\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx\) [123]

Optimal result

Integrand size = 38, antiderivative size = 242 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {2048 (A-3 B) c^3 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {512 (A-3 B) c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {64 (A-3 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f} \] Output:

-2048/15*(A-3*B)*c^3*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/f+512/5*(A-3* 
B)*c^2*sec(f*x+e)^3*(c-c*sin(f*x+e))^(5/2)/a^3/f-64/5*(A-3*B)*c*sec(f*x+e) 
^3*(c-c*sin(f*x+e))^(7/2)/a^3/f-16/15*(A-3*B)*sec(f*x+e)^3*(c-c*sin(f*x+e) 
)^(9/2)/a^3/f-1/5*(A-3*B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(11/2)/a^3/c/f-1/5 
*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e))^(15/2)/a^3/c^3/f
 

Mathematica [A] (verified)

Time = 13.76 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {c^4 (-1+\sin (e+f x))^4 \sqrt {c-c \sin (e+f x)} (11298 A-33516 B-40 (137 A-402 B) \cos (2 (e+f x))-10 (A-6 B) \cos (4 (e+f x))+15600 A \sin (e+f x)-47430 B \sin (e+f x)-400 A \sin (3 (e+f x))+1335 B \sin (3 (e+f x))-3 B \sin (5 (e+f x)))}{120 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \] Input:

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(9/2))/(a + a*Sin[e + 
 f*x])^3,x]
 

Output:

-1/120*(c^4*(-1 + Sin[e + f*x])^4*Sqrt[c - c*Sin[e + f*x]]*(11298*A - 3351 
6*B - 40*(137*A - 402*B)*Cos[2*(e + f*x)] - 10*(A - 6*B)*Cos[4*(e + f*x)] 
+ 15600*A*Sin[e + f*x] - 47430*B*Sin[e + f*x] - 400*A*Sin[3*(e + f*x)] + 1 
335*B*Sin[3*(e + f*x)] - 3*B*Sin[5*(e + f*x)]))/(a^3*f*(Cos[(e + f*x)/2] - 
 Sin[(e + f*x)/2])^9*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)
 

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.91, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 3446, 3042, 3334, 3042, 3153, 3042, 3153, 3042, 3153, 3042, 3153, 3042, 3152}

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.

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{9/2} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{9/2} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^3}dx\)

\(\Big \downarrow \) 3446

\(\displaystyle \frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{15/2}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{15/2}}{\cos (e+f x)^6}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3334

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{13/2}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \int \frac {(c-c \sin (e+f x))^{13/2}}{\cos (e+f x)^4}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {16}{5} c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{11/2}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {16}{5} c \int \frac {(c-c \sin (e+f x))^{11/2}}{\cos (e+f x)^4}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {16}{5} c \left (4 c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {16}{5} c \left (4 c \int \frac {(c-c \sin (e+f x))^{9/2}}{\cos (e+f x)^4}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {16}{5} c \left (4 c \left (8 c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {16}{5} c \left (4 c \left (8 c \int \frac {(c-c \sin (e+f x))^{7/2}}{\cos (e+f x)^4}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {16}{5} c \left (4 c \left (8 c \left (-4 c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2}dx-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {16}{5} c \left (4 c \left (8 c \left (-4 c \int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^4}dx-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {16}{5} c \left (4 c \left (8 c \left (\frac {8 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 f}}{a^3 c^3}\)

Input:

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(9/2))/(a + a*Sin[e + f*x]) 
^3,x]
 

Output:

(-1/5*((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(15/2))/f - ((A - 3*B)* 
c*((2*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(11/2))/(5*f) + (16*c*((2*c*Se 
c[e + f*x]^3*(c - c*Sin[e + f*x])^(9/2))/(3*f) + 4*c*((2*c*Sec[e + f*x]^3* 
(c - c*Sin[e + f*x])^(7/2))/f + 8*c*((8*c^2*Sec[e + f*x]^3*(c - c*Sin[e + 
f*x])^(3/2))/(3*f) - (2*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(5/2))/f)))) 
/5))/2)/(a^3*c^3)
 

Defintions of rubi rules used

rule 3042

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

rule 3152

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

rule 3153

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 - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p))   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] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && 
NeQ[m + p, 0]
 

rule 3334

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

rule 3446

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si 
mp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin 
[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* 
d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] 
&& GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [A] (verified)

Time = 126.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.59

method	result	size
default	\(-\frac {2 c^{5} \left (\sin \left (f x +e \right )-1\right ) \left (3 B \sin \left (f x +e \right ) \cos \left (f x +e \right )^{4}+\left (5 A -30 B \right ) \cos \left (f x +e \right )^{4}+\left (100 A -336 B \right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+\left (680 A -1980 B \right ) \cos \left (f x +e \right )^{2}+\left (-1000 A +3048 B \right ) \sin \left (f x +e \right )-1048 A +3096 B \right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(143\)

Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x,method=_R 
ETURNVERBOSE)
 

Output:

-2/15*c^5/a^3*(sin(f*x+e)-1)/(1+sin(f*x+e))^2*(3*B*sin(f*x+e)*cos(f*x+e)^4 
+(5*A-30*B)*cos(f*x+e)^4+(100*A-336*B)*cos(f*x+e)^2*sin(f*x+e)+(680*A-1980 
*B)*cos(f*x+e)^2+(-1000*A+3048*B)*sin(f*x+e)-1048*A+3096*B)/cos(f*x+e)/(c- 
c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (5 \, {\left (A - 6 \, B\right )} c^{4} \cos \left (f x + e\right )^{4} + 20 \, {\left (34 \, A - 99 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} - 8 \, {\left (131 \, A - 387 \, B\right )} c^{4} + {\left (3 \, B c^{4} \cos \left (f x + e\right )^{4} + 4 \, {\left (25 \, A - 84 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} - 8 \, {\left (125 \, A - 381 \, B\right )} c^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x, al 
gorithm="fricas")
 

Output:

-2/15*(5*(A - 6*B)*c^4*cos(f*x + e)^4 + 20*(34*A - 99*B)*c^4*cos(f*x + e)^ 
2 - 8*(131*A - 387*B)*c^4 + (3*B*c^4*cos(f*x + e)^4 + 4*(25*A - 84*B)*c^4* 
cos(f*x + e)^2 - 8*(125*A - 381*B)*c^4)*sin(f*x + e))*sqrt(-c*sin(f*x + e) 
 + c)/(a^3*f*cos(f*x + e)^3 - 2*a^3*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*f* 
cos(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(9/2)/(a+a*sin(f*x+e))**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 945 vs. \(2 (218) = 436\).

Time = 0.15 (sec) , antiderivative size = 945, normalized size of antiderivative = 3.90 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x, al 
gorithm="maxima")
 

Output:

2/15*((363*c^(9/2) + 1800*c^(9/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 5301*c 
^(9/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 11600*c^(9/2)*sin(f*x + e)^3/ 
(cos(f*x + e) + 1)^3 + 21343*c^(9/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 
 30200*c^(9/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 40065*c^(9/2)*sin(f*x 
 + e)^6/(cos(f*x + e) + 1)^6 + 40800*c^(9/2)*sin(f*x + e)^7/(cos(f*x + e) 
+ 1)^7 + 40065*c^(9/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 30200*c^(9/2) 
*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 21343*c^(9/2)*sin(f*x + e)^10/(cos( 
f*x + e) + 1)^10 + 11600*c^(9/2)*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 5 
301*c^(9/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 1800*c^(9/2)*sin(f*x + 
 e)^13/(cos(f*x + e) + 1)^13 + 363*c^(9/2)*sin(f*x + e)^14/(cos(f*x + e) + 
 1)^14)*A/((a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + 
 e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 
5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + 
e) + 1)^5)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(9/2)) - 6*(181*c^(9/ 
2) + 905*c^(9/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 2627*c^(9/2)*sin(f*x + 
e)^2/(cos(f*x + e) + 1)^2 + 5870*c^(9/2)*sin(f*x + e)^3/(cos(f*x + e) + 1) 
^3 + 10521*c^(9/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 15351*c^(9/2)*sin 
(f*x + e)^5/(cos(f*x + e) + 1)^5 + 19695*c^(9/2)*sin(f*x + e)^6/(cos(f*x + 
 e) + 1)^6 + 20772*c^(9/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 19695*c^( 
9/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 15351*c^(9/2)*sin(f*x + e)^9...
 

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x, al 
gorithm="giac")
 

Output:

-1024/15*sqrt(2)*(A*c^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*B*c^4*sgn( 
sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*A*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) 
- 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) 
 + 1)^2 + 15*B*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi 
+ 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 10*A*c^4*(cos 
(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(co 
s(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4 - 30*B*c^4*(cos(-1/4*pi + 1/2*f*x + 1/ 
2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1 
/2*e) + 1)^4 - 6*A*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^5*sgn(sin(-1/4 
*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^5 - 6*B*c^4*( 
cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^5*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/ 
(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^5)*sqrt(c)/(a^3*f*((cos(-1/4*pi + 1/2 
*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 1)^5)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3} \,d x \] Input:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(9/2))/(a + a*sin(e + f*x)) 
^3,x)
 

Output:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(9/2))/(a + a*sin(e + f*x)) 
^3, x)
 

Reduce [F]

\[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {\sqrt {c}\, c^{4} \left (\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) a +\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{5}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) b +\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) a -4 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) b -4 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) a +6 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) b +6 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) a -4 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) b -4 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) a +\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) b \right )}{a^{3}} \] Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x)
 

Output:

(sqrt(c)*c**4*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**3 + 3*sin(e + 
f*x)**2 + 3*sin(e + f*x) + 1),x)*a + int((sqrt( - sin(e + f*x) + 1)*sin(e 
+ f*x)**5)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*b 
 + int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**4)/(sin(e + f*x)**3 + 3*si 
n(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*a - 4*int((sqrt( - sin(e + f*x) + 1 
)*sin(e + f*x)**4)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 
 1),x)*b - 4*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3)/(sin(e + f*x) 
**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*a + 6*int((sqrt( - sin(e 
+ f*x) + 1)*sin(e + f*x)**3)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin( 
e + f*x) + 1),x)*b + 6*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(si 
n(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*a - 4*int((sqrt 
( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**3 + 3*sin(e + f*x)** 
2 + 3*sin(e + f*x) + 1),x)*b - 4*int((sqrt( - sin(e + f*x) + 1)*sin(e + f* 
x))/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*a + int( 
(sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**3 + 3*sin(e + f*x) 
**2 + 3*sin(e + f*x) + 1),x)*b))/a**3