\(\int \frac {1+b x^2}{\sqrt {1-b^2 x^4}} \, dx\) [14]

Optimal result

Integrand size = 22, antiderivative size = 16 \[ \int \frac {1+b x^2}{\sqrt {1-b^2 x^4}} \, dx=\frac {E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}} \]

[Out]

EllipticE(x*b^(1/2),I)/b^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1213, 435} \[ \int \frac {1+b x^2}{\sqrt {1-b^2 x^4}} \, dx=\frac {E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}} \]

[In]

Int[(1 + b*x^2)/Sqrt[1 - b^2*x^4],x]

[Out]

EllipticE[ArcSin[Sqrt[b]*x], -1]/Sqrt[b]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 1213

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt {1+b x^2}}{\sqrt {1-b x^2}} \, dx \\ & = \frac {E\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.81 \[ \int \frac {1+b x^2}{\sqrt {1-b^2 x^4}} \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},b^2 x^4\right )+\frac {1}{3} b x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},b^2 x^4\right ) \]

[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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 1.89 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.25

method	result	size
meijerg	\(\frac {b \,x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};b^{2} x^{4}\right )}{3}+x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};b^{2} x^{4}\right )\)	\(36\)
default	\(\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, F\left (\sqrt {b}\, x , i\right )}{\sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}-\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (F\left (\sqrt {b}\, x , i\right )-E\left (\sqrt {b}\, x , i\right )\right )}{\sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\)	\(100\)
elliptic	\(\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, F\left (\sqrt {b}\, x , i\right )}{\sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}-\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (F\left (\sqrt {b}\, x , i\right )-E\left (\sqrt {b}\, x , i\right )\right )}{\sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\)	\(100\)

[In]

int((b*x^2+1)/(-b^2*x^4+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/3*b*x^3*hypergeom([1/2,3/4],[7/4],b^2*x^4)+x*hypergeom([1/4,1/2],[5/4],b^2*x^4)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (11) = 22\).

Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 4.44 \[ \int \frac {1+b x^2}{\sqrt {1-b^2 x^4}} \, dx=\frac {\frac {\sqrt {-b^{2}} {\left (b + 1\right )} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} - \frac {\sqrt {-b^{2}} x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} - \sqrt {-b^{2} x^{4} + 1} b}{b^{2} x} \]

[In]

integrate((b*x^2+1)/(-b^2*x^4+1)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (12) = 24\).

Time = 0.85 (sec) , antiderivative size = 70, normalized size of antiderivative = 4.38 \[ \int \frac {1+b x^2}{\sqrt {1-b^2 x^4}} \, dx=\frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

[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(2*I*pi))/(4*gamma(7/4)) + x*gamma(1/4)*hyper((
1/4, 1/2), (5/4,), b**2*x**4*exp_polar(2*I*pi))/(4*gamma(5/4))

Maxima [F]

\[ \int \frac {1+b x^2}{\sqrt {1-b^2 x^4}} \, dx=\int { \frac {b x^{2} + 1}{\sqrt {-b^{2} x^{4} + 1}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1+b x^2}{\sqrt {1-b^2 x^4}} \, dx=\int { \frac {b x^{2} + 1}{\sqrt {-b^{2} x^{4} + 1}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1+b x^2}{\sqrt {1-b^2 x^4}} \, dx=\int \frac {b\,x^2+1}{\sqrt {1-b^2\,x^4}} \,d x \]

[In]

int((b*x^2 + 1)/(1 - b^2*x^4)^(1/2),x)

[Out]

int((b*x^2 + 1)/(1 - b^2*x^4)^(1/2), x)