\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^2} \, dx\) [115]

Optimal result

Integrand size = 38, antiderivative size = 242 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {2048 (7 A-13 B) c^4 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{105 a^2 f}-\frac {512 (7 A-13 B) c^3 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{105 a^2 f}-\frac {64 (7 A-13 B) c^2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{105 a^2 f}-\frac {16 (7 A-13 B) c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{105 a^2 f}-\frac {(7 A-13 B) \sec (e+f x) (c-c \sin (e+f x))^{9/2}}{21 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 a^2 c^2 f} \] Output:

2048/105*(7*A-13*B)*c^4*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f-512/105*(7 
*A-13*B)*c^3*sec(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a^2/f-64/105*(7*A-13*B)*c^2 
*sec(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a^2/f-16/105*(7*A-13*B)*c*sec(f*x+e)*(c 
-c*sin(f*x+e))^(7/2)/a^2/f-1/21*(7*A-13*B)*sec(f*x+e)*(c-c*sin(f*x+e))^(9/ 
2)/a^2/f-1/3*(A-B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(13/2)/a^2/c^2/f
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(953\) vs. \(2(242)=484\).

Time = 16.46 (sec) , antiderivative size = 953, normalized size of antiderivative = 3.94 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(-32*(A - B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^(9 
/2))/(3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a + a*Sin[e + f*x])^2) 
+ (32*(2*A - 3*B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(c - c*Sin[e + f 
*x])^(9/2))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a + a*Sin[e + f*x] 
)^2) + ((164*A - 351*B)*Cos[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x) 
/2])^4*(c - c*Sin[e + f*x])^(9/2))/(4*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/ 
2])^9*(a + a*Sin[e + f*x])^2) + ((26*A - 83*B)*Cos[(3*(e + f*x))/2]*(Cos[( 
e + f*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(9/2))/(12*f*(Cos[( 
e + f*x)/2] - Sin[(e + f*x)/2])^9*(a + a*Sin[e + f*x])^2) - ((2*A - 13*B)* 
Cos[(5*(e + f*x))/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e 
+ f*x])^(9/2))/(20*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a + a*Sin[e 
+ f*x])^2) + (B*Cos[(7*(e + f*x))/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) 
^4*(c - c*Sin[e + f*x])^(9/2))/(28*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) 
^9*(a + a*Sin[e + f*x])^2) + ((164*A - 351*B)*Sin[(e + f*x)/2]*(Cos[(e + f 
*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(9/2))/(4*f*(Cos[(e + f* 
x)/2] - Sin[(e + f*x)/2])^9*(a + a*Sin[e + f*x])^2) - ((26*A - 83*B)*(Cos[ 
(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(9/2)*Sin[(3*(e + 
f*x))/2])/(12*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a + a*Sin[e + f*x 
])^2) - ((2*A - 13*B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e 
 + f*x])^(9/2)*Sin[(5*(e + f*x))/2])/(20*f*(Cos[(e + f*x)/2] - Sin[(e +...
 

Rubi [A] (verified)

Time = 1.38 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.90, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{11/2}dx-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \int \frac {(c-c \sin (e+f x))^{11/2}}{\cos (e+f x)^2}dx-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \left (\frac {16}{7} c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{9/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \left (\frac {16}{7} c \int \frac {(c-c \sin (e+f x))^{9/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \left (\frac {16}{7} c \left (\frac {12}{5} c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{7/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \left (\frac {16}{7} c \left (\frac {12}{5} c \int \frac {(c-c \sin (e+f x))^{7/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \left (\frac {16}{7} c \left (\frac {12}{5} c \left (\frac {8}{3} c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \left (\frac {16}{7} c \left (\frac {12}{5} c \left (\frac {8}{3} c \int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \left (\frac {16}{7} c \left (\frac {12}{5} c \left (\frac {8}{3} c \left (4 c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \left (\frac {16}{7} c \left (\frac {12}{5} c \left (\frac {8}{3} c \left (4 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {-\frac {1}{6} c (7 A-13 B) \left (\frac {16}{7} c \left (\frac {12}{5} c \left (\frac {8}{3} c \left (\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}-\frac {8 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{13/2}}{3 f}}{a^2 c^2}\)

Input:

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

Output:

(-1/3*((A - B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(13/2))/f - ((7*A - 13* 
B)*c*((2*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(7*f) + (16*c*((2*c*Se 
c[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(5*f) + (12*c*((2*c*Sec[e + f*x]*(c 
 - c*Sin[e + f*x])^(5/2))/(3*f) + (8*c*((-8*c^2*Sec[e + f*x]*Sqrt[c - c*Si 
n[e + f*x]])/f + (2*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/f))/3))/5)) 
/7))/6)/(a^2*c^2)
 

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 = 20.97 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.59

method	result	size
default	\(-\frac {2 c^{5} \left (\sin \left (f x +e \right )-1\right ) \left (15 B \sin \left (f x +e \right ) \cos \left (f x +e \right )^{4}+\left (21 A -114 B \right ) \cos \left (f x +e \right )^{4}+\left (196 A -544 B \right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+\left (-1848 A +3732 B \right ) \cos \left (f x +e \right )^{2}+\left (7448 A -13592 B \right ) \sin \left (f x +e \right )+6888 A -13032 B \right )}{105 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(143\)

Input:

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

Output:

-2/105*c^5/a^2*(sin(f*x+e)-1)/(1+sin(f*x+e))*(15*B*sin(f*x+e)*cos(f*x+e)^4 
+(21*A-114*B)*cos(f*x+e)^4+(196*A-544*B)*cos(f*x+e)^2*sin(f*x+e)+(-1848*A+ 
3732*B)*cos(f*x+e)^2+(7448*A-13592*B)*sin(f*x+e)+6888*A-13032*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 = 153, normalized size of antiderivative = 0.63 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {2 \, {\left (3 \, {\left (7 \, A - 38 \, B\right )} c^{4} \cos \left (f x + e\right )^{4} - 12 \, {\left (154 \, A - 311 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} + 24 \, {\left (287 \, A - 543 \, B\right )} c^{4} + {\left (15 \, B c^{4} \cos \left (f x + e\right )^{4} + 4 \, {\left (49 \, A - 136 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} + 8 \, {\left (931 \, A - 1699 \, B\right )} c^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{105 \, {\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}} \] Input:

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

Output:

2/105*(3*(7*A - 38*B)*c^4*cos(f*x + e)^4 - 12*(154*A - 311*B)*c^4*cos(f*x 
+ e)^2 + 24*(287*A - 543*B)*c^4 + (15*B*c^4*cos(f*x + e)^4 + 4*(49*A - 136 
*B)*c^4*cos(f*x + e)^2 + 8*(931*A - 1699*B)*c^4)*sin(f*x + e))*sqrt(-c*sin 
(f*x + e) + c)/(a^2*f*cos(f*x + e)*sin(f*x + e) + a^2*f*cos(f*x + e))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 762 vs. \(2 (218) = 436\).

Time = 0.15 (sec) , antiderivative size = 762, normalized size of antiderivative = 3.15 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-2/105*(7*(723*c^(9/2) + 2184*c^(9/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 53 
70*c^(9/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10696*c^(9/2)*sin(f*x + e 
)^3/(cos(f*x + e) + 1)^3 + 15021*c^(9/2)*sin(f*x + e)^4/(cos(f*x + e) + 1) 
^4 + 21168*c^(9/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 20748*c^(9/2)*sin 
(f*x + e)^6/(cos(f*x + e) + 1)^6 + 21168*c^(9/2)*sin(f*x + e)^7/(cos(f*x + 
 e) + 1)^7 + 15021*c^(9/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 10696*c^( 
9/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 5370*c^(9/2)*sin(f*x + e)^10/(c 
os(f*x + e) + 1)^10 + 2184*c^(9/2)*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 
 723*c^(9/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12)*A/((a^2 + 3*a^2*sin(f* 
x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^ 
2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^ 
2 + 1)^(9/2)) - 2*(4707*c^(9/2) + 14121*c^(9/2)*sin(f*x + e)/(cos(f*x + e) 
 + 1) + 35250*c^(9/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 68549*c^(9/2)* 
sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 99549*c^(9/2)*sin(f*x + e)^4/(cos(f* 
x + e) + 1)^4 + 134802*c^(9/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 13801 
2*c^(9/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 134802*c^(9/2)*sin(f*x + e 
)^7/(cos(f*x + e) + 1)^7 + 99549*c^(9/2)*sin(f*x + e)^8/(cos(f*x + e) + 1) 
^8 + 68549*c^(9/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 35250*c^(9/2)*sin 
(f*x + e)^10/(cos(f*x + e) + 1)^10 + 14121*c^(9/2)*sin(f*x + e)^11/(cos(f* 
x + e) + 1)^11 + 4707*c^(9/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12)*B/...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 978 vs. \(2 (218) = 436\).

Time = 0.45 (sec) , antiderivative size = 978, normalized size of antiderivative = 4.04 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-16/105*sqrt(2)*sqrt(c)*(35*(11*A*c^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) 
- 17*B*c^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 24*A*c^4*(cos(-1/4*pi + 1 
/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/ 
2*f*x + 1/2*e) + 1) - 36*B*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(si 
n(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 9*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)/(a^2*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4* 
pi + 1/2*f*x + 1/2*e) + 1) + 1)^3) - (511*A*c^4*sgn(sin(-1/4*pi + 1/2*f*x 
+ 1/2*e)) - 1069*B*c^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 3262*A*c^4*(c 
os(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*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) + 6958*B*c^4*(cos(-1/4*pi + 1/2*f*x + 1/ 
2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2 
*e) + 1) + 8421*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 - 18459*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 - 10780*A*c^4*(cos(-1/4*pi + 1/2 
*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/ 
2*f*x + 1/2*e) + 1)^3 + 24220*B*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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