3.320 \(\int \frac{\sqrt{c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=29 \[ -\frac{2 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a f} \]

[Out]

(-2*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a*f)

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Rubi [A]  time = 0.12751, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {2736, 2673} \[ -\frac{2 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - c*Sin[e + f*x]]/(a + a*Sin[e + f*x]),x]

[Out]

(-2*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a*f)

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps

\begin{align*} \int \frac{\sqrt{c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx &=\frac{\int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{a c}\\ &=-\frac{2 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a f}\\ \end{align*}

Mathematica [A]  time = 0.0964544, size = 29, normalized size = 1. \[ -\frac{2 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c - c*Sin[e + f*x]]/(a + a*Sin[e + f*x]),x]

[Out]

(-2*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a*f)

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Maple [A]  time = 0.408, size = 39, normalized size = 1.3 \begin{align*} 2\,{\frac{c \left ( -1+\sin \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) a\sqrt{c-c\sin \left ( fx+e \right ) }f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c/a*(-1+sin(f*x+e))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [B]  time = 2.24396, size = 103, normalized size = 3.55 \begin{align*} \frac{2 \,{\left (\sqrt{c} + \frac{\sqrt{c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} f \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(sqrt(c) + sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)/((a + a*sin(f*x + e)/(cos(f*x + e) + 1))*f*sqrt(sin(
f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))

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Fricas [A]  time = 1.00828, size = 66, normalized size = 2.28 \begin{align*} -\frac{2 \, \sqrt{-c \sin \left (f x + e\right ) + c}}{a f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(-c*sin(f*x + e) + c)/(a*f*cos(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-c*sin(e + f*x) + c)/(sin(e + f*x) + 1), x)/a

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Giac [B]  time = 1.50303, size = 262, normalized size = 9.03 \begin{align*} -\frac{\frac{\sqrt{2} \sqrt{c} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{\sqrt{2} a - a} - \frac{4 \,{\left ({\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )} c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) - c^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )\right )}}{{\left ({\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{2} + 2 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )} \sqrt{c} - c\right )} a}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(2)*sqrt(c)*sgn(tan(1/2*f*x + 1/2*e) - 1)/(sqrt(2)*a - a) - 4*((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*ta
n(1/2*f*x + 1/2*e)^2 + c))*c*sgn(tan(1/2*f*x + 1/2*e) - 1) - c^(3/2)*sgn(tan(1/2*f*x + 1/2*e) - 1))/(((sqrt(c)
*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2 + 2*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1
/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)*a))/f