\(\int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx\) [111]

Optimal result

Integrand size = 38, antiderivative size = 73 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=-\frac {(A-3 B) c \cos (e+f x)}{a f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a c f} \]

[Out]

-(A-B)*sec(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a/c/f-(A-3*B)*c*cos(f*x+e)/a/f/(c-c*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {3046, 2934, 2725} \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=-\frac {c (A-3 B) \cos (e+f x)}{a f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a c f} \]

[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 2725

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]

Rule 2934

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 3046

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]))

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 1.63 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=\frac {2 \sec (e+f x) (-A+2 B+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a f} \]

[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)

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.73

method	result	size
default	\(\frac {2 c \left (\sin \left (f x +e \right )-1\right ) \left (-B \sin \left (f x +e \right )+A -2 B \right )}{a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(53\)

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=\frac {2 \, {\left (B \sin \left (f x + e\right ) - A + 2 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{a f \cos \left (f x + e\right )} \]

[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))

Sympy [F]

\[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=\frac {\int \frac {A \sqrt {- c \sin {\left (e + f x \right )} + c}}{\sin {\left (e + f x \right )} + 1}\, dx + \int \frac {B \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}}{\sin {\left (e + f x \right )} + 1}\, dx}{a} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (69) = 138\).

Time = 0.31 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.38 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=-\frac {2 \, {\left (\frac {2 \, B {\left (\sqrt {c} + \frac {\sqrt {c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} - \frac {A {\left (\sqrt {c} + \frac {\sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{f} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (69) = 138\).

Time = 0.35 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.40 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=-\frac {2 \, \sqrt {2} {\left (A \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, B \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {A {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - \frac {B {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )} \sqrt {c}}{a f {\left (\frac {{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - 1\right )}} \]

[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]

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

Mupad [B] (verification not implemented)

Time = 15.14 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.75 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=\frac {2\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (2\,B\,\sin \left (2\,e+2\,f\,x\right )-2\,A\,\sin \left (2\,e+2\,f\,x\right )-4\,A\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+7\,B\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+B\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )\right )}{a\,f\,\left (4\,{\sin \left (e+f\,x\right )}^2+\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )-4\right )} \]

[In]

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

[Out]

(2*(-c*(sin(e + f*x) - 1))^(1/2)*(2*B*sin(2*e + 2*f*x) - 2*A*sin(2*e + 2*f*x) - 4*A*(2*sin(e/2 + (f*x)/2)^2 -
1) + 7*B*(2*sin(e/2 + (f*x)/2)^2 - 1) + B*(2*sin((3*e)/2 + (3*f*x)/2)^2 - 1)))/(a*f*(sin(e + f*x) + sin(3*e +
3*f*x) + 4*sin(e + f*x)^2 - 4))