3.3.42 \(\int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx\) [242]

3.3.42.1 Optimal result
3.3.42.2 Mathematica [A] (verified)
3.3.42.3 Rubi [A] (verified)
3.3.42.4 Maple [A] (verified)
3.3.42.5 Fricas [B] (verification not implemented)
3.3.42.6 Sympy [B] (verification not implemented)
3.3.42.7 Maxima [B] (verification not implemented)
3.3.42.8 Giac [A] (verification not implemented)
3.3.42.9 Mupad [B] (verification not implemented)

3.3.42.1 Optimal result

Integrand size = 26, antiderivative size = 72 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx=\frac {a^2 x}{c^2}+\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac {2 a^2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )} \]

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

Time = 6.72 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.68 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-3 (8+3 e+3 f x) \cos \left (\frac {1}{2} (e+f x)\right )+(16+3 e+3 f x) \cos \left (\frac {3}{2} (e+f x)\right )+6 (2 (2+e+f x)+(e+f x) \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{6 c^2 f (-1+\sin (e+f x))^2} \]

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

Time = 0.42 (sec) , antiderivative size = 79, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3159

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3159

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

\(\Big \downarrow \) 24

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

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

3.3.42.3.1 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]))
 
3.3.42.4 Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.74

method result size
derivativedivides \(\frac {2 a^{2} \left (-\frac {8}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{2}}\) \(53\)
default \(\frac {2 a^{2} \left (-\frac {8}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{2}}\) \(53\)
risch \(\frac {a^{2} x}{c^{2}}-\frac {8 \left (-3 i a^{2} {\mathrm e}^{i \left (f x +e \right )}+3 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-2 a^{2}\right )}{3 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,c^{2}}\) \(67\)
parallelrisch \(\frac {a^{2} \left (3 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) f x -9 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) x f +9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) f x -3 f x -24 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+8\right )}{3 f \,c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\) \(85\)
norman \(\frac {\frac {a^{2} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {a^{2} x}{c}+\frac {8 a^{2}}{3 c f}+\frac {3 a^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}-\frac {5 a^{2} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {7 a^{2} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {7 a^{2} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {5 a^{2} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {3 a^{2} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {8 a^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}+\frac {16 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {16 a^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {8 a^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {8 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\) \(299\)

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

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (72) = 144\).

Time = 0.27 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.19 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx=-\frac {6 \, a^{2} f x - {\left (3 \, a^{2} f x + 8 \, a^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, a^{2} + {\left (3 \, a^{2} f x - 4 \, a^{2}\right )} \cos \left (f x + e\right ) - {\left (6 \, a^{2} f x - 4 \, a^{2} + {\left (3 \, a^{2} f x - 8 \, a^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

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

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

Time = 2.03 (sec) , antiderivative size = 473, normalized size of antiderivative = 6.57 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx=\begin {cases} \frac {3 a^{2} f x \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} - \frac {9 a^{2} f x \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} + \frac {9 a^{2} f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} - \frac {3 a^{2} f x}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} - \frac {24 a^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} + \frac {8 a^{2}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} & \text {for}\: f \neq 0 \\\frac {x \left (a \sin {\left (e \right )} + a\right )^{2}}{\left (- c \sin {\left (e \right )} + c\right )^{2}} & \text {otherwise} \end {cases} \]

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

Leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (72) = 144\).

Time = 0.32 (sec) , antiderivative size = 364, normalized size of antiderivative = 5.06 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx=\frac {2 \, {\left (a^{2} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 4}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{2}}\right )} - \frac {a^{2} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {2 \, a^{2} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \]

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

Time = 0.34 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.79 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx=\frac {\frac {3 \, {\left (f x + e\right )} a^{2}}{c^{2}} - \frac {8 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a^{2}\right )}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}}{3 \, f} \]

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

Time = 6.41 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx=\frac {a^2\,x}{c^2}-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,a^2\,\left (e+f\,x\right )-\frac {a^2\,\left (9\,e+9\,f\,x-24\right )}{3}\right )-a^2\,\left (e+f\,x\right )+\frac {a^2\,\left (3\,e+3\,f\,x-8\right )}{3}}{c^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^3} \]

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