3.1.81 \(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx\) [81]

3.1.81.1 Optimal result
3.1.81.2 Mathematica [A] (verified)
3.1.81.3 Rubi [A] (verified)
3.1.81.4 Maple [A] (verified)
3.1.81.5 Fricas [A] (verification not implemented)
3.1.81.6 Sympy [F(-1)]
3.1.81.7 Maxima [F]
3.1.81.8 Giac [A] (verification not implemented)
3.1.81.9 Mupad [F(-1)]

3.1.81.1 Optimal result

Integrand size = 36, antiderivative size = 198 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {256 a (11 A-5 B) c^5 \cos ^3(e+f x)}{3465 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a (11 A-5 B) c^4 \cos ^3(e+f x)}{1155 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f} \]

output
256/3465*a*(11*A-5*B)*c^5*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+2/99*a*(11 
*A-5*B)*c^2*cos(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/f-2/11*a*B*c*cos(f*x+e)^3* 
(c-c*sin(f*x+e))^(5/2)/f+64/1155*a*(11*A-5*B)*c^4*cos(f*x+e)^3/f/(c-c*sin( 
f*x+e))^(1/2)+8/231*a*(11*A-5*B)*c^3*cos(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/f
 
3.1.81.2 Mathematica [A] (verified)

Time = 4.16 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.75 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c-c \sin (e+f x)} (-35332 A+27085 B+60 (121 A-202 B) \cos (2 (e+f x))+315 B \cos (4 (e+f x))+30558 A \sin (e+f x)-31530 B \sin (e+f x)-770 A \sin (3 (e+f x))+2870 B \sin (3 (e+f x)))}{13860 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

input
Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^( 
7/2),x]
 
output
-1/13860*(a*c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + 
 f*x]]*(-35332*A + 27085*B + 60*(121*A - 202*B)*Cos[2*(e + f*x)] + 315*B*C 
os[4*(e + f*x)] + 30558*A*Sin[e + f*x] - 31530*B*Sin[e + f*x] - 770*A*Sin[ 
3*(e + f*x)] + 2870*B*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f 
*x)/2]))
 
3.1.81.3 Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3446, 3042, 3335, 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.

\(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^{7/2} (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3446

\(\displaystyle a c \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \int \cos (e+f x)^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}dx\)

\(\Big \downarrow \) 3335

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3153

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3153

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3153

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3152

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

input
Int[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2),x 
]
 
output
a*c*((-2*B*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(5/2))/(11*f) + ((11*A - 5* 
B)*((2*c*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(9*f) + (4*c*((2*c*Cos 
[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(7*f) + (8*c*((8*c^2*Cos[e + f*x]^3) 
/(15*f*(c - c*Sin[e + f*x])^(3/2)) + (2*c*Cos[e + f*x]^3)/(5*f*Sqrt[c - c* 
Sin[e + f*x]])))/7))/3))/11)
 

3.1.81.3.1 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 3335
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* 
(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S 
imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1))   Int[(g*Cos[e + f*x])^p*(a + 
 b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ 
a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1, 0]
 

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]))
 
3.1.81.4 Maple [A] (verified)

Time = 7.46 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.60

method result size
default \(\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right )^{2} a \left (315 B \left (\cos ^{4}\left (f x +e \right )\right )+\left (-385 A +1435 B \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (1815 A -3345 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (3916 A -4300 B \right ) \sin \left (f x +e \right )-5324 A +4940 B \right )}{3465 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) \(119\)
parts \(\frac {2 a A \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right ) \left (5 \left (\sin ^{3}\left (f x +e \right )\right )-27 \left (\sin ^{2}\left (f x +e \right )\right )+71 \sin \left (f x +e \right )-177\right )}{35 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 B a \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right ) \left (315 \left (\sin ^{5}\left (f x +e \right )\right )-1505 \left (\sin ^{4}\left (f x +e \right )\right )+3205 \left (\sin ^{3}\left (f x +e \right )\right )-4539 \left (\sin ^{2}\left (f x +e \right )\right )+6052 \sin \left (f x +e \right )-12104\right )}{3465 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a \left (A +B \right ) \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right ) \left (5 \left (\sin ^{4}\left (f x +e \right )\right )-25 \left (\sin ^{3}\left (f x +e \right )\right )+57 \left (\sin ^{2}\left (f x +e \right )\right )-91 \sin \left (f x +e \right )+182\right )}{45 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) \(265\)

input
int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x,method=_RET 
URNVERBOSE)
 
output
2/3465*(sin(f*x+e)-1)*c^4*(1+sin(f*x+e))^2*a*(315*B*cos(f*x+e)^4+(-385*A+1 
435*B)*cos(f*x+e)^2*sin(f*x+e)+(1815*A-3345*B)*cos(f*x+e)^2+(3916*A-4300*B 
)*sin(f*x+e)-5324*A+4940*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 
3.1.81.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.45 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {2 \, {\left (315 \, B a c^{3} \cos \left (f x + e\right )^{6} - 35 \, {\left (11 \, A - 32 \, B\right )} a c^{3} \cos \left (f x + e\right )^{5} + 5 \, {\left (209 \, A - 221 \, B\right )} a c^{3} \cos \left (f x + e\right )^{4} + 2 \, {\left (1243 \, A - 1195 \, B\right )} a c^{3} \cos \left (f x + e\right )^{3} - 32 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right )^{2} + 128 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right ) + 256 \, {\left (11 \, A - 5 \, B\right )} a c^{3} - {\left (315 \, B a c^{3} \cos \left (f x + e\right )^{5} + 35 \, {\left (11 \, A - 23 \, B\right )} a c^{3} \cos \left (f x + e\right )^{4} + 10 \, {\left (143 \, A - 191 \, B\right )} a c^{3} \cos \left (f x + e\right )^{3} - 96 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right )^{2} - 128 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right ) - 256 \, {\left (11 \, A - 5 \, B\right )} a c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3465 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

input
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x, algo 
rithm="fricas")
 
output
2/3465*(315*B*a*c^3*cos(f*x + e)^6 - 35*(11*A - 32*B)*a*c^3*cos(f*x + e)^5 
 + 5*(209*A - 221*B)*a*c^3*cos(f*x + e)^4 + 2*(1243*A - 1195*B)*a*c^3*cos( 
f*x + e)^3 - 32*(11*A - 5*B)*a*c^3*cos(f*x + e)^2 + 128*(11*A - 5*B)*a*c^3 
*cos(f*x + e) + 256*(11*A - 5*B)*a*c^3 - (315*B*a*c^3*cos(f*x + e)^5 + 35* 
(11*A - 23*B)*a*c^3*cos(f*x + e)^4 + 10*(143*A - 191*B)*a*c^3*cos(f*x + e) 
^3 - 96*(11*A - 5*B)*a*c^3*cos(f*x + e)^2 - 128*(11*A - 5*B)*a*c^3*cos(f*x 
 + e) - 256*(11*A - 5*B)*a*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f 
*cos(f*x + e) - f*sin(f*x + e) + f)
 
3.1.81.6 Sympy [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\text {Timed out} \]

input
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2),x)
 
output
Timed out
 
3.1.81.7 Maxima [F]

\[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \,d x } \]

input
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x, algo 
rithm="maxima")
 
output
integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^ 
(7/2), x)
 
3.1.81.8 Giac [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.49 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {\sqrt {2} {\left (6930 \, B a c^{3} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 315 \, B a c^{3} \cos \left (-\frac {11}{4} \, \pi + \frac {11}{2} \, f x + \frac {11}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 48510 \, {\left (2 \, A a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 693 \, {\left (16 \, A a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, B a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) + 495 \, {\left (10 \, A a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 9 \, B a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) - 385 \, {\left (2 \, A a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, B a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right )\right )} \sqrt {c}}{55440 \, f} \]

input
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x, algo 
rithm="giac")
 
output
-1/55440*sqrt(2)*(6930*B*a*c^3*cos(-3/4*pi + 3/2*f*x + 3/2*e)*sgn(sin(-1/4 
*pi + 1/2*f*x + 1/2*e)) - 315*B*a*c^3*cos(-11/4*pi + 11/2*f*x + 11/2*e)*sg 
n(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 48510*(2*A*a*c^3*sgn(sin(-1/4*pi + 1/2 
*f*x + 1/2*e)) - B*a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-1/4*pi 
+ 1/2*f*x + 1/2*e) - 693*(16*A*a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 
 5*B*a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-5/4*pi + 5/2*f*x + 5/ 
2*e) + 495*(10*A*a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 9*B*a*c^3*sgn 
(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-7/4*pi + 7/2*f*x + 7/2*e) - 385*(2* 
A*a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*B*a*c^3*sgn(sin(-1/4*pi + 
1/2*f*x + 1/2*e)))*cos(-9/4*pi + 9/2*f*x + 9/2*e))*sqrt(c)/f
 
3.1.81.9 Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2} \,d x \]

input
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(7/2),x 
)
 
output
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(7/2), 
x)