\(\int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx\) [308]

Optimal result

Integrand size = 28, antiderivative size = 73 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=\frac {8 a^2 c^4 \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^2 c^3 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}} \] Output:

8/35*a^2*c^4*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)+2/7*a^2*c^3*cos(f*x+e)^ 
5/f/(c-c*sin(f*x+e))^(3/2)
 

Mathematica [A] (verified)

Time = 8.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 a^2 c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (-9+5 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{35 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

(-2*a^2*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(-9 + 5*Sin[e + f*x])*Sq 
rt[c - c*Sin[e + f*x]])/(35*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3153

\(\displaystyle a^2 c^2 \left (\frac {4}{7} c \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx+\frac {2 c \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {4}{7} c \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{3/2}}dx+\frac {2 c \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 3152

\(\displaystyle a^2 c^2 \left (\frac {8 c^2 \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}+\frac {2 c \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\right )\)

Input:

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

Output:

a^2*c^2*((8*c^2*Cos[e + f*x]^5)/(35*f*(c - c*Sin[e + f*x])^(5/2)) + (2*c*C 
os[e + f*x]^5)/(7*f*(c - c*Sin[e + f*x])^(3/2)))
 

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 = 1.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {2 \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right )^{3} a^{2} \left (5 \sin \left (f x +e \right )-9\right )}{35 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(61\)
parts	\(\frac {2 a^{2} \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-5\right )}{3 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a^{2} \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (15 \sin \left (f x +e \right )^{3}-39 \sin \left (f x +e \right )^{2}+52 \sin \left (f x +e \right )-104\right )}{105 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {4 a^{2} \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )^{2}-3 \sin \left (f x +e \right )+6\right )}{5 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(202\)

Input:

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

Output:

2/35*(-1+sin(f*x+e))*c^2*(1+sin(f*x+e))^3*a^2*(5*sin(f*x+e)-9)/cos(f*x+e)/ 
(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (65) = 130\).

Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.08 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 \, {\left (5 \, a^{2} c \cos \left (f x + e\right )^{4} - a^{2} c \cos \left (f x + e\right )^{3} + 2 \, a^{2} c \cos \left (f x + e\right )^{2} - 8 \, a^{2} c \cos \left (f x + e\right ) - 16 \, a^{2} c - {\left (5 \, a^{2} c \cos \left (f x + e\right )^{3} + 6 \, a^{2} c \cos \left (f x + e\right )^{2} + 8 \, a^{2} c \cos \left (f x + e\right ) + 16 \, a^{2} c\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{35 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \] Input:

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

Output:

-2/35*(5*a^2*c*cos(f*x + e)^4 - a^2*c*cos(f*x + e)^3 + 2*a^2*c*cos(f*x + e 
)^2 - 8*a^2*c*cos(f*x + e) - 16*a^2*c - (5*a^2*c*cos(f*x + e)^3 + 6*a^2*c* 
cos(f*x + e)^2 + 8*a^2*c*cos(f*x + e) + 16*a^2*c)*sin(f*x + e))*sqrt(-c*si 
n(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (65) = 130\).

Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.86 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=-\frac {\sqrt {2} {\left (105 \, a^{2} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 35 \, a^{2} c \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 7 \, a^{2} c \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, a^{2} c \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{140 \, f} \] Input:

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

Output:

-1/140*sqrt(2)*(105*a^2*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)*sgn(sin(-1/4*pi + 
 1/2*f*x + 1/2*e)) + 35*a^2*c*cos(-3/4*pi + 3/2*f*x + 3/2*e)*sgn(sin(-1/4* 
pi + 1/2*f*x + 1/2*e)) - 7*a^2*c*cos(-5/4*pi + 5/2*f*x + 5/2*e)*sgn(sin(-1 
/4*pi + 1/2*f*x + 1/2*e)) - 5*a^2*c*cos(-7/4*pi + 7/2*f*x + 7/2*e)*sgn(sin 
(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(c)/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(c)*a**2*c*(int(sqrt( - sin(e + f*x) + 1),x) - int(sqrt( - sin(e + f*x 
) + 1)*sin(e + f*x)**3,x) - int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2, 
x) + int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x),x))