\(\int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx\) [134]

Optimal result

Integrand size = 40, antiderivative size = 92 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a (A+B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}+\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 c f \sqrt {a+a \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {3050, 2817} \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 c f \sqrt {a \sin (e+f x)+a}}-\frac {a (A+B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}} \]

[In]

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

[Out]

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

Rule 2817

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rule 3050

Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[B/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x
] - Dist[(B*c - A*d)/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.62 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {\sec (e+f x) \sqrt {a (1+\sin (e+f x))} (A+B \sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}}{2 B f} \]

[In]

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

[Out]

(Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*(A + B*Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])/(2*B*f)

Maple [A] (verified)

Time = 2.88 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.53

method	result	size
default	\(\frac {\tan \left (f x +e \right ) \left (B \sin \left (f x +e \right )+2 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{2 f}\)	\(49\)
parts	\(\frac {A \tan \left (f x +e \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{f}+\frac {B \sin \left (f x +e \right ) \tan \left (f x +e \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{2 f}\)	\(81\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {{\left (B \cos \left (f x + e\right )^{2} - 2 \, A \sin \left (f x + e\right ) - B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, f \cos \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c), x)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.52 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left (B \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - A \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )} \sqrt {a} \sqrt {c}}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (B\,\cos \left (e+f\,x\right )+B\,\cos \left (3\,e+3\,f\,x\right )-4\,A\,\sin \left (2\,e+2\,f\,x\right )\right )}{4\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

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