\(\int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx\) [133]

Optimal result

Integrand size = 40, antiderivative size = 94 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 c f \sqrt {a+a \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 94, 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} \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 c f \sqrt {a \sin (e+f x)+a}}-\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}} \]

[In]

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

[Out]

-1/2*(a*(A + B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(f*Sqrt[a + a*Sin[e + f*x]]) + (a*B*Cos[e + f*x]*(c -
c*Sin[e + f*x])^(5/2))/(3*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)} (c-c \sin (e+f x))^{3/2} \, dx-\frac {B \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx}{c} \\ & = -\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 c f \sqrt {a+a \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.55 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.89 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {c \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (2 (6 A-B) \sin (e+f x)+\cos (2 (e+f x)) (3 A-3 B+2 B \sin (e+f x)))}{12 f} \]

[In]

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

[Out]

(c*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 - 3*B + 2*B*Sin[e + f*x])))/(12*f)

Maple [A] (verified)

Time = 3.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76

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

[In]

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

[Out]

-1/6*c/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 = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {{\left (3 \, {\left (A - B\right )} c \cos \left (f x + e\right )^{2} - 3 \, {\left (A - B\right )} c + 2 \, {\left (B c \cos \left (f x + e\right )^{2} + {\left (3 \, A - B\right )} c\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, f \cos \left (f x + e\right )} \]

[In]

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

[Out]

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

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.53 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 \, {\left (4 \, B c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 3 \, A c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3 \, B c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}\right )} \sqrt {a} \sqrt {c}}{3 \, f} \]

[In]

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

[Out]

-2/3*(4*B*c*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)^6 - 3*A*c*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 - 3*B*c*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)*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 2.00 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.30 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {c\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (3\,A\,\cos \left (e+f\,x\right )-3\,B\,\cos \left (e+f\,x\right )+3\,A\,\cos \left (3\,e+3\,f\,x\right )-3\,B\,\cos \left (3\,e+3\,f\,x\right )+12\,A\,\sin \left (2\,e+2\,f\,x\right )-2\,B\,\sin \left (2\,e+2\,f\,x\right )+B\,\sin \left (4\,e+4\,f\,x\right )\right )}{12\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

(c*(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))