3.3.57 \(\int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^5} \, dx\) [257]

3.3.57.1 Optimal result
3.3.57.2 Mathematica [A] (verified)
3.3.57.3 Rubi [A] (verified)
3.3.57.4 Maple [C] (verified)
3.3.57.5 Fricas [B] (verification not implemented)
3.3.57.6 Sympy [B] (verification not implemented)
3.3.57.7 Maxima [B] (verification not implemented)
3.3.57.8 Giac [B] (verification not implemented)
3.3.57.9 Mupad [B] (verification not implemented)

3.3.57.1 Optimal result

Integrand size = 26, antiderivative size = 69 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^5} \, dx=\frac {a^3 c^3 \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^8}+\frac {a^3 c^2 \cos ^7(e+f x)}{63 f (c-c \sin (e+f x))^7} \]

output
1/9*a^3*c^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^8+1/63*a^3*c^2*cos(f*x+e)^7/f/ 
(c-c*sin(f*x+e))^7
 
3.3.57.2 Mathematica [A] (verified)

Time = 2.41 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.96 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^5} \, dx=\frac {a^3 \left (315 \cos \left (\frac {1}{2} (e+f x)\right )-189 \cos \left (\frac {3}{2} (e+f x)\right )-63 \cos \left (\frac {5}{2} (e+f x)\right )+9 \cos \left (\frac {7}{2} (e+f x)\right )+189 \sin \left (\frac {1}{2} (e+f x)\right )+105 \sin \left (\frac {3}{2} (e+f x)\right )-27 \sin \left (\frac {5}{2} (e+f x)\right )-\sin \left (\frac {9}{2} (e+f x)\right )\right )}{504 c^5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9} \]

input
Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^5,x]
 
output
(a^3*(315*Cos[(e + f*x)/2] - 189*Cos[(3*(e + f*x))/2] - 63*Cos[(5*(e + f*x 
))/2] + 9*Cos[(7*(e + f*x))/2] + 189*Sin[(e + f*x)/2] + 105*Sin[(3*(e + f* 
x))/2] - 27*Sin[(5*(e + f*x))/2] - Sin[(9*(e + f*x))/2]))/(504*c^5*f*(Cos[ 
(e + f*x)/2] - Sin[(e + f*x)/2])^9)
 
3.3.57.3 Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97, 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, 3151, 3042, 3150}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

\(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^8}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^8}dx\)

\(\Big \downarrow \) 3151

\(\displaystyle a^3 c^3 \left (\frac {\int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^7}dx}{9 c}+\frac {\cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^8}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \left (\frac {\int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^7}dx}{9 c}+\frac {\cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^8}\right )\)

\(\Big \downarrow \) 3150

\(\displaystyle a^3 c^3 \left (\frac {\cos ^7(e+f x)}{63 c f (c-c \sin (e+f x))^7}+\frac {\cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^8}\right )\)

input
Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^5,x]
 
output
a^3*c^3*(Cos[e + f*x]^7/(9*f*(c - c*Sin[e + f*x])^8) + Cos[e + f*x]^7/(63* 
c*f*(c - c*Sin[e + f*x])^7))
 

3.3.57.3.1 Defintions of rubi rules used

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

rule 3150
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/(a*f*g*m)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] 
 && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]
 

rule 3151
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/(a*f*g*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl 
ify[2*m + p + 1])   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] && ILtQ[Simpli 
fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 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]))
 
3.3.57.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.59

method result size
risch \(\frac {2 i a^{3} \left (105 i {\mathrm e}^{6 i \left (f x +e \right )}+63 \,{\mathrm e}^{7 i \left (f x +e \right )}-189 i {\mathrm e}^{4 i \left (f x +e \right )}-315 \,{\mathrm e}^{5 i \left (f x +e \right )}+27 i {\mathrm e}^{2 i \left (f x +e \right )}+189 \,{\mathrm e}^{3 i \left (f x +e \right )}+i-9 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{63 f \,c^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{9}}\) \(110\)
parallelrisch \(-\frac {2 \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )-\left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {23 \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}-5 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+11 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-3 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {25 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{7}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{7}+\frac {8}{63}\right ) a^{3}}{f \,c^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{9}}\) \(127\)
derivativedivides \(\frac {2 a^{3} \left (-\frac {928}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {128}{9 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{9}}-\frac {7}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {64}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{8}}-\frac {136}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {76}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {496}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {86}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\right )}{f \,c^{5}}\) \(148\)
default \(\frac {2 a^{3} \left (-\frac {928}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {128}{9 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{9}}-\frac {7}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {64}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{8}}-\frac {136}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {76}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {496}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {86}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\right )}{f \,c^{5}}\) \(148\)
norman \(\frac {\frac {2 a^{3} \left (\tan ^{13}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {16 a^{3} \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {42 a^{3} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {16 a^{3}}{63 c f}-\frac {2 a^{3} \left (\tan ^{14}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {2 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{7 c f}+\frac {48 a^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{7 c f}-\frac {64 a^{3} \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {74 a^{3} \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {166 a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{21 c f}+\frac {202 a^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{7 c f}+\frac {352 a^{3} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{7 c f}-\frac {848 a^{3} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{7 c f}-\frac {928 a^{3} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{21 c f}-\frac {6490 a^{3} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{63 c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3} c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{9}}\) \(351\)

input
int((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x,method=_RETURNVERBOSE)
 
output
2/63*I*a^3*(105*I*exp(6*I*(f*x+e))+63*exp(7*I*(f*x+e))-189*I*exp(4*I*(f*x+ 
e))-315*exp(5*I*(f*x+e))+27*I*exp(2*I*(f*x+e))+189*exp(3*I*(f*x+e))+I-9*ex 
p(I*(f*x+e)))/f/c^5/(exp(I*(f*x+e))-I)^9
 
3.3.57.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (67) = 134\).

Time = 0.26 (sec) , antiderivative size = 276, normalized size of antiderivative = 4.00 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^5} \, dx=-\frac {a^{3} \cos \left (f x + e\right )^{5} - 4 \, a^{3} \cos \left (f x + e\right )^{4} + 19 \, a^{3} \cos \left (f x + e\right )^{3} + 52 \, a^{3} \cos \left (f x + e\right )^{2} - 28 \, a^{3} \cos \left (f x + e\right ) - 56 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{4} + 5 \, a^{3} \cos \left (f x + e\right )^{3} + 24 \, a^{3} \cos \left (f x + e\right )^{2} - 28 \, a^{3} \cos \left (f x + e\right ) - 56 \, a^{3}\right )} \sin \left (f x + e\right )}{63 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} + 5 \, c^{5} f \cos \left (f x + e\right )^{4} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} - 20 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f - {\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} - 12 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f\right )} \sin \left (f x + e\right )\right )}} \]

input
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x, algorithm="fricas")
 
output
-1/63*(a^3*cos(f*x + e)^5 - 4*a^3*cos(f*x + e)^4 + 19*a^3*cos(f*x + e)^3 + 
 52*a^3*cos(f*x + e)^2 - 28*a^3*cos(f*x + e) - 56*a^3 + (a^3*cos(f*x + e)^ 
4 + 5*a^3*cos(f*x + e)^3 + 24*a^3*cos(f*x + e)^2 - 28*a^3*cos(f*x + e) - 5 
6*a^3)*sin(f*x + e))/(c^5*f*cos(f*x + e)^5 + 5*c^5*f*cos(f*x + e)^4 - 8*c^ 
5*f*cos(f*x + e)^3 - 20*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f*x + e) + 16*c 
^5*f - (c^5*f*cos(f*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 - 12*c^5*f*cos(f*x + 
 e)^2 + 8*c^5*f*cos(f*x + e) + 16*c^5*f)*sin(f*x + e))
 
3.3.57.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1717 vs. \(2 (60) = 120\).

Time = 25.07 (sec) , antiderivative size = 1717, normalized size of antiderivative = 24.88 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^5} \, dx=\text {Too large to display} \]

input
integrate((a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**5,x)
 
output
Piecewise((-126*a**3*tan(e/2 + f*x/2)**8/(63*c**5*f*tan(e/2 + f*x/2)**9 - 
567*c**5*f*tan(e/2 + f*x/2)**8 + 2268*c**5*f*tan(e/2 + f*x/2)**7 - 5292*c* 
*5*f*tan(e/2 + f*x/2)**6 + 7938*c**5*f*tan(e/2 + f*x/2)**5 - 7938*c**5*f*t 
an(e/2 + f*x/2)**4 + 5292*c**5*f*tan(e/2 + f*x/2)**3 - 2268*c**5*f*tan(e/2 
 + f*x/2)**2 + 567*c**5*f*tan(e/2 + f*x/2) - 63*c**5*f) + 126*a**3*tan(e/2 
 + f*x/2)**7/(63*c**5*f*tan(e/2 + f*x/2)**9 - 567*c**5*f*tan(e/2 + f*x/2)* 
*8 + 2268*c**5*f*tan(e/2 + f*x/2)**7 - 5292*c**5*f*tan(e/2 + f*x/2)**6 + 7 
938*c**5*f*tan(e/2 + f*x/2)**5 - 7938*c**5*f*tan(e/2 + f*x/2)**4 + 5292*c* 
*5*f*tan(e/2 + f*x/2)**3 - 2268*c**5*f*tan(e/2 + f*x/2)**2 + 567*c**5*f*ta 
n(e/2 + f*x/2) - 63*c**5*f) - 966*a**3*tan(e/2 + f*x/2)**6/(63*c**5*f*tan( 
e/2 + f*x/2)**9 - 567*c**5*f*tan(e/2 + f*x/2)**8 + 2268*c**5*f*tan(e/2 + f 
*x/2)**7 - 5292*c**5*f*tan(e/2 + f*x/2)**6 + 7938*c**5*f*tan(e/2 + f*x/2)* 
*5 - 7938*c**5*f*tan(e/2 + f*x/2)**4 + 5292*c**5*f*tan(e/2 + f*x/2)**3 - 2 
268*c**5*f*tan(e/2 + f*x/2)**2 + 567*c**5*f*tan(e/2 + f*x/2) - 63*c**5*f) 
+ 630*a**3*tan(e/2 + f*x/2)**5/(63*c**5*f*tan(e/2 + f*x/2)**9 - 567*c**5*f 
*tan(e/2 + f*x/2)**8 + 2268*c**5*f*tan(e/2 + f*x/2)**7 - 5292*c**5*f*tan(e 
/2 + f*x/2)**6 + 7938*c**5*f*tan(e/2 + f*x/2)**5 - 7938*c**5*f*tan(e/2 + f 
*x/2)**4 + 5292*c**5*f*tan(e/2 + f*x/2)**3 - 2268*c**5*f*tan(e/2 + f*x/2)* 
*2 + 567*c**5*f*tan(e/2 + f*x/2) - 63*c**5*f) - 1386*a**3*tan(e/2 + f*x/2) 
**4/(63*c**5*f*tan(e/2 + f*x/2)**9 - 567*c**5*f*tan(e/2 + f*x/2)**8 + 2...
 
3.3.57.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1389 vs. \(2 (67) = 134\).

Time = 0.24 (sec) , antiderivative size = 1389, normalized size of antiderivative = 20.13 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^5} \, dx=\text {Too large to display} \]

input
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x, algorithm="maxima")
 
output
-2/315*(a^3*(432*sin(f*x + e)/(cos(f*x + e) + 1) - 1728*sin(f*x + e)^2/(co 
s(f*x + e) + 1)^2 + 3612*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 5418*sin(f* 
x + e)^4/(cos(f*x + e) + 1)^4 + 5040*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 
 3360*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 1260*sin(f*x + e)^7/(cos(f*x + 
 e) + 1)^7 - 315*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 83)/(c^5 - 9*c^5*si 
n(f*x + e)/(cos(f*x + e) + 1) + 36*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 
 - 84*c^5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 126*c^5*sin(f*x + e)^4/(co 
s(f*x + e) + 1)^4 - 126*c^5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 84*c^5*s 
in(f*x + e)^6/(cos(f*x + e) + 1)^6 - 36*c^5*sin(f*x + e)^7/(cos(f*x + e) + 
 1)^7 + 9*c^5*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - c^5*sin(f*x + e)^9/(co 
s(f*x + e) + 1)^9) - 15*a^3*(45*sin(f*x + e)/(cos(f*x + e) + 1) - 117*sin( 
f*x + e)^2/(cos(f*x + e) + 1)^2 + 273*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 
- 315*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 315*sin(f*x + e)^5/(cos(f*x + 
e) + 1)^5 - 147*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 63*sin(f*x + e)^7/(c 
os(f*x + e) + 1)^7 - 5)/(c^5 - 9*c^5*sin(f*x + e)/(cos(f*x + e) + 1) + 36* 
c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 84*c^5*sin(f*x + e)^3/(cos(f*x + 
 e) + 1)^3 + 126*c^5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 126*c^5*sin(f*x 
 + e)^5/(cos(f*x + e) + 1)^5 + 84*c^5*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 
- 36*c^5*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 9*c^5*sin(f*x + e)^8/(cos(f 
*x + e) + 1)^8 - c^5*sin(f*x + e)^9/(cos(f*x + e) + 1)^9) - 10*a^3*(9*s...
 
3.3.57.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (67) = 134\).

Time = 0.34 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.22 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^5} \, dx=-\frac {2 \, {\left (63 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 63 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 483 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 315 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 693 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 189 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 225 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8 \, a^{3}\right )}}{63 \, c^{5} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{9}} \]

input
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x, algorithm="giac")
 
output
-2/63*(63*a^3*tan(1/2*f*x + 1/2*e)^8 - 63*a^3*tan(1/2*f*x + 1/2*e)^7 + 483 
*a^3*tan(1/2*f*x + 1/2*e)^6 - 315*a^3*tan(1/2*f*x + 1/2*e)^5 + 693*a^3*tan 
(1/2*f*x + 1/2*e)^4 - 189*a^3*tan(1/2*f*x + 1/2*e)^3 + 225*a^3*tan(1/2*f*x 
 + 1/2*e)^2 - 9*a^3*tan(1/2*f*x + 1/2*e) + 8*a^3)/(c^5*f*(tan(1/2*f*x + 1/ 
2*e) - 1)^9)
 
3.3.57.9 Mupad [B] (verification not implemented)

Time = 7.89 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.75 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^5} \, dx=\frac {\sqrt {2}\,a^3\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {37\,\cos \left (3\,e+3\,f\,x\right )}{8}-\frac {63\,\sin \left (e+f\,x\right )}{2}-\frac {113\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {257\,\cos \left (e+f\,x\right )}{8}+\frac {7\,\cos \left (4\,e+4\,f\,x\right )}{16}+\frac {63\,\sin \left (2\,e+2\,f\,x\right )}{8}+\frac {9\,\sin \left (3\,e+3\,f\,x\right )}{2}-\frac {9\,\sin \left (4\,e+4\,f\,x\right )}{16}+\frac {1013}{16}\right )}{1008\,c^5\,f\,{\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f\,x}{2}\right )}^9} \]

input
int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^5,x)
 
output
(2^(1/2)*a^3*cos(e/2 + (f*x)/2)*((37*cos(3*e + 3*f*x))/8 - (63*sin(e + f*x 
))/2 - (113*cos(2*e + 2*f*x))/4 - (257*cos(e + f*x))/8 + (7*cos(4*e + 4*f* 
x))/16 + (63*sin(2*e + 2*f*x))/8 + (9*sin(3*e + 3*f*x))/2 - (9*sin(4*e + 4 
*f*x))/16 + 1013/16))/(1008*c^5*f*cos(e/2 + pi/4 + (f*x)/2)^9)