\(\int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [150]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 16, antiderivative size = 51 \[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^(1/2)/f/(a+b*sin(f*x+e
)^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3262, 3261} \[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )}{f \sqrt {a+b \sin ^2(e+f x)}} \]

[In]

Int[1/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(EllipticF[e + f*x, -(b/a)]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])/(f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 3261

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(Sqrt[a]*f))*EllipticF[e + f*x, -b/a]
, x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3262

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[1 + b*(Sin[e + f*x]^2/a)]/Sqrt[a +
b*Sin[e + f*x]^2], Int[1/Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+\frac {b \sin ^2(e+f x)}{a}} \int \frac {1}{\sqrt {1+\frac {b \sin ^2(e+f x)}{a}}} \, dx}{\sqrt {a+b \sin ^2(e+f x)}} \\ & = \frac {\operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )}{f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]

[In]

Integrate[1/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)])/(f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

Maple [C] (verified)

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

Time = 0.44 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02

method result size
default \(\frac {\sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \operatorname {am}^{-1}\left (f x +e \bigg | \frac {i \sqrt {b}}{\sqrt {a}}\right )}{f \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}\) \(52\)

[In]

int(1/(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/f/(a+b*sin(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*InverseJacobiAM(f*x+e,I/a^(1/2)*b^(1/2))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 305, normalized size of antiderivative = 5.98 \[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {{\left (2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} + {\left (-2 i \, a - i \, b\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (-2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} + {\left (2 i \, a + i \, b\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}})}{b^{2} f} \]

[In]

integrate(1/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-((2*I*sqrt(-b)*b*sqrt((a^2 + a*b)/b^2) + (-2*I*a - I*b)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/
b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 +
8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (-2*I*sqrt(-b)*b*sqrt((a^2 + a*b)/b^2) + (2*I*a +
I*b)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2)
 + 2*a + b)/b)*(cos(f*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))
/b^2))/(b^2*f)

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {1}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]

[In]

integrate(1/(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(1/sqrt(a + b*sin(e + f*x)**2), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]

[In]

integrate(1/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(b*sin(f*x + e)^2 + a), x)

Giac [F]

\[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]

[In]

integrate(1/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {1}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]

[In]

int(1/(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(1/(a + b*sin(e + f*x)^2)^(1/2), x)