\(\int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^4} \, dx\) [252]

Optimal result

Integrand size = 26, antiderivative size = 67 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^4} \, dx=\frac {a^2 c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}+\frac {a^2 c \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^5} \] Output:

1/7*a^2*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^6+1/35*a^2*c*cos(f*x+e)^5/f/(c 
-c*sin(f*x+e))^5
 

Mathematica [A] (verified)

Time = 3.92 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.75 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^4} \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-35 \cos \left (\frac {1}{2} (e+f x)\right )+14 \cos \left (\frac {3}{2} (e+f x)\right )+\cos \left (\frac {7}{2} (e+f x)\right )-70 \sin \left (\frac {1}{2} (e+f x)\right )-35 \sin \left (\frac {3}{2} (e+f x)\right )+7 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{140 c^4 f (-1+\sin (e+f x))^4} \] Input:

Integrate[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^4,x]
 

Output:

-1/140*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(-35*Cos[(e + f*x)/2] + 
14*Cos[(3*(e + f*x))/2] + Cos[(7*(e + f*x))/2] - 70*Sin[(e + f*x)/2] - 35* 
Sin[(3*(e + f*x))/2] + 7*Sin[(5*(e + f*x))/2]))/(c^4*f*(-1 + Sin[e + f*x]) 
^4)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

\(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^6}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^6}dx\)

\(\Big \downarrow \) 3151

\(\displaystyle a^2 c^2 \left (\frac {\int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^5}dx}{7 c}+\frac {\cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {\int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^5}dx}{7 c}+\frac {\cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}\right )\)

\(\Big \downarrow \) 3150

\(\displaystyle a^2 c^2 \left (\frac {\cos ^5(e+f x)}{35 c f (c-c \sin (e+f x))^5}+\frac {\cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}\right )\)

Input:

Int[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^4,x]
 

Output:

a^2*c^2*(Cos[e + f*x]^5/(7*f*(c - c*Sin[e + f*x])^6) + Cos[e + f*x]^5/(35* 
c*f*(c - c*Sin[e + f*x])^5))
 

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]))
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.91 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.30

method	result	size
risch	\(-\frac {2 i a^{2} \left (35 i {\mathrm e}^{4 i \left (f x +e \right )}+35 \,{\mathrm e}^{5 i \left (f x +e \right )}-14 i {\mathrm e}^{2 i \left (f x +e \right )}-70 \,{\mathrm e}^{3 i \left (f x +e \right )}-i+7 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{35 f \,c^{4} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{7}}\)	\(87\)
parallelrisch	\(-\frac {2 a^{2} \left (35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+140 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-70 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+91 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+6\right )}{35 f \,c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\)	\(103\)
derivativedivides	\(\frac {2 a^{2} \left (-\frac {5}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {32}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {16}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {14}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {128}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {24}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}\right )}{f \,c^{4}}\)	\(118\)
default	\(\frac {2 a^{2} \left (-\frac {5}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {32}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {16}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {14}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {128}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {24}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}\right )}{f \,c^{4}}\)	\(118\)
norman	\(\frac {\frac {2 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f c}-\frac {12 a^{2}}{35 c f}-\frac {2 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{f c}+\frac {2 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 c f}+\frac {8 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f c}-\frac {12 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{f c}+\frac {24 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{5 c f}+\frac {52 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5 c f}-\frac {116 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{5 f c}-\frac {206 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{35 c f}-\frac {656 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{35 c f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2} c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\)	\(263\)

Input:

int((a+sin(f*x+e)*a)^2/(c-c*sin(f*x+e))^4,x,method=_RETURNVERBOSE)
 

Output:

-2/35*I*a^2*(35*I*exp(4*I*(f*x+e))+35*exp(5*I*(f*x+e))-14*I*exp(2*I*(f*x+e 
))-70*exp(3*I*(f*x+e))-I+7*exp(I*(f*x+e)))/f/c^4/(exp(I*(f*x+e))-I)^7
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (65) = 130\).

Time = 0.08 (sec) , antiderivative size = 222, normalized size of antiderivative = 3.31 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^4} \, dx=-\frac {a^{2} \cos \left (f x + e\right )^{4} + 4 \, a^{2} \cos \left (f x + e\right )^{3} + 13 \, a^{2} \cos \left (f x + e\right )^{2} - 10 \, a^{2} \cos \left (f x + e\right ) - 20 \, a^{2} - {\left (a^{2} \cos \left (f x + e\right )^{3} - 3 \, a^{2} \cos \left (f x + e\right )^{2} + 10 \, a^{2} \cos \left (f x + e\right ) + 20 \, a^{2}\right )} \sin \left (f x + e\right )}{35 \, {\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f + {\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algorithm="fricas")
 

Output:

-1/35*(a^2*cos(f*x + e)^4 + 4*a^2*cos(f*x + e)^3 + 13*a^2*cos(f*x + e)^2 - 
 10*a^2*cos(f*x + e) - 20*a^2 - (a^2*cos(f*x + e)^3 - 3*a^2*cos(f*x + e)^2 
 + 10*a^2*cos(f*x + e) + 20*a^2)*sin(f*x + e))/(c^4*f*cos(f*x + e)^4 - 3*c 
^4*f*cos(f*x + e)^3 - 8*c^4*f*cos(f*x + e)^2 + 4*c^4*f*cos(f*x + e) + 8*c^ 
4*f + (c^4*f*cos(f*x + e)^3 + 4*c^4*f*cos(f*x + e)^2 - 4*c^4*f*cos(f*x + e 
) - 8*c^4*f)*sin(f*x + e))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1074 vs. \(2 (58) = 116\).

Time = 8.78 (sec) , antiderivative size = 1074, normalized size of antiderivative = 16.03 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**4,x)
 

Output:

Piecewise((-70*a**2*tan(e/2 + f*x/2)**6/(35*c**4*f*tan(e/2 + f*x/2)**7 - 2 
45*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4 
*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan( 
e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) + 70*a**2*tan(e 
/2 + f*x/2)**5/(35*c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2 
)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 + 
1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2)**2 + 245*c** 
4*f*tan(e/2 + f*x/2) - 35*c**4*f) - 280*a**2*tan(e/2 + f*x/2)**4/(35*c**4* 
f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e/ 
2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/2 + f* 
x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 3 
5*c**4*f) + 140*a**2*tan(e/2 + f*x/2)**3/(35*c**4*f*tan(e/2 + f*x/2)**7 - 
245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c** 
4*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan 
(e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) - 182*a**2*tan 
(e/2 + f*x/2)**2/(35*c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x 
/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 
+ 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2)**2 + 245*c 
**4*f*tan(e/2 + f*x/2) - 35*c**4*f) + 14*a**2*tan(e/2 + f*x/2)/(35*c**4*f* 
tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 816 vs. \(2 (65) = 130\).

Time = 0.06 (sec) , antiderivative size = 816, normalized size of antiderivative = 12.18 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algorithm="maxima")
 

Output:

2/105*(2*a^2*(91*sin(f*x + e)/(cos(f*x + e) + 1) - 168*sin(f*x + e)^2/(cos 
(f*x + e) + 1)^2 + 280*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 175*sin(f*x + 
 e)^4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 13) 
/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos 
(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin 
(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1 
)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos( 
f*x + e) + 1)^7) - 3*a^2*(49*sin(f*x + e)/(cos(f*x + e) + 1) - 147*sin(f*x 
 + e)^2/(cos(f*x + e) + 1)^2 + 210*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 2 
10*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) 
+ 1)^5 - 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 12)/(c^4 - 7*c^4*sin(f*x 
 + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35 
*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x 
+ e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7*c^4*sin(f*x + 
 e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) - 4* 
a^2*(14*sin(f*x + e)/(cos(f*x + e) + 1) - 42*sin(f*x + e)^2/(cos(f*x + e) 
+ 1)^2 + 35*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 35*sin(f*x + e)^4/(cos(f 
*x + e) + 1)^4 - 2)/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4* 
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(cos(f*x + e) 
+ 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + ...
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.81 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^4} \, dx=-\frac {2 \, {\left (35 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 35 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 140 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 70 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 91 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 7 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, a^{2}\right )}}{35 \, c^{4} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}} \] Input:

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algorithm="giac")
 

Output:

-2/35*(35*a^2*tan(1/2*f*x + 1/2*e)^6 - 35*a^2*tan(1/2*f*x + 1/2*e)^5 + 140 
*a^2*tan(1/2*f*x + 1/2*e)^4 - 70*a^2*tan(1/2*f*x + 1/2*e)^3 + 91*a^2*tan(1 
/2*f*x + 1/2*e)^2 - 7*a^2*tan(1/2*f*x + 1/2*e) + 6*a^2)/(c^4*f*(tan(1/2*f* 
x + 1/2*e) - 1)^7)
 

Mupad [B] (verification not implemented)

Time = 17.93 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.48 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^4} \, dx=\frac {\sqrt {2}\,a^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {5\,\cos \left (3\,e+3\,f\,x\right )}{8}-\frac {105\,\sin \left (e+f\,x\right )}{8}-\frac {27\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {121\,\cos \left (e+f\,x\right )}{8}+\frac {7\,\sin \left (2\,e+2\,f\,x\right )}{2}+\frac {7\,\sin \left (3\,e+3\,f\,x\right )}{8}+\frac {109}{4}\right )}{280\,c^4\,f\,{\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f\,x}{2}\right )}^7} \] Input:

int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^4,x)
 

Output:

(2^(1/2)*a^2*cos(e/2 + (f*x)/2)*((5*cos(3*e + 3*f*x))/8 - (105*sin(e + f*x 
))/8 - (27*cos(2*e + 2*f*x))/4 - (121*cos(e + f*x))/8 + (7*sin(2*e + 2*f*x 
))/2 + (7*sin(3*e + 3*f*x))/8 + 109/4))/(280*c^4*f*cos(e/2 + pi/4 + (f*x)/ 
2)^7)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.69 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^4} \, dx=\frac {2 a^{2} \left (-5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}-70 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-105 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+14 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-28 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{35 c^{4} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}-7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )} \] Input:

int((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x)
 

Output:

(2*a**2*( - 5*tan((e + f*x)/2)**7 - 70*tan((e + f*x)/2)**5 + 35*tan((e + f 
*x)/2)**4 - 105*tan((e + f*x)/2)**3 + 14*tan((e + f*x)/2)**2 - 28*tan((e + 
 f*x)/2) - 1))/(35*c**4*f*(tan((e + f*x)/2)**7 - 7*tan((e + f*x)/2)**6 + 2 
1*tan((e + f*x)/2)**5 - 35*tan((e + f*x)/2)**4 + 35*tan((e + f*x)/2)**3 - 
21*tan((e + f*x)/2)**2 + 7*tan((e + f*x)/2) - 1))