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

Optimal. Leaf size=73 \[ -\frac{c (A-3 B) \cos (e+f x)}{a f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a c f} \]

[Out]

-(((A - 3*B)*c*Cos[e + f*x])/(a*f*Sqrt[c - c*Sin[e + f*x]])) - ((A - B)*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2
))/(a*c*f)

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Rubi [A]  time = 0.274729, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2967, 2855, 2646} \[ -\frac{c (A-3 B) \cos (e+f x)}{a f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A - 3*B)*c*Cos[e + f*x])/(a*f*Sqrt[c - c*Sin[e + f*x]])) - ((A - B)*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2
))/(a*c*f)

Rule 2967

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

Rule 2855

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

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

Mathematica [A]  time = 0.207427, size = 44, normalized size = 0.6 \[ \frac{2 \sec (e+f x) \sqrt{c-c \sin (e+f x)} (-A+B \sin (e+f x)+2 B)}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.644, size = 53, normalized size = 0.7 \begin{align*} 2\,{\frac{c \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( -B\sin \left ( fx+e \right ) +A-2\,B \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((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e)),x)

[Out]

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

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Maxima [B]  time = 1.50121, size = 235, normalized size = 3.22 \begin{align*} -\frac{2 \,{\left (\frac{2 \, B{\left (\sqrt{c} + \frac{\sqrt{c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \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 )} \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} - \frac{A{\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 )} \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.37219, size = 101, normalized size = 1.38 \begin{align*} \frac{2 \,{\left (B \sin \left (f x + e\right ) - A + 2 \, B\right )} \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((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

2*(B*sin(f*x + e) - A + 2*B)*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{A \sqrt{- c \sin{\left (e + f x \right )} + c}}{\sin{\left (e + f x \right )} + 1}\, dx + \int \frac{B \sqrt{- c \sin{\left (e + f x \right )} + c} \sin{\left (e + f x \right )}}{\sin{\left (e + f x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.57594, size = 475, normalized size = 6.51 \begin{align*} -\frac{\frac{{\left (\sqrt{2} A \sqrt{c} + \sqrt{2} B \sqrt{c} - 4 \, B \sqrt{c}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{\sqrt{2} a - a} + \frac{2 \,{\left (\frac{B c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a} + \frac{B c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{a}\right )}}{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}} - \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 )} A c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) -{\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 )} B c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) - A c^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) + B 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((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

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