\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx\) [124]

Optimal result

Integrand size = 38, antiderivative size = 209 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {128 (3 A-13 B) c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {32 (3 A-13 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {4 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac {(3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f} \] Output:

-128/15*(3*A-13*B)*c^2*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/f+32/5*(3*A 
-13*B)*c*sec(f*x+e)^3*(c-c*sin(f*x+e))^(5/2)/a^3/f-4/5*(3*A-13*B)*sec(f*x+ 
e)^3*(c-c*sin(f*x+e))^(7/2)/a^3/f-1/15*(3*A-13*B)*sec(f*x+e)^3*(c-c*sin(f* 
x+e))^(9/2)/a^3/c/f-1/5*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e))^(13/2)/a^3/c^3 
/f
 

Mathematica [A] (verified)

Time = 12.18 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.76 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} (1092 A-4557 B+(-540 A+2200 B) \cos (2 (e+f x))+5 B \cos (4 (e+f x))+1410 A \sin (e+f x)-6390 B \sin (e+f x)-30 A \sin (3 (e+f x))+170 B \sin (3 (e+f x)))}{60 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3} \] Input:

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + 
 f*x])^3,x]
 

Output:

-1/60*(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]* 
(1092*A - 4557*B + (-540*A + 2200*B)*Cos[2*(e + f*x)] + 5*B*Cos[4*(e + f*x 
)] + 1410*A*Sin[e + f*x] - 6390*B*Sin[e + f*x] - 30*A*Sin[3*(e + f*x)] + 1 
70*B*Sin[3*(e + f*x)]))/(a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + 
Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.89, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

\(\displaystyle \frac {-\frac {1}{10} c (3 A-13 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{11/2}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{10} c (3 A-13 B) \int \frac {(c-c \sin (e+f x))^{11/2}}{\cos (e+f x)^4}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{10} c (3 A-13 B) \left (4 c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{10} c (3 A-13 B) \left (4 c \int \frac {(c-c \sin (e+f x))^{9/2}}{\cos (e+f x)^4}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{10} c (3 A-13 B) \left (4 c \left (8 c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{10} c (3 A-13 B) \left (4 c \left (8 c \int \frac {(c-c \sin (e+f x))^{7/2}}{\cos (e+f x)^4}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{10} c (3 A-13 B) \left (4 c \left (8 c \left (-4 c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2}dx-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{10} c (3 A-13 B) \left (4 c \left (8 c \left (-4 c \int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^4}dx-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {-\frac {1}{10} c (3 A-13 B) \left (4 c \left (8 c \left (\frac {8 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 f}}{a^3 c^3}\)

Input:

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x]) 
^3,x]
 

Output:

(-1/5*((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(13/2))/f - ((3*A - 13* 
B)*c*((2*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(9/2))/(3*f) + 4*c*((2*c*Se 
c[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))))/10)/(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 = 125.86 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {2 c^{4} \left (\sin \left (f x +e \right )-1\right ) \left (5 B \cos \left (f x +e \right )^{4}+\left (-15 A +85 B \right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+\left (-135 A +545 B \right ) \cos \left (f x +e \right )^{2}+\left (180 A -820 B \right ) \sin \left (f x +e \right )+204 A -844 B \right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(121\)

Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x,method=_R 
ETURNVERBOSE)
 

Output:

2/15*c^4/a^3*(sin(f*x+e)-1)/(1+sin(f*x+e))^2*(5*B*cos(f*x+e)^4+(-15*A+85*B 
)*cos(f*x+e)^2*sin(f*x+e)+(-135*A+545*B)*cos(f*x+e)^2+(180*A-820*B)*sin(f* 
x+e)+204*A-844*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 = 148, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {2 \, {\left (5 \, B c^{3} \cos \left (f x + e\right )^{4} - 5 \, {\left (27 \, A - 109 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} + 4 \, {\left (51 \, A - 211 \, B\right )} c^{3} - 5 \, {\left ({\left (3 \, A - 17 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 4 \, {\left (9 \, A - 41 \, B\right )} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x, al 
gorithm="fricas")
 

Output:

2/15*(5*B*c^3*cos(f*x + e)^4 - 5*(27*A - 109*B)*c^3*cos(f*x + e)^2 + 4*(51 
*A - 211*B)*c^3 - 5*((3*A - 17*B)*c^3*cos(f*x + e)^2 - 4*(9*A - 41*B)*c^3) 
*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(a^3*f*cos(f*x + e)^3 - 2*a^3*f*c 
os(f*x + e)*sin(f*x + e) - 2*a^3*f*cos(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2)/(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. 854 vs. \(2 (189) = 378\).

Time = 0.15 (sec) , antiderivative size = 854, normalized size of antiderivative = 4.09 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x, al 
gorithm="maxima")
 

Output:

2/15*(3*(23*c^(7/2) + 110*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 318*c^ 
(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 590*c^(7/2)*sin(f*x + e)^3/(co 
s(f*x + e) + 1)^3 + 1065*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 122 
0*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 1540*c^(7/2)*sin(f*x + e)^ 
6/(cos(f*x + e) + 1)^6 + 1220*c^(7/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 
+ 1065*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 590*c^(7/2)*sin(f*x + 
 e)^9/(cos(f*x + e) + 1)^9 + 318*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1 
)^10 + 110*c^(7/2)*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 23*c^(7/2)*sin( 
f*x + e)^12/(cos(f*x + e) + 1)^12)*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)^(7/2)) - 2*(147*c^(7/2) + 735*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1 
) + 1992*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4015*c^(7/2)*sin(f* 
x + e)^3/(cos(f*x + e) + 1)^3 + 6605*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) 
+ 1)^4 + 8370*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 9520*c^(7/2)*s 
in(f*x + e)^6/(cos(f*x + e) + 1)^6 + 8370*c^(7/2)*sin(f*x + e)^7/(cos(f*x 
+ e) + 1)^7 + 6605*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 4015*c^(7 
/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 1992*c^(7/2)*sin(f*x + e)^10/(co 
s(f*x + e) + 1)^10 + 735*c^(7/2)*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 ...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 774 vs. \(2 (189) = 378\).

Time = 0.36 (sec) , antiderivative size = 774, normalized size of antiderivative = 3.70 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x, al 
gorithm="giac")
 

Output:

-4/15*sqrt(2)*sqrt(c)*(5*(3*A*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 19 
*B*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 6*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) + 42*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 - 15*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^3*((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) - (33*A*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2* 
e)) - 113*B*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 150*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*p 
i + 1/2*f*x + 1/2*e) + 1) - 490*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) 
+ 240*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 - 740*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 + 90*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 - 390*B*c^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*...
 

Mupad [F(-1)]

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

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(7/2))/(a + a*sin(e + f*x)) 
^3,x)
 

Output:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(7/2))/(a + a*sin(e + f*x)) 
^3, x)
 

Reduce [F]

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

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x)
 

Output:

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