\(\int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\) [162]

Optimal result

Integrand size = 40, antiderivative size = 192 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {(7 A+B) c^3 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 (7 A+B) c^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{105 f}+\frac {(7 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{42 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f} \]

[Out]

1/42*(7*A+B)*c*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(3/2)/f-1/7*B*cos(f*x+e)*(a+a*sin(f*x+e))^(7
/2)*(c-c*sin(f*x+e))^(5/2)/f+1/105*(7*A+B)*c^3*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/f/(c-c*sin(f*x+e))^(1/2)+2/10
5*(7*A+B)*c^2*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {3052, 2819, 2817} \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^3 (7 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 c^2 (7 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{105 f}+\frac {c (7 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{42 f}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f} \]

[In]

Int[(a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

((7*A + B)*c^3*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(105*f*Sqrt[c - c*Sin[e + f*x]]) + (2*(7*A + B)*c^2*Co
s[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]])/(105*f) + ((7*A + B)*c*Cos[e + f*x]*(a + a*Sin
[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(3/2))/(42*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e
+ f*x])^(5/2))/(7*f)

Rule 2817

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 2819

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] + Dist[a*((2*m - 1)/(
m + n)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[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 3052

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*
(m + n + 1))), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Si
n[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f}+\frac {1}{7} (7 A+B) \int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx \\ & = \frac {(7 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{42 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f}+\frac {1}{21} (2 (7 A+B) c) \int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx \\ & = \frac {2 (7 A+B) c^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{105 f}+\frac {(7 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{42 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f}+\frac {1}{105} \left (4 (7 A+B) c^2\right ) \int (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)} \, dx \\ & = \frac {(7 A+B) c^3 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 (7 A+B) c^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{105 f}+\frac {(7 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{42 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.90 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.21 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {a^3 c^2 (-1+\sin (e+f x))^2 (1+\sin (e+f x))^3 \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (-525 (A+B) \cos (2 (e+f x))-210 (A+B) \cos (4 (e+f x))-35 A \cos (6 (e+f x))-35 B \cos (6 (e+f x))+4200 A \sin (e+f x)+525 B \sin (e+f x)+700 A \sin (3 (e+f x))-35 B \sin (3 (e+f x))+84 A \sin (5 (e+f x))-63 B \sin (5 (e+f x))-15 B \sin (7 (e+f x)))}{6720 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7} \]

[In]

Integrate[(a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(a^3*c^2*(-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x])^3*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-525*
(A + B)*Cos[2*(e + f*x)] - 210*(A + B)*Cos[4*(e + f*x)] - 35*A*Cos[6*(e + f*x)] - 35*B*Cos[6*(e + f*x)] + 4200
*A*Sin[e + f*x] + 525*B*Sin[e + f*x] + 700*A*Sin[3*(e + f*x)] - 35*B*Sin[3*(e + f*x)] + 84*A*Sin[5*(e + f*x)]
- 63*B*Sin[5*(e + f*x)] - 15*B*Sin[7*(e + f*x)]))/(6720*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*
x)/2] + Sin[(e + f*x)/2])^7)

Maple [A] (verified)

Time = 75.32 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {a^{3} c^{2} \tan \left (f x +e \right ) \left (30 B \left (\cos ^{4}\left (f x +e \right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )+35 A \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )-35 B \left (\cos ^{2}\left (f x +e \right )\right ) \left (\sin ^{3}\left (f x +e \right )\right )+42 A \left (\cos ^{4}\left (f x +e \right )\right )+24 B \left (\sin ^{2}\left (f x +e \right )\right ) \left (\cos ^{2}\left (f x +e \right )\right )+35 A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-70 B \left (\sin ^{3}\left (f x +e \right )\right )+56 A \left (\cos ^{2}\left (f x +e \right )\right )+16 B \left (\sin ^{2}\left (f x +e \right )\right )+35 A \sin \left (f x +e \right )+105 B \sin \left (f x +e \right )+112 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{210 f}\)	\(200\)
parts	\(\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, a^{3} c^{2} \left (-5 \left (\cos ^{5}\left (f x +e \right )\right )+6 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+8 \cos \left (f x +e \right ) \sin \left (f x +e \right )+16 \tan \left (f x +e \right )+5 \sec \left (f x +e \right )\right )}{30 f}-\frac {B \sec \left (f x +e \right ) \left (30 \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )+35 \left (\cos ^{4}\left (f x +e \right )\right )+24 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+35 \left (\cos ^{2}\left (f x +e \right )\right )+16 \sin \left (f x +e \right )+35\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, c^{2} a^{3} \left (\cos ^{2}\left (f x +e \right )-1\right )}{210 f}\)	\(211\)

[In]

int((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/210*a^3*c^2/f*tan(f*x+e)*(30*B*cos(f*x+e)^4*sin(f*x+e)^2+35*A*cos(f*x+e)^4*sin(f*x+e)-35*B*cos(f*x+e)^2*sin(
f*x+e)^3+42*A*cos(f*x+e)^4+24*B*sin(f*x+e)^2*cos(f*x+e)^2+35*A*sin(f*x+e)*cos(f*x+e)^2-70*B*sin(f*x+e)^3+56*A*
cos(f*x+e)^2+16*B*sin(f*x+e)^2+35*A*sin(f*x+e)+105*B*sin(f*x+e)+112*A)*(-c*(sin(f*x+e)-1))^(1/2)*(a*(1+sin(f*x
+e)))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.78 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {{\left (35 \, {\left (A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{6} - 35 \, {\left (A + B\right )} a^{3} c^{2} + 2 \, {\left (15 \, B a^{3} c^{2} \cos \left (f x + e\right )^{6} - 3 \, {\left (7 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{4} - 4 \, {\left (7 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{2} - 8 \, {\left (7 \, A + B\right )} a^{3} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{210 \, f \cos \left (f x + e\right )} \]

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/210*(35*(A + B)*a^3*c^2*cos(f*x + e)^6 - 35*(A + B)*a^3*c^2 + 2*(15*B*a^3*c^2*cos(f*x + e)^6 - 3*(7*A + B)*
a^3*c^2*cos(f*x + e)^4 - 4*(7*A + B)*a^3*c^2*cos(f*x + e)^2 - 8*(7*A + B)*a^3*c^2)*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. \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(7/2)*(-c*sin(f*x + e) + c)^(5/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 453 vs. \(2 (168) = 336\).

Time = 0.56 (sec) , antiderivative size = 453, normalized size of antiderivative = 2.36 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {16 \, {\left (120 \, B a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{14} - 70 \, A a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 490 \, B a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} + 252 \, A a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 756 \, B a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 315 \, A a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 525 \, B a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} + 140 \, A a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 140 \, B a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}\right )} \sqrt {a} \sqrt {c}}{105 \, f} \]

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

16/105*(120*B*a^3*c^2*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^2*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 - 490*B*a^3*c^2*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 + 252*A*a^3*c^2*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 + 756*B*a^3*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(s
in(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^10 - 315*A*a^3*c^2*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 - 525*B*a^3*c^2*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 + 140*A*a^3*c^2*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)^6 + 140*B*a^3*
c^2*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)^6)*
sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 17.66 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.99 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\mathrm {e}}^{-e\,7{}\mathrm {i}-f\,x\,7{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a^3\,c^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\left (A\,1{}\mathrm {i}+B\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,5{}\mathrm {i}}{32\,f}+\frac {a^3\,c^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\left (A\,1{}\mathrm {i}+B\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{16\,f}+\frac {a^3\,c^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\left (A\,1{}\mathrm {i}+B\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{96\,f}+\frac {a^3\,c^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\left (4\,A-3\,B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{160\,f}+\frac {a^3\,c^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\left (20\,A-B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{96\,f}+\frac {5\,a^3\,c^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\left (8\,A+B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}-\frac {B\,a^3\,c^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{224\,f}\right )}{2\,\cos \left (e+f\,x\right )} \]

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(7/2)*(c - c*sin(e + f*x))^(5/2),x)

[Out]

(exp(- e*7i - f*x*7i)*(c - c*sin(e + f*x))^(1/2)*((a^3*c^2*exp(e*7i + f*x*7i)*cos(2*e + 2*f*x)*(A*1i + B*1i)*(
a + a*sin(e + f*x))^(1/2)*5i)/(32*f) + (a^3*c^2*exp(e*7i + f*x*7i)*cos(4*e + 4*f*x)*(A*1i + B*1i)*(a + a*sin(e
+ f*x))^(1/2)*1i)/(16*f) + (a^3*c^2*exp(e*7i + f*x*7i)*cos(6*e + 6*f*x)*(A*1i + B*1i)*(a + a*sin(e + f*x))^(1
/2)*1i)/(96*f) + (a^3*c^2*exp(e*7i + f*x*7i)*sin(5*e + 5*f*x)*(4*A - 3*B)*(a + a*sin(e + f*x))^(1/2))/(160*f)
+ (a^3*c^2*exp(e*7i + f*x*7i)*sin(3*e + 3*f*x)*(20*A - B)*(a + a*sin(e + f*x))^(1/2))/(96*f) + (5*a^3*c^2*exp(
e*7i + f*x*7i)*sin(e + f*x)*(8*A + B)*(a + a*sin(e + f*x))^(1/2))/(32*f) - (B*a^3*c^2*exp(e*7i + f*x*7i)*sin(7
*e + 7*f*x)*(a + a*sin(e + f*x))^(1/2))/(224*f)))/(2*cos(e + f*x))