\(\int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx\) [142]

Optimal result

Integrand size = 40, antiderivative size = 96 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {(A-B) c \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 a f \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/2*(A-B)*c*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(1/2)+1/3*B*c*cos(f*x+e)*(a+a*sin(f*x+e))^(5/
2)/a/f/(c-c*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 96, 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} \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

((A - B)*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(2*f*Sqrt[c - c*Sin[e + f*x]]) + (B*c*Cos[e + f*x]*(a + a*
Sin[e + f*x])^(5/2))/(3*a*f*Sqrt[c - c*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}& = \frac {B \int (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx}{a}-(-A+B) \int (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx \\ & = \frac {(A-B) c \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 a f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.39 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (-2 (6 A+B) \sin (e+f x)+\cos (2 (e+f x)) (3 (A+B)+2 B \sin (e+f x)))}{12 f} \]

[In]

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

[Out]

-1/12*(a*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-2*(6*A + B)*Sin[e + f*x] + Cos[2*(
e + f*x)]*(3*(A + B) + 2*B*Sin[e + f*x])))/f

Maple [A] (verified)

Time = 3.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {a \tan \left (f x +e \right ) \left (2 B \left (\sin ^{2}\left (f x +e \right )\right )+3 A \sin \left (f x +e \right )+3 B \sin \left (f x +e \right )+6 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{6 f}\)	\(71\)
parts	\(-\frac {A a \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )-2 \tan \left (f x +e \right )-\sec \left (f x +e \right )\right )}{2 f}-\frac {B \sec \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) \left (2 \sin \left (f x +e \right )+3\right ) a \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{6 f}\)	\(121\)

[In]

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*a/f*tan(f*x+e)*(2*B*sin(f*x+e)^2+3*A*sin(f*x+e)+3*B*sin(f*x+e)+6*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 = 87, normalized size of antiderivative = 0.91 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {{\left (3 \, {\left (A + B\right )} a \cos \left (f x + e\right )^{2} - 3 \, {\left (A + B\right )} a + 2 \, {\left (B a \cos \left (f x + e\right )^{2} - {\left (3 \, A + B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, f \cos \left (f x + e\right )} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/6*(3*(A + B)*a*cos(f*x + e)^2 - 3*(A + B)*a + 2*(B*a*cos(f*x + e)^2 - (3*A + B)*a)*sin(f*x + e))*sqrt(a*sin
(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))

Sympy [F]

\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]

[In]

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

[Out]

Integral((a*(sin(e + f*x) + 1))**(3/2)*sqrt(-c*(sin(e + f*x) - 1))*(A + B*sin(e + f*x)), x)

Maxima [F]

\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {-c \sin \left (f x + e\right ) + c} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)*sqrt(-c*sin(f*x + e) + c), x)

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.50 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left (4 \, B a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, A a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, B a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{3 \, f} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-2/3*(4*B*a*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1
/2*e)) + 3*A*a*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x
+ 1/2*e)) - 3*B*a*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f
*x + 1/2*e)))*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 14.14 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.27 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (3\,A\,\cos \left (e+f\,x\right )+3\,B\,\cos \left (e+f\,x\right )+3\,A\,\cos \left (3\,e+3\,f\,x\right )+3\,B\,\cos \left (3\,e+3\,f\,x\right )-12\,A\,\sin \left (2\,e+2\,f\,x\right )-2\,B\,\sin \left (2\,e+2\,f\,x\right )+B\,\sin \left (4\,e+4\,f\,x\right )\right )}{12\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

-(a*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(3*A*cos(e + f*x) + 3*B*cos(e + f*x) + 3*A*cos(
3*e + 3*f*x) + 3*B*cos(3*e + 3*f*x) - 12*A*sin(2*e + 2*f*x) - 2*B*sin(2*e + 2*f*x) + B*sin(4*e + 4*f*x)))/(12*
f*(cos(2*e + 2*f*x) + 1))