\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx\) [116]

Optimal result

Integrand size = 38, antiderivative size = 201 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {128 (5 A-11 B) c^3 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{15 a^2 f}-\frac {32 (5 A-11 B) c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}-\frac {4 (5 A-11 B) c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f}-\frac {(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f} \] Output:

128/15*(5*A-11*B)*c^3*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f-32/15*(5*A-1 
1*B)*c^2*sec(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a^2/f-4/15*(5*A-11*B)*c*sec(f*x 
+e)*(c-c*sin(f*x+e))^(5/2)/a^2/f-1/15*(5*A-11*B)*sec(f*x+e)*(c-c*sin(f*x+e 
))^(7/2)/a^2/f-1/3*(A-B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(11/2)/a^2/c^2/f
 

Mathematica [A] (verified)

Time = 10.92 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.79 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} (-2100 A+4725 B+12 (25 A-62 B) \cos (2 (e+f x))+3 B \cos (4 (e+f x))-2730 A \sin (e+f x)+5838 B \sin (e+f x)-10 A \sin (3 (e+f x))+46 B \sin (3 (e+f x)))}{60 a^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2} \] Input:

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

Output:

-1/60*(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]* 
(-2100*A + 4725*B + 12*(25*A - 62*B)*Cos[2*(e + f*x)] + 3*B*Cos[4*(e + f*x 
)] - 2730*A*Sin[e + f*x] + 5838*B*Sin[e + f*x] - 10*A*Sin[3*(e + f*x)] + 4 
6*B*Sin[3*(e + f*x)]))/(a^2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + S 
in[e + f*x])^2)
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.91, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3446, 3042, 3334, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

\(\displaystyle \frac {\int \sec ^4(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{11/2}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

\(\displaystyle \frac {-\frac {1}{6} c (5 A-11 B) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{9/2}dx-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (5 A-11 B) \int \frac {(c-c \sin (e+f x))^{9/2}}{\cos (e+f x)^2}dx-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{6} c (5 A-11 B) \left (\frac {12}{5} c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{7/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (5 A-11 B) \left (\frac {12}{5} c \int \frac {(c-c \sin (e+f x))^{7/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{6} c (5 A-11 B) \left (\frac {12}{5} c \left (\frac {8}{3} c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (5 A-11 B) \left (\frac {12}{5} c \left (\frac {8}{3} c \int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{6} c (5 A-11 B) \left (\frac {12}{5} c \left (\frac {8}{3} c \left (4 c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (5 A-11 B) \left (\frac {12}{5} c \left (\frac {8}{3} c \left (4 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {-\frac {1}{6} c (5 A-11 B) \left (\frac {12}{5} c \left (\frac {8}{3} c \left (\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}-\frac {8 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 f}}{a^2 c^2}\)

Input:

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

Output:

(-1/3*((A - B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(11/2))/f - ((5*A - 11* 
B)*c*((2*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(5*f) + (12*c*((2*c*Se 
c[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*Sin[e + f*x]])/f + (2*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2) 
)/f))/3))/5))/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 = 20.73 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {2 c^{4} \left (\sin \left (f x +e \right )-1\right ) \left (3 B \cos \left (f x +e \right )^{4}+\left (-5 A +23 B \right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+\left (75 A -189 B \right ) \cos \left (f x +e \right )^{2}+\left (-340 A +724 B \right ) \sin \left (f x +e \right )-300 A +684 B \right )}{15 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(121\)

Input:

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

Output:

2/15*c^4/a^2*(sin(f*x+e)-1)/(1+sin(f*x+e))*(3*B*cos(f*x+e)^4+(-5*A+23*B)*c 
os(f*x+e)^2*sin(f*x+e)+(75*A-189*B)*cos(f*x+e)^2+(-340*A+724*B)*sin(f*x+e) 
-300*A+684*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.66 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (3 \, B c^{3} \cos \left (f x + e\right )^{4} + 3 \, {\left (25 \, A - 63 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 12 \, {\left (25 \, A - 57 \, B\right )} c^{3} - {\left ({\left (5 \, A - 23 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} + 4 \, {\left (85 \, A - 181 \, B\right )} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}} \] Input:

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

Output:

-2/15*(3*B*c^3*cos(f*x + e)^4 + 3*(25*A - 63*B)*c^3*cos(f*x + e)^2 - 12*(2 
5*A - 57*B)*c^3 - ((5*A - 23*B)*c^3*cos(f*x + e)^2 + 4*(85*A - 181*B)*c^3) 
*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. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2)/(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. 670 vs. \(2 (181) = 362\).

Time = 0.15 (sec) , antiderivative size = 670, normalized size of antiderivative = 3.33 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-2/15*(5*(45*c^(7/2) + 138*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 285*c 
^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 544*c^(7/2)*sin(f*x + e)^3/(c 
os(f*x + e) + 1)^3 + 630*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 812 
*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 630*c^(7/2)*sin(f*x + e)^6/ 
(cos(f*x + e) + 1)^6 + 544*c^(7/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 2 
85*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 138*c^(7/2)*sin(f*x + e)^ 
9/(cos(f*x + e) + 1)^9 + 45*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10) 
*A/((a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(c 
os(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)^(7/2)) - 2*(249*c^(7/2) + 747*c^(7/2)*sin(f* 
x + e)/(cos(f*x + e) + 1) + 1611*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1) 
^2 + 2896*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3612*c^(7/2)*sin(f 
*x + e)^4/(cos(f*x + e) + 1)^4 + 4298*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) 
 + 1)^5 + 3612*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 2896*c^(7/2)* 
sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 1611*c^(7/2)*sin(f*x + e)^8/(cos(f*x 
 + e) + 1)^8 + 747*c^(7/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 249*c^(7/ 
2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*B/((a^2 + 3*a^2*sin(f*x + e)/(co 
s(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)^(7/ 
2)))/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 774 vs. \(2 (181) = 362\).

Time = 0.40 (sec) , antiderivative size = 774, normalized size of antiderivative = 3.85 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-16/15*sqrt(2)*sqrt(c)*(5*(4*A*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 7 
*B*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 9*A*c^3*(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) - 15*B*c^3*(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) + 3*A*c^3*(c 
os(-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 - 6*B*c^3*(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) - (20*A*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e 
)) - 53*B*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 85*A*c^3*(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) + 235*B*c^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sg 
n(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1 
25*A*c^3*(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 - 365*B*c^3*(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 - 75*A*c^3*(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 + 165*B*c^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi ...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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