\(\int (3+3 \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx\) [351]

Optimal result

Integrand size = 30, antiderivative size = 43 \[ \int (3+3 \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=\frac {c \cos (e+f x) (3+3 \sin (e+f x))^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {2817} \[ \int (3+3 \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(2*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)]

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

Mathematica [A] (verified)

Time = 5.53 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.44 \[ \int (3+3 \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {3 \sqrt {3} \sec (e+f x) (\cos (2 (e+f x))-4 \sin (e+f x)) \sqrt {1+\sin (e+f x)} \sqrt {c-c \sin (e+f x)}}{4 f} \]

[In]

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

[Out]

(-3*Sqrt[3]*Sec[e + f*x]*(Cos[2*(e + f*x)] - 4*Sin[e + f*x])*Sqrt[1 + Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])/
(4*f)

Maple [A] (verified)

Time = 2.90 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28

method	result	size
default	\(-\frac {a \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, \left (\cos \left (f x +e \right )-2 \tan \left (f x +e \right )-\sec \left (f x +e \right )\right )}{2 f}\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.42 \[ \int (3+3 \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {{\left (a \cos \left (f x + e\right )^{2} - 2 \, a \sin \left (f x + e\right ) - a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, f \cos \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate((a+a*sin(f*x+e))**(3/2)*(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)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.19 \[ \int (3+3 \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \, a^{\frac {3}{2}} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \]

[In]

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

[Out]

-2*a^(3/2)*sqrt(c)*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))/f

Mupad [B] (verification not implemented)

Time = 7.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.65 \[ \int (3+3 \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (\cos \left (e+f\,x\right )+\cos \left (3\,e+3\,f\,x\right )-4\,\sin \left (2\,e+2\,f\,x\right )\right )}{4\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

int((a + 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)*(cos(e + f*x) + cos(3*e + 3*f*x) - 4*sin(2*e +
2*f*x)))/(4*f*(cos(2*e + 2*f*x) + 1))