\(\int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx\) [300]

Optimal result

Integrand size = 26, antiderivative size = 69 \[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {8 a c^3 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a c^2 \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}} \] Output:

8/15*a*c^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+2/5*a*c^2*cos(f*x+e)^3/f/ 
(c-c*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 a c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (-7+3 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

(-2*a*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-7 + 3*Sin[e + f*x])*Sqrt 
[c - c*Sin[e + f*x]])/(15*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3215, 3042, 3153, 3042, 3152}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^{3/2}dx\)

\(\Big \downarrow \) 3215

\(\displaystyle a c \int \cos ^2(e+f x) \sqrt {c-c \sin (e+f x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \int \cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}dx\)

\(\Big \downarrow \) 3153

\(\displaystyle a c \left (\frac {4}{5} c \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \left (\frac {4}{5} c \int \frac {\cos (e+f x)^2}{\sqrt {c-c \sin (e+f x)}}dx+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )\)

\(\Big \downarrow \) 3152

\(\displaystyle a c \left (\frac {8 c^2 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )\)

Input:

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

Output:

a*c*((8*c^2*Cos[e + f*x]^3)/(15*f*(c - c*Sin[e + f*x])^(3/2)) + (2*c*Cos[e 
 + f*x]^3)/(5*f*Sqrt[c - c*Sin[e + f*x]]))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3152

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x 
])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 
 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
 

rule 3153

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p))   Int[(g* 
Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && 
NeQ[m + p, 0]
 

rule 3215

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + 
 d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((Lt 
Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {2 \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right )^{2} a \left (3 \sin \left (f x +e \right )-7\right )}{15 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(59\)
parts	\(\frac {2 a \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-5\right )}{3 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )^{2}-3 \sin \left (f x +e \right )+6\right )}{5 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(120\)

Input:

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

Output:

2/15*(-1+sin(f*x+e))*c^2*(1+sin(f*x+e))^2*a*(3*sin(f*x+e)-7)/cos(f*x+e)/(c 
-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.58 \[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {2 \, {\left (3 \, a c \cos \left (f x + e\right )^{3} - a c \cos \left (f x + e\right )^{2} + 4 \, a c \cos \left (f x + e\right ) + 8 \, a c + {\left (3 \, a c \cos \left (f x + e\right )^{2} + 4 \, a c \cos \left (f x + e\right ) + 8 \, a c\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \] Input:

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

Output:

2/15*(3*a*c*cos(f*x + e)^3 - a*c*cos(f*x + e)^2 + 4*a*c*cos(f*x + e) + 8*a 
*c + (3*a*c*cos(f*x + e)^2 + 4*a*c*cos(f*x + e) + 8*a*c)*sin(f*x + e))*sqr 
t(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)
 

Sympy [F]

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

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

Output:

a*(Integral(c*sqrt(-c*sin(e + f*x) + c), x) + Integral(-c*sqrt(-c*sin(e + 
f*x) + c)*sin(e + f*x)**2, x))
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.43 \[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {\sqrt {2} {\left (30 \, a c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, a c \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, a c \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{30 \, f} \] Input:

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

Output:

-1/30*sqrt(2)*(30*a*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)*sgn(sin(-1/4*pi + 1/2 
*f*x + 1/2*e)) + 5*a*c*cos(-3/4*pi + 3/2*f*x + 3/2*e)*sgn(sin(-1/4*pi + 1/ 
2*f*x + 1/2*e)) - 3*a*c*cos(-5/4*pi + 5/2*f*x + 5/2*e)*sgn(sin(-1/4*pi + 1 
/2*f*x + 1/2*e)))*sqrt(c)/f
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\int \left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(c)*a*c*(int(sqrt( - sin(e + f*x) + 1),x) - int(sqrt( - sin(e + f*x) + 
 1)*sin(e + f*x)**2,x))