3.3.72 \(\int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^2} \, dx\) [272]

3.3.72.1 Optimal result
3.3.72.2 Mathematica [A] (verified)
3.3.72.3 Rubi [A] (verified)
3.3.72.4 Maple [A] (verified)
3.3.72.5 Fricas [B] (verification not implemented)
3.3.72.6 Sympy [B] (verification not implemented)
3.3.72.7 Maxima [B] (verification not implemented)
3.3.72.8 Giac [A] (verification not implemented)
3.3.72.9 Mupad [B] (verification not implemented)

3.3.72.1 Optimal result

Integrand size = 26, antiderivative size = 70 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^2} \, dx=\frac {c^2 x}{a^2}-\frac {2 a c^2 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac {2 c^2 \cos (e+f x)}{f \left (a^2+a^2 \sin (e+f x)\right )} \]

output
c^2*x/a^2-2/3*a*c^2*cos(f*x+e)^3/f/(a+a*sin(f*x+e))^3+2*c^2*cos(f*x+e)/f/( 
a^2+a^2*sin(f*x+e))
 
3.3.72.2 Mathematica [A] (verified)

Time = 1.63 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.70 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^2} \, dx=\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 (-8+3 e+3 f x) \cos \left (\frac {1}{2} (e+f x)\right )+(16-3 e-3 f x) \cos \left (\frac {3}{2} (e+f x)\right )+6 (2 (-2+e+f x)+(e+f x) \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{6 a^2 f (1+\sin (e+f x))^2} \]

input
Integrate[(c - c*Sin[e + f*x])^2/(a + a*Sin[e + f*x])^2,x]
 
output
(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(3*(-8 + 3*e + 3*f*x)*Cos[(e + 
f*x)/2] + (16 - 3*e - 3*f*x)*Cos[(3*(e + f*x))/2] + 6*(2*(-2 + e + f*x) + 
(e + f*x)*Cos[e + f*x])*Sin[(e + f*x)/2]))/(6*a^2*f*(1 + Sin[e + f*x])^2)
 
3.3.72.3 Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 3215, 3042, 3159, 3042, 3159, 24}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3159

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3159

\(\displaystyle a^2 c^2 \left (-\frac {-\frac {\int 1dx}{a^2}-\frac {2 \cos (e+f x)}{f \left (a^2 \sin (e+f x)+a^2\right )}}{a^2}-\frac {2 \cos ^3(e+f x)}{3 a f (a \sin (e+f x)+a)^3}\right )\)

\(\Big \downarrow \) 24

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

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

3.3.72.3.1 Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3159
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f 
*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 
)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; 
FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & 
& NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
 

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]))
 
3.3.72.4 Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.76

method result size
derivativedivides \(\frac {2 c^{2} \left (\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {8}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}\right )}{f \,a^{2}}\) \(53\)
default \(\frac {2 c^{2} \left (\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {8}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}\right )}{f \,a^{2}}\) \(53\)
risch \(\frac {c^{2} x}{a^{2}}+\frac {8 i c^{2} {\mathrm e}^{i \left (f x +e \right )}+8 c^{2} {\mathrm e}^{2 i \left (f x +e \right )}-\frac {16 c^{2}}{3}}{f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3}}\) \(67\)
parallelrisch \(\frac {c^{2} \left (3 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) f x +9 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) x f +9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) f x +3 f x +24 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+8\right )}{3 f \,a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\) \(85\)
norman \(\frac {\frac {c^{2} x}{a}+\frac {c^{2} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {8 c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f a}+\frac {8 c^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {16 c^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {8 c^{2}}{3 f a}+\frac {3 c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}+\frac {5 c^{2} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {7 c^{2} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {7 c^{2} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {5 c^{2} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {3 c^{2} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {16 c^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}+\frac {8 c^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2} a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\) \(298\)

input
int((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^2,x,method=_RETURNVERBOSE)
 
output
2/f*c^2/a^2*(arctan(tan(1/2*f*x+1/2*e))-8/3/(tan(1/2*f*x+1/2*e)+1)^3+4/(ta 
n(1/2*f*x+1/2*e)+1)^2)
 
3.3.72.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (68) = 136\).

Time = 0.32 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.26 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^2} \, dx=-\frac {6 \, c^{2} f x - {\left (3 \, c^{2} f x - 8 \, c^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, c^{2} + {\left (3 \, c^{2} f x + 4 \, c^{2}\right )} \cos \left (f x + e\right ) + {\left (6 \, c^{2} f x + 4 \, c^{2} + {\left (3 \, c^{2} f x + 8 \, c^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

input
integrate((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^2,x, algorithm="fricas")
 
output
-1/3*(6*c^2*f*x - (3*c^2*f*x - 8*c^2)*cos(f*x + e)^2 - 4*c^2 + (3*c^2*f*x 
+ 4*c^2)*cos(f*x + e) + (6*c^2*f*x + 4*c^2 + (3*c^2*f*x + 8*c^2)*cos(f*x + 
 e))*sin(f*x + e))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - 
(a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))
 
3.3.72.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (65) = 130\).

Time = 2.06 (sec) , antiderivative size = 473, normalized size of antiderivative = 6.76 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^2} \, dx=\begin {cases} \frac {3 c^{2} f x \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} + \frac {9 c^{2} f x \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} + \frac {9 c^{2} f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} + \frac {3 c^{2} f x}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} + \frac {24 c^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} + \frac {8 c^{2}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} & \text {for}\: f \neq 0 \\\frac {x \left (- c \sin {\left (e \right )} + c\right )^{2}}{\left (a \sin {\left (e \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

input
integrate((c-c*sin(f*x+e))**2/(a+a*sin(f*x+e))**2,x)
 
output
Piecewise((3*c**2*f*x*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**3 + 
9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 9*c 
**2*f*x*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e 
/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 9*c**2*f*x*tan(e/ 
2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 
9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 3*c**2*f*x/(3*a**2*f*tan(e/2 + f*x 
/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2 
*f) + 24*c**2*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*ta 
n(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 8*c**2/(3*a**2 
*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + 
 f*x/2) + 3*a**2*f), Ne(f, 0)), (x*(-c*sin(e) + c)**2/(a*sin(e) + a)**2, T 
rue))
 
3.3.72.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (68) = 136\).

Time = 0.29 (sec) , antiderivative size = 361, normalized size of antiderivative = 5.16 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^2} \, dx=\frac {2 \, {\left (c^{2} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 4}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} - \frac {c^{2} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {2 \, c^{2} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \]

input
integrate((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^2,x, algorithm="maxima")
 
output
2/3*(c^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + 
 e) + 1)^2 + 4)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f 
*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) 
+ 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - c^2*(3*sin(f*x + e)/(co 
s(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a^2 + 3*a^2* 
sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^ 
2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 2*c^2*(3*sin(f*x + e)/(cos( 
f*x + e) + 1) + 1)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*si 
n(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^ 
3))/f
 
3.3.72.8 Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.79 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^2} \, dx=\frac {\frac {3 \, {\left (f x + e\right )} c^{2}}{a^{2}} + \frac {8 \, {\left (3 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c^{2}\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \]

input
integrate((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^2,x, algorithm="giac")
 
output
1/3*(3*(f*x + e)*c^2/a^2 + 8*(3*c^2*tan(1/2*f*x + 1/2*e) + c^2)/(a^2*(tan( 
1/2*f*x + 1/2*e) + 1)^3))/f
 
3.3.72.9 Mupad [B] (verification not implemented)

Time = 6.53 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.27 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^2} \, dx=\frac {c^2\,x}{a^2}-\frac {c^2\,\left (e+f\,x\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,c^2\,\left (e+f\,x\right )-\frac {c^2\,\left (9\,e+9\,f\,x+24\right )}{3}\right )-\frac {c^2\,\left (3\,e+3\,f\,x+8\right )}{3}}{a^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^3} \]

input
int((c - c*sin(e + f*x))^2/(a + a*sin(e + f*x))^2,x)
 
output
(c^2*x)/a^2 - (c^2*(e + f*x) + tan(e/2 + (f*x)/2)*(3*c^2*(e + f*x) - (c^2* 
(9*e + 9*f*x + 24))/3) - (c^2*(3*e + 3*f*x + 8))/3)/(a^2*f*(tan(e/2 + (f*x 
)/2) + 1)^3)