\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx\) [123]

Optimal result

Integrand size = 38, antiderivative size = 242 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {2048 (A-3 B) c^3 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {512 (A-3 B) c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {64 (A-3 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f} \]

[Out]

-2048/15*(A-3*B)*c^3*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/f+512/5*(A-3*B)*c^2*sec(f*x+e)^3*(c-c*sin(f*x+e))
^(5/2)/a^3/f-64/5*(A-3*B)*c*sec(f*x+e)^3*(c-c*sin(f*x+e))^(7/2)/a^3/f-16/15*(A-3*B)*sec(f*x+e)^3*(c-c*sin(f*x+
e))^(9/2)/a^3/f-1/5*(A-3*B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(11/2)/a^3/c/f-1/5*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e
))^(15/2)/a^3/c^3/f

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3046, 2934, 2753, 2752} \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}-\frac {2048 c^3 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {512 c^2 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {64 c (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f} \]

[In]

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(9/2))/(a + a*Sin[e + f*x])^3,x]

[Out]

(-2048*(A - 3*B)*c^3*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*f) + (512*(A - 3*B)*c^2*Sec[e + f*x]^3
*(c - c*Sin[e + f*x])^(5/2))/(5*a^3*f) - (64*(A - 3*B)*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(5*a^3*f)
- (16*(A - 3*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(9/2))/(15*a^3*f) - ((A - 3*B)*Sec[e + f*x]^3*(c - c*Sin[e
+ f*x])^(11/2))/(5*a^3*c*f) - ((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(15/2))/(5*a^3*c^3*f)

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2753

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] + Dist[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 2934

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] + Dist[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 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[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] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{15/2} \, dx}{a^3 c^3} \\ & = -\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}-\frac {(A-3 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{13/2} \, dx}{2 a^3 c^2} \\ & = -\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}-\frac {(8 (A-3 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{11/2} \, dx}{5 a^3 c} \\ & = -\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}-\frac {(32 (A-3 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{5 a^3} \\ & = -\frac {64 (A-3 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}-\frac {(256 (A-3 B) c) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a^3} \\ & = \frac {512 (A-3 B) c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {64 (A-3 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}+\frac {\left (1024 (A-3 B) c^2\right ) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{5 a^3} \\ & = -\frac {2048 (A-3 B) c^3 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {512 (A-3 B) c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {64 (A-3 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 9.44 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {c^4 (-1+\sin (e+f x))^4 \sqrt {c-c \sin (e+f x)} (11298 A-33516 B-40 (137 A-402 B) \cos (2 (e+f x))-10 (A-6 B) \cos (4 (e+f x))+15600 A \sin (e+f x)-47430 B \sin (e+f x)-400 A \sin (3 (e+f x))+1335 B \sin (3 (e+f x))-3 B \sin (5 (e+f x)))}{120 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(9/2))/(a + a*Sin[e + f*x])^3,x]

[Out]

-1/120*(c^4*(-1 + Sin[e + f*x])^4*Sqrt[c - c*Sin[e + f*x]]*(11298*A - 33516*B - 40*(137*A - 402*B)*Cos[2*(e +
f*x)] - 10*(A - 6*B)*Cos[4*(e + f*x)] + 15600*A*Sin[e + f*x] - 47430*B*Sin[e + f*x] - 400*A*Sin[3*(e + f*x)] +
1335*B*Sin[3*(e + f*x)] - 3*B*Sin[5*(e + f*x)]))/(a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(Cos[(e + f*x
)/2] + Sin[(e + f*x)/2])^5)

Maple [A] (verified)

Time = 120.73 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.59

method	result	size
default	\(-\frac {2 c^{5} \left (\sin \left (f x +e \right )-1\right ) \left (3 B \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (5 A -30 B \right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (100 A -336 B \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (680 A -1980 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (-1000 A +3048 B \right ) \sin \left (f x +e \right )-1048 A +3096 B \right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(143\)

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

-2/15*c^5/a^3*(sin(f*x+e)-1)/(1+sin(f*x+e))^2*(3*B*cos(f*x+e)^4*sin(f*x+e)+(5*A-30*B)*cos(f*x+e)^4+(100*A-336*
B)*cos(f*x+e)^2*sin(f*x+e)+(680*A-1980*B)*cos(f*x+e)^2+(-1000*A+3048*B)*sin(f*x+e)-1048*A+3096*B)/cos(f*x+e)/(
c-c*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (5 \, {\left (A - 6 \, B\right )} c^{4} \cos \left (f x + e\right )^{4} + 20 \, {\left (34 \, A - 99 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} - 8 \, {\left (131 \, A - 387 \, B\right )} c^{4} + {\left (3 \, B c^{4} \cos \left (f x + e\right )^{4} + 4 \, {\left (25 \, A - 84 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} - 8 \, {\left (125 \, A - 381 \, B\right )} c^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \]

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-2/15*(5*(A - 6*B)*c^4*cos(f*x + e)^4 + 20*(34*A - 99*B)*c^4*cos(f*x + e)^2 - 8*(131*A - 387*B)*c^4 + (3*B*c^4
*cos(f*x + e)^4 + 4*(25*A - 84*B)*c^4*cos(f*x + e)^2 - 8*(125*A - 381*B)*c^4)*sin(f*x + e))*sqrt(-c*sin(f*x +
e) + c)/(a^3*f*cos(f*x + e)^3 - 2*a^3*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*f*cos(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(9/2)/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 945 vs. \(2 (218) = 436\).

Time = 0.33 (sec) , antiderivative size = 945, normalized size of antiderivative = 3.90 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

2/15*((363*c^(9/2) + 1800*c^(9/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 5301*c^(9/2)*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + 11600*c^(9/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 21343*c^(9/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4
+ 30200*c^(9/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 40065*c^(9/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 408
00*c^(9/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 40065*c^(9/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 30200*c^(
9/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 21343*c^(9/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 11600*c^(9/2)
*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 5301*c^(9/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 1800*c^(9/2)*sin
(f*x + e)^13/(cos(f*x + e) + 1)^13 + 363*c^(9/2)*sin(f*x + e)^14/(cos(f*x + e) + 1)^14)*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)^(9/2)) - 6*(181*c^(9/2) + 905*c^(9/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 2627*c^(9/2)*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5870*c^(9/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 10521*c^(9/2)*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 + 15351*c^(9/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 19695*c^(9/2)*sin(f*x + e)^
6/(cos(f*x + e) + 1)^6 + 20772*c^(9/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 19695*c^(9/2)*sin(f*x + e)^8/(cos
(f*x + e) + 1)^8 + 15351*c^(9/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 10521*c^(9/2)*sin(f*x + e)^10/(cos(f*x
+ e) + 1)^10 + 5870*c^(9/2)*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 2627*c^(9/2)*sin(f*x + e)^12/(cos(f*x + e)
+ 1)^12 + 905*c^(9/2)*sin(f*x + e)^13/(cos(f*x + e) + 1)^13 + 181*c^(9/2)*sin(f*x + e)^14/(cos(f*x + e) + 1)^
14)*B/((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)^(9/2)))/f

Giac [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {1024 \, \sqrt {2} {\left (A c^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, B c^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {5 \, A c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {15 \, B c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {10 \, A c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}} - \frac {30 \, B c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}} - \frac {6 \, A c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} - \frac {6 \, B c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}\right )} \sqrt {c}}{15 \, a^{3} f {\left (\frac {{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - 1\right )}^{5}} \]

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1024/15*sqrt(2)*(A*c^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*B*c^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*
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 + 10*A*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))
/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4 - 30*B*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/
2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4 - 6*A*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^5*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)^5 - 6*B*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e)
- 1)^5*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^5)*sqrt(c)/(a^3*f*((cos(-1/4*
pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 1)^5)

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]

[In]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(9/2))/(a + a*sin(e + f*x))^3,x)

[Out]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(9/2))/(a + a*sin(e + f*x))^3, x)