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

Optimal. Leaf size=51 \[ \frac{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{f \sqrt{a+b \sin ^2(e+f x)}} \]

[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])

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Rubi [A]  time = 0.0328417, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3183, 3182} \[ \frac{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{f \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[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 3183

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]

Rule 3182

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+b \sin ^2(e+f x)}} \, dx &=\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{F\left (e+f x\left |-\frac{b}{a}\right .\right ) \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}{f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0846094, size = 60, normalized size = 1.18 \[ \frac{\sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{f \sqrt{2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[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)]])

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Maple [C]  time = 0.266, size = 60, normalized size = 1.2 \begin{align*}{\frac{1}{f}\sqrt{-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}-a-b}{a}}}{\it InverseJacobiAM} \left ( fx+e,{i\sqrt{b}{\frac{1}{\sqrt{a}}}} \right ){\frac{1}{\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sin \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{b \cos \left (f x + e\right )^{2} - a - b}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-b*cos(f*x + e)^2 + a + b)/(b*cos(f*x + e)^2 - a - b), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sin \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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