\(\int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx\) [17]

Optimal result

Integrand size = 22, antiderivative size = 89 \[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=-\frac {\sqrt {1-b^2 x^4} E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}}+\frac {2 \sqrt {1-b^2 x^4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}} \]

[Out]

-EllipticE(x*b^(1/2),I)*(-b^2*x^4+1)^(1/2)/b^(1/2)/(b^2*x^4-1)^(1/2)+2*EllipticF(x*b^(1/2),I)*(-b^2*x^4+1)^(1/
2)/b^(1/2)/(b^2*x^4-1)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1214, 1213, 434, 435, 254, 227} \[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\frac {2 \sqrt {1-b^2 x^4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b} \sqrt {b^2 x^4-1}}-\frac {\sqrt {1-b^2 x^4} E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {b^2 x^4-1}} \]

[In]

Int[(1 - b*x^2)/Sqrt[-1 + b^2*x^4],x]

[Out]

-((Sqrt[1 - b^2*x^4]*EllipticE[ArcSin[Sqrt[b]*x], -1])/(Sqrt[b]*Sqrt[-1 + b^2*x^4])) + (2*Sqrt[1 - b^2*x^4]*El
lipticF[ArcSin[Sqrt[b]*x], -1])/(Sqrt[b]*Sqrt[-1 + b^2*x^4])

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 254

Int[((a1_.) + (b1_.)*(x_)^(n_))^(p_.)*((a2_.) + (b2_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[(a1*a2 + b1*b2*x^(2*
n))^p, x] /; FreeQ[{a1, b1, a2, b2, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a
2, 0]))

Rule 434

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[b/d, Int[Sqrt[c + d*x^2]/Sqrt[a + b
*x^2], x], x] - Dist[(b*c - a*d)/d, Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x]
&& PosQ[d/c] && NegQ[b/a]

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]

Rule 1214

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + c*(x^4/a)]/Sqrt[a + c*x^4], In
t[(d + e*x^2)/Sqrt[1 + c*(x^4/a)], 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}& = \frac {\sqrt {1-b^2 x^4} \int \frac {1-b x^2}{\sqrt {1-b^2 x^4}} \, dx}{\sqrt {-1+b^2 x^4}} \\ & = \frac {\sqrt {1-b^2 x^4} \int \frac {\sqrt {1-b x^2}}{\sqrt {1+b x^2}} \, dx}{\sqrt {-1+b^2 x^4}} \\ & = -\frac {\sqrt {1-b^2 x^4} \int \frac {\sqrt {1+b x^2}}{\sqrt {1-b x^2}} \, dx}{\sqrt {-1+b^2 x^4}}+\frac {\left (2 \sqrt {1-b^2 x^4}\right ) \int \frac {1}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx}{\sqrt {-1+b^2 x^4}} \\ & = -\frac {\sqrt {1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}}+\frac {\left (2 \sqrt {1-b^2 x^4}\right ) \int \frac {1}{\sqrt {1-b^2 x^4}} \, dx}{\sqrt {-1+b^2 x^4}} \\ & = -\frac {\sqrt {1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}}+\frac {2 \sqrt {1-b^2 x^4} F\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\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 = 10.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=-\frac {\sqrt {1-b^2 x^4} \left (-3 x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},b^2 x^4\right )+b x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},b^2 x^4\right )\right )}{3 \sqrt {-1+b^2 x^4}} \]

[In]

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

[Out]

-1/3*(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]))/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 = 1.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99

method	result	size
meijerg	\(-\frac {b \sqrt {-\operatorname {signum}\left (b^{2} x^{4}-1\right )}\, x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};b^{2} x^{4}\right )}{3 \sqrt {\operatorname {signum}\left (b^{2} x^{4}-1\right )}}+\frac {\sqrt {-\operatorname {signum}\left (b^{2} x^{4}-1\right )}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};b^{2} x^{4}\right )}{\sqrt {\operatorname {signum}\left (b^{2} x^{4}-1\right )}}\)	\(88\)
default	\(\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, F\left (x \sqrt {-b}, i\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (F\left (x \sqrt {-b}, i\right )-E\left (x \sqrt {-b}, i\right )\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\)	\(108\)
elliptic	\(\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, F\left (x \sqrt {-b}, i\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (F\left (x \sqrt {-b}, i\right )-E\left (x \sqrt {-b}, i\right )\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\)	\(108\)

[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 = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=-\frac {\frac {{\left (b - 1\right )} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} + \frac {x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} + \sqrt {b^{2} x^{4} - 1}}{b x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.67 \[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

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