\(\int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx\) [21]

Optimal result

Integrand size = 22, antiderivative size = 156 \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=-\frac {x \sqrt {-1-b^2 x^4}}{1+b x^2}-\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {-1-b^2 x^4}}+\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-1-b^2 x^4}} \]

[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)))*Elliptic
E(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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1212, 226, 1210} \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}-\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}-\frac {x \sqrt {-b^2 x^4-1}}{b x^2+1} \]

[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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

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*} \text {integral}& = 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*}

Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.49 \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=\frac {\sqrt {1+b^2 x^4} \left (3 x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-b^2 x^4\right )+b x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-b^2 x^4\right )\right )}{3 \sqrt {-1-b^2 x^4}} \]

[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 [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.80 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.58

method	result	size
meijerg	\(\frac {b \sqrt {\operatorname {signum}\left (b^{2} x^{4}+1\right )}\, x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-b^{2} x^{4}\right )}{3 \sqrt {-\operatorname {signum}\left (b^{2} x^{4}+1\right )}}+\frac {\sqrt {\operatorname {signum}\left (b^{2} x^{4}+1\right )}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-b^{2} x^{4}\right )}{\sqrt {-\operatorname {signum}\left (b^{2} x^{4}+1\right )}}\)	\(90\)
default	\(\frac {\sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, F\left (x \sqrt {-i b}, i\right )}{\sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}-\frac {i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (F\left (x \sqrt {-i b}, i\right )-E\left (x \sqrt {-i b}, i\right )\right )}{\sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\)	\(122\)
elliptic	\(\frac {\sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, F\left (x \sqrt {-i b}, i\right )}{\sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}-\frac {i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (F\left (x \sqrt {-i b}, i\right )-E\left (x \sqrt {-i b}, i\right )\right )}{\sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\)	\(122\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.54 \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=-\frac {\sqrt {-b^{2}} {\left (b - 1\right )} x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + \sqrt {-b^{2}} x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + \sqrt {-b^{2} x^{4} - 1}}{b x} \]

[In]

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

[Out]

-(sqrt(-b^2)*(b - 1)*x*(-1/b^2)^(3/4)*elliptic_f(arcsin((-1/b^2)^(1/4)/x), -1) + sqrt(-b^2)*x*(-1/b^2)^(3/4)*e
lliptic_e(arcsin((-1/b^2)^(1/4)/x), -1) + sqrt(-b^2*x^4 - 1))/(b*x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.85 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.46 \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=- \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

[In]

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

[Out]

-I*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)) - I*x*gamma(1/4)*hype
r((1/4, 1/2), (5/4,), b**2*x**4*exp_polar(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. \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=\int \frac {b\,x^2+1}{\sqrt {-b^2\,x^4-1}} \,d x \]

[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)