\(\int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx\) [262]

Optimal result

Integrand size = 26, antiderivative size = 92 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\frac {5 a^3 x}{c^2}-\frac {5 a^3 \cos (e+f x)}{c^2 f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \] Output:

5*a^3*x/c^2-5*a^3*cos(f*x+e)/c^2/f+2/3*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x 
+e))^4-10/3*a^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^2
 

Mathematica [A] (verified)

Time = 9.55 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.62 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 (23+15 e+15 f x) \cos \left (\frac {1}{2} (e+f x)\right )-(121+30 e+30 f x) \cos \left (\frac {3}{2} (e+f x)\right )+3 \cos \left (\frac {5}{2} (e+f x)\right )-6 (31+20 e+20 f x+2 (-2+5 e+5 f x) \cos (e+f x)-\cos (2 (e+f x))) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{12 c^2 f (-1+\sin (e+f x))^2} \] Input:

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

Output:

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(6*(23 + 15*e + 15*f*x)*Cos[(e 
+ f*x)/2] - (121 + 30*e + 30*f*x)*Cos[(3*(e + f*x))/2] + 3*Cos[(5*(e + f*x 
))/2] - 6*(31 + 20*e + 20*f*x + 2*(-2 + 5*e + 5*f*x)*Cos[e + f*x] - Cos[2* 
(e + f*x)])*Sin[(e + f*x)/2]))/(12*c^2*f*(-1 + Sin[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3215, 3042, 3159, 3042, 3159, 3042, 3161, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

\(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^5}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^5}dx\)

\(\Big \downarrow \) 3159

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3159

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3161

\(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^4}-\frac {5 \left (\frac {2 \cos ^3(e+f x)}{c f (c-c \sin (e+f x))^2}-\frac {3 \left (\frac {\int 1dx}{c}-\frac {\cos (e+f x)}{c f}\right )}{c^2}\right )}{3 c^2}\right )\)

\(\Big \downarrow \) 24

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

Input:

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

Output:

a^3*c^3*((2*Cos[e + f*x]^5)/(3*c*f*(c - c*Sin[e + f*x])^4) - (5*((-3*(x/c 
- Cos[e + f*x]/(c*f)))/c^2 + (2*Cos[e + f*x]^3)/(c*f*(c - c*Sin[e + f*x])^ 
2)))/(3*c^2))
 

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 3161

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

Maple [A] (verified)

Time = 1.79 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {1}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+5 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {16}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f \,c^{2}}\)	\(87\)
default	\(\frac {2 a^{3} \left (-\frac {1}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+5 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {16}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f \,c^{2}}\)	\(87\)
risch	\(\frac {5 a^{3} x}{c^{2}}-\frac {a^{3} {\mathrm e}^{i \left (f x +e \right )}}{2 c^{2} f}-\frac {a^{3} {\mathrm e}^{-i \left (f x +e \right )}}{2 c^{2} f}-\frac {8 \left (-12 i a^{3} {\mathrm e}^{i \left (f x +e \right )}+9 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}-7 a^{3}\right )}{3 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,c^{2}}\)	\(108\)
parallelrisch	\(\frac {a^{3} \left (15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} x f -45 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} x f +60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} x f +24 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} x f -102 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+45 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) f x +82 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-15 f x -114 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+46\right )}{3 f \,c^{2} \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\)	\(169\)
norman	\(\frac {\frac {8 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{c f}-\frac {5 a^{3} x}{c}+\frac {46 a^{3}}{3 c f}+\frac {15 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}-\frac {30 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c}+\frac {50 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c}-\frac {60 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c}+\frac {60 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c}-\frac {50 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c}+\frac {30 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c}-\frac {15 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{c}+\frac {5 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{c}-\frac {110 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c f}+\frac {130 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 c f}+\frac {78 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c f}-\frac {34 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c f}+\frac {58 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c f}-\frac {38 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c f}-\frac {106 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\)	\(406\)

Input:

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

Output:

2/f*a^3/c^2*(-1/(1+tan(1/2*f*x+1/2*e)^2)+5*arctan(tan(1/2*f*x+1/2*e))-16/3 
/(tan(1/2*f*x+1/2*e)-1)^3-8/(tan(1/2*f*x+1/2*e)-1)^2+4/(tan(1/2*f*x+1/2*e) 
-1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (90) = 180\).

Time = 0.09 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.00 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=-\frac {3 \, a^{3} \cos \left (f x + e\right )^{3} + 30 \, a^{3} f x + 8 \, a^{3} - {\left (15 \, a^{3} f x + 31 \, a^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (15 \, a^{3} f x - 26 \, a^{3}\right )} \cos \left (f x + e\right ) - {\left (30 \, a^{3} f x - 3 \, a^{3} \cos \left (f x + e\right )^{2} - 8 \, a^{3} + {\left (15 \, a^{3} f x - 34 \, a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

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

Output:

-1/3*(3*a^3*cos(f*x + e)^3 + 30*a^3*f*x + 8*a^3 - (15*a^3*f*x + 31*a^3)*co 
s(f*x + e)^2 + (15*a^3*f*x - 26*a^3)*cos(f*x + e) - (30*a^3*f*x - 3*a^3*co 
s(f*x + e)^2 - 8*a^3 + (15*a^3*f*x - 34*a^3)*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))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1282 vs. \(2 (87) = 174\).

Time = 4.03 (sec) , antiderivative size = 1282, normalized size of antiderivative = 13.93 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((15*a**3*f*x*tan(e/2 + f*x/2)**5/(3*c**2*f*tan(e/2 + f*x/2)**5 - 
 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f* 
tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 45*a**3*f*x* 
tan(e/2 + f*x/2)**4/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x 
/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9 
*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 60*a**3*f*x*tan(e/2 + f*x/2)**3/(3* 
c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan( 
e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2 
) - 3*c**2*f) - 60*a**3*f*x*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2) 
**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c* 
*2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 45*a**3 
*f*x*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f 
*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 
 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 15*a**3*f*x/(3*c**2*f*tan(e/2 + f 
*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 
12*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**3*tan(e/2 + f*x/2)**4/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 
 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)* 
*2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 102*a**3*tan(e/2 + f*x/2)**3/ 
(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 594 vs. \(2 (90) = 180\).

Time = 0.13 (sec) , antiderivative size = 594, normalized size of antiderivative = 6.46 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

2/3*(2*a^3*((12*sin(f*x + e)/(cos(f*x + e) + 1) - 11*sin(f*x + e)^2/(cos(f 
*x + e) + 1)^2 + 9*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 3*sin(f*x + e)^4/ 
(cos(f*x + e) + 1)^4 - 5)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4 
*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4*c^2*sin(f*x + e)^3/(cos(f*x + 
 e) + 1)^3 + 3*c^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^2*sin(f*x + e)^ 
5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) + 
 3*a^3*((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^3*(3*sin(f*x + e)/(cos( 
f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c^2 - 3*c^2*si 
n(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*a^3*(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*sin( 
f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) 
)/f
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\frac {\frac {15 \, {\left (f x + e\right )} a^{3}}{c^{2}} - \frac {6 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )} c^{2}} + \frac {8 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 12 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, a^{3}\right )}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}}{3 \, f} \] Input:

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

Output:

1/3*(15*(f*x + e)*a^3/c^2 - 6*a^3/((tan(1/2*f*x + 1/2*e)^2 + 1)*c^2) + 8*( 
3*a^3*tan(1/2*f*x + 1/2*e)^2 - 12*a^3*tan(1/2*f*x + 1/2*e) + 5*a^3)/(c^2*( 
tan(1/2*f*x + 1/2*e) - 1)^3))/f
 

Mupad [B] (verification not implemented)

Time = 18.63 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.37 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\frac {5\,a^3\,x}{c^2}+\frac {5\,a^3\,\left (e+f\,x\right )-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (15\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (45\,e+45\,f\,x-114\right )}{3}\right )-\frac {a^3\,\left (15\,e+15\,f\,x-46\right )}{3}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (15\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (45\,e+45\,f\,x-24\right )}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (20\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (60\,e+60\,f\,x-82\right )}{3}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (20\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (60\,e+60\,f\,x-102\right )}{3}\right )}{c^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^3\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \] Input:

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

Output:

(5*a^3*x)/c^2 + (5*a^3*(e + f*x) - tan(e/2 + (f*x)/2)*(15*a^3*(e + f*x) - 
(a^3*(45*e + 45*f*x - 114))/3) - (a^3*(15*e + 15*f*x - 46))/3 + tan(e/2 + 
(f*x)/2)^4*(15*a^3*(e + f*x) - (a^3*(45*e + 45*f*x - 24))/3) + tan(e/2 + ( 
f*x)/2)^2*(20*a^3*(e + f*x) - (a^3*(60*e + 60*f*x - 82))/3) - tan(e/2 + (f 
*x)/2)^3*(20*a^3*(e + f*x) - (a^3*(60*e + 60*f*x - 102))/3))/(c^2*f*(tan(e 
/2 + (f*x)/2) - 1)^3*(tan(e/2 + (f*x)/2)^2 + 1))
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.87 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\frac {a^{3} \left (-3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}+15 \cos \left (f x +e \right ) \sin \left (f x +e \right ) f x +19 \cos \left (f x +e \right ) \sin \left (f x +e \right )-15 \cos \left (f x +e \right ) f x -8 \cos \left (f x +e \right )+3 \sin \left (f x +e \right )^{3}+15 \sin \left (f x +e \right )^{2} f x -46 \sin \left (f x +e \right )^{2}-30 \sin \left (f x +e \right ) f x +19 \sin \left (f x +e \right )+15 f x +8\right )}{3 c^{2} f \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )-\cos \left (f x +e \right )+\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1\right )} \] Input:

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

Output:

(a**3*( - 3*cos(e + f*x)*sin(e + f*x)**2 + 15*cos(e + f*x)*sin(e + f*x)*f* 
x + 19*cos(e + f*x)*sin(e + f*x) - 15*cos(e + f*x)*f*x - 8*cos(e + f*x) + 
3*sin(e + f*x)**3 + 15*sin(e + f*x)**2*f*x - 46*sin(e + f*x)**2 - 30*sin(e 
 + f*x)*f*x + 19*sin(e + f*x) + 15*f*x + 8))/(3*c**2*f*(cos(e + f*x)*sin(e 
 + f*x) - cos(e + f*x) + sin(e + f*x)**2 - 2*sin(e + f*x) + 1))