\(\int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx\) [161]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 40, antiderivative size = 226 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {2 a^4 A \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{35 f \sqrt {a+a \sin (e+f x)}}-\frac {4 a^3 A \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{35 f}-\frac {a^2 A \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{7 f}-\frac {a A \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f} \] Output:

-2/35*a^4*A*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/f/(a+a*sin(f*x+e))^(1/2)-4/3 
5*a^3*A*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(7/2)/f-1/7*a^2 
*A*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2)/f-1/7*a*A*cos( 
f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2)/f-1/8*B*cos(f*x+e)*(a 
+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(7/2)/f
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 7.03 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.61 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a^3 c^3 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (1225 B+1960 B \cos (2 (e+f x))+980 B \cos (4 (e+f x))+280 B \cos (6 (e+f x))+35 B \cos (8 (e+f x))-19600 A \sin (e+f x)-3920 A \sin (3 (e+f x))-784 A \sin (5 (e+f x))-80 A \sin (7 (e+f x)))}{35840 f} \] Input:

Integrate[(a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f 
*x])^(7/2),x]
 

Output:

-1/35840*(a^3*c^3*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e 
 + f*x]]*(1225*B + 1960*B*Cos[2*(e + f*x)] + 980*B*Cos[4*(e + f*x)] + 280* 
B*Cos[6*(e + f*x)] + 35*B*Cos[8*(e + f*x)] - 19600*A*Sin[e + f*x] - 3920*A 
*Sin[3*(e + f*x)] - 784*A*Sin[5*(e + f*x)] - 80*A*Sin[7*(e + f*x)]))/f
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3452, 3042, 3219, 3042, 3219, 3042, 3219, 3042, 3217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3452

\(\displaystyle A \int (\sin (e+f x) a+a)^{7/2} (c-c \sin (e+f x))^{7/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle A \int (\sin (e+f x) a+a)^{7/2} (c-c \sin (e+f x))^{7/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\)

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3217

\(\displaystyle A \left (\frac {6}{7} a \left (\frac {2}{3} a \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\)

Input:

Int[(a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^( 
7/2),x]
 

Output:

-1/8*(B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(7/2) 
)/f + A*(-1/7*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f* 
x])^(7/2))/f + (6*a*(-1/6*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - 
c*Sin[e + f*x])^(7/2))/f + (2*a*(-1/10*(a^2*Cos[e + f*x]*(c - c*Sin[e + f* 
x])^(7/2))/(f*Sqrt[a + a*Sin[e + f*x]]) - (a*Cos[e + f*x]*Sqrt[a + a*Sin[e 
 + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(5*f)))/3))/7)
 

Defintions of rubi rules used

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

rule 3217
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f 
_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^ 
n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f, n 
}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
 

rule 3219
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
(m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n 
))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre 
eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I 
GtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m]) &&  !( 
ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
 

rule 3452
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] - Simp[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1))   Int[( 
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, 
e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, 
 -2^(-1)] && NeQ[m + n + 1, 0]
 
Maple [A] (verified)

Time = 90.03 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.14

method result size
default \(\frac {1280 c^{3} a^{3} \left (A \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}+\frac {\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2}+\frac {\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5}+\frac {1}{20}\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}+\frac {7 \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {1}{2}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} B \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}-2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\frac {3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2}-\frac {\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2}+\frac {1}{8}\right )}{4}\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{70 f \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-35 f}\) \(258\)
parts \(\frac {16 A \sqrt {4}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (20 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}+10 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a^{3} c^{3} \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{35 f}+\frac {2 B \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, \left (16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{12}-48 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}+64 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}-48 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+22 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-6 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a^{3} c^{3}}{f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}\) \(313\)

Input:

int((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x,metho 
d=_RETURNVERBOSE)
 

Output:

1280*c^3*a^3*(A*(cos(1/4*Pi+1/2*f*x+1/2*e)^6+1/2*cos(1/4*Pi+1/2*f*x+1/2*e) 
^4+1/5*cos(1/4*Pi+1/2*f*x+1/2*e)^2+1/20)*(cos(1/2*f*x+1/2*e)^2-1/2)*tan(1/ 
4*Pi+1/2*f*x+1/2*e)*sin(1/4*Pi+1/2*f*x+1/2*e)^6+7/4*(cos(1/2*f*x+1/2*e)^4- 
cos(1/2*f*x+1/2*e)^2+1/2)*sin(1/2*f*x+1/2*e)^2*B*cos(1/2*f*x+1/2*e)^2*(cos 
(1/2*f*x+1/2*e)^8-2*cos(1/2*f*x+1/2*e)^6+3/2*cos(1/2*f*x+1/2*e)^4-1/2*cos( 
1/2*f*x+1/2*e)^2+1/8))*(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*(a*sin(1/4*Pi 
+1/2*f*x+1/2*e)^2)^(1/2)/(70*f*cos(1/2*f*x+1/2*e)^2-35*f)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.59 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {{\left (35 \, B a^{3} c^{3} \cos \left (f x + e\right )^{8} - 35 \, B a^{3} c^{3} - 8 \, {\left (5 \, A a^{3} c^{3} \cos \left (f x + e\right )^{6} + 6 \, A a^{3} c^{3} \cos \left (f x + e\right )^{4} + 8 \, A a^{3} c^{3} \cos \left (f x + e\right )^{2} + 16 \, A a^{3} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{280 \, f \cos \left (f x + e\right )} \] Input:

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x 
, algorithm="fricas")
 

Output:

-1/280*(35*B*a^3*c^3*cos(f*x + e)^8 - 35*B*a^3*c^3 - 8*(5*A*a^3*c^3*cos(f* 
x + e)^6 + 6*A*a^3*c^3*cos(f*x + e)^4 + 8*A*a^3*c^3*cos(f*x + e)^2 + 16*A* 
a^3*c^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/ 
(f*cos(f*x + e))
 

Sympy [F(-1)]

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

integrate((a+a*sin(f*x+e))**(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2) 
,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x 
, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 453 vs. \(2 (196) = 392\).

Time = 0.29 (sec) , antiderivative size = 453, normalized size of antiderivative = 2.00 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx =\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x 
, algorithm="giac")
 

Output:

32/35*(35*B*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 
1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^16 - 20*A*a^3*c^3*sgn(cos 
(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4* 
pi + 1/2*f*x + 1/2*e)^14 - 140*B*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e 
))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^14 + 
 70*A*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f* 
x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^12 + 210*B*a^3*c^3*sgn(cos(-1/4 
*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 
1/2*f*x + 1/2*e)^12 - 84*A*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn 
(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^10 - 140*B 
*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1 
/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^10 + 35*A*a^3*c^3*sgn(cos(-1/4*pi + 
1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f* 
x + 1/2*e)^8 + 35*B*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1 
/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8)*sqrt(a)*sqrt(c 
)/f
 

Mupad [B] (verification not implemented)

Time = 40.76 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.70 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\mathrm {e}}^{-e\,8{}\mathrm {i}-f\,x\,8{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {35\,A\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}-\frac {7\,B\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}-\frac {7\,B\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{128\,f}-\frac {B\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}-\frac {B\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (8\,e+8\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{512\,f}+\frac {7\,A\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {7\,A\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{160\,f}+\frac {A\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{224\,f}\right )}{2\,\cos \left (e+f\,x\right )} \] Input:

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

Output:

(exp(- e*8i - f*x*8i)*(c - c*sin(e + f*x))^(1/2)*((35*A*a^3*c^3*exp(e*8i + 
 f*x*8i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2))/(32*f) - (7*B*a^3*c^3*ex 
p(e*8i + f*x*8i)*cos(2*e + 2*f*x)*(a + a*sin(e + f*x))^(1/2))/(64*f) - (7* 
B*a^3*c^3*exp(e*8i + f*x*8i)*cos(4*e + 4*f*x)*(a + a*sin(e + f*x))^(1/2))/ 
(128*f) - (B*a^3*c^3*exp(e*8i + f*x*8i)*cos(6*e + 6*f*x)*(a + a*sin(e + f* 
x))^(1/2))/(64*f) - (B*a^3*c^3*exp(e*8i + f*x*8i)*cos(8*e + 8*f*x)*(a + a* 
sin(e + f*x))^(1/2))/(512*f) + (7*A*a^3*c^3*exp(e*8i + f*x*8i)*sin(3*e + 3 
*f*x)*(a + a*sin(e + f*x))^(1/2))/(32*f) + (7*A*a^3*c^3*exp(e*8i + f*x*8i) 
*sin(5*e + 5*f*x)*(a + a*sin(e + f*x))^(1/2))/(160*f) + (A*a^3*c^3*exp(e*8 
i + f*x*8i)*sin(7*e + 7*f*x)*(a + a*sin(e + f*x))^(1/2))/(224*f)))/(2*cos( 
e + f*x))
 

Reduce [F]

\[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\sqrt {c}\, \sqrt {a}\, a^{3} c^{3} \left (-\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{7}d x \right ) b -\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{6}d x \right ) a +3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{5}d x \right ) b +3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right ) a -3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b -3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}d x \right ) a \right ) \] Input:

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

Output:

sqrt(c)*sqrt(a)*a**3*c**3*( - int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f 
*x) + 1)*sin(e + f*x)**7,x)*b - int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + 
 f*x) + 1)*sin(e + f*x)**6,x)*a + 3*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin 
(e + f*x) + 1)*sin(e + f*x)**5,x)*b + 3*int(sqrt(sin(e + f*x) + 1)*sqrt( - 
 sin(e + f*x) + 1)*sin(e + f*x)**4,x)*a - 3*int(sqrt(sin(e + f*x) + 1)*sqr 
t( - sin(e + f*x) + 1)*sin(e + f*x)**3,x)*b - 3*int(sqrt(sin(e + f*x) + 1) 
*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2,x)*a + int(sqrt(sin(e + f*x) + 
1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x),x)*b + int(sqrt(sin(e + f*x) + 1 
)*sqrt( - sin(e + f*x) + 1),x)*a)