.00, 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 = {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 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]

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]

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*}

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Mathematica [C]  time = 0.01, size = 47, normalized size = 0.31 \[ x \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-b^2 x^4\right )+\frac {1}{3} b x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-b^2 x^4\right ) \]

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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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{2} + 1}{\sqrt {b^{2} x^{4} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+1)/(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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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 [C]  time = 0.00, size = 120, normalized size = 0.79 \[ \frac {\sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \EllipticF \left (\sqrt {i b}\, x , i\right )}{\sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}+\frac {i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (-\EllipticE \left (\sqrt {i b}\, x , i\right )+\EllipticF \left (\sqrt {i b}\, x , i\right )\right )}{\sqrt {i b}\, \sqrt {b^{2} x^{4}+1}} \]

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]

I/(I*b)^(1/2)*(-I*b*x^2+1)^(1/2)*(I*b*x^2+1)^(1/2)/(b^2*x^4+1)^(1/2)*(EllipticF((I*b)^(1/2)*x,I)-EllipticE((I*
b)^(1/2)*x,I))+1/(I*b)^(1/2)*(-I*b*x^2+1)^(1/2)*(I*b*x^2+1)^(1/2)/(b^2*x^4+1)^(1/2)*EllipticF((I*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 [C]  time = 2.60, 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*gamma(5/4))

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