\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx\) [124]
Optimal result
Integrand size = 38, antiderivative size = 209 \[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {128 (3 A-13 B) c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {32 (3 A-13 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {4 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac {(3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f}
\]
[Out]
-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
Rubi [A] (verified)
Time = 0.40 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of
steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3046, 2934, 2753, 2752}
\[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f}-\frac {128 c^2 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}-\frac {(3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 c f}-\frac {4 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}+\frac {32 c (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}
\]
[In]
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^3,x]
[Out]
(-128*(3*A - 13*B)*c^2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*f) + (32*(3*A - 13*B)*c*Sec[e + f*x]
^3*(c - c*Sin[e + f*x])^(5/2))/(5*a^3*f) - (4*(3*A - 13*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(5*a^3*f
) - ((3*A - 13*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(9/2))/(15*a^3*c*f) - ((A - B)*Sec[e + f*x]^5*(c - c*Sin
[e + f*x])^(13/2))/(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))^{13/2} \, dx}{a^3 c^3} \\ & = -\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f}-\frac {(3 A-13 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{11/2} \, dx}{10 a^3 c^2} \\ & = -\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}-\frac {(2 (3 A-13 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{5 a^3 c} \\ & = -\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}-\frac {(16 (3 A-13 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a^3} \\ & = \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}+\frac {(64 (3 A-13 B) c) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{5 a^3} \\ & = -\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} \\
\end{align*}
Mathematica [A] (verified)
Time = 8.77 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.76
\[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} (1092 A-4557 B+(-540 A+2200 B) \cos (2 (e+f x))+5 B \cos (4 (e+f x))+1410 A \sin (e+f x)-6390 B \sin (e+f x)-30 A \sin (3 (e+f x))+170 B \sin (3 (e+f x)))}{60 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3}
\]
[In]
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^3,x]
[Out]
-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)] +
170*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)
Maple [A] (verified)
Time = 121.35 (sec) , antiderivative size = 121, normalized size of antiderivative =
0.58
| | |
| method | result | size |
| | |
| default |
\(\frac {2 c^{4} \left (\sin \left (f x +e \right )-1\right ) \left (5 B \left (\cos ^{4}\left (f x +e \right )\right )+\left (-15 A +85 B \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-135 A +545 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (180 A -820 B \right ) \sin \left (f x +e \right )+204 A -844 B \right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) |
\(121\) |
| | |
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[In]
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x,method=_RETURNVERBOSE)
[Out]
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+54
5*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)
none
Time = 0.26 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.71
\[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {2 \, {\left (5 \, B c^{3} \cos \left (f x + e\right )^{4} - 5 \, {\left (27 \, A - 109 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} + 4 \, {\left (51 \, A - 211 \, B\right )} c^{3} - 5 \, {\left ({\left (3 \, A - 17 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 4 \, {\left (9 \, A - 41 \, B\right )} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}}
\]
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
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
*cos(f*x + e)*sin(f*x + e) - 2*a^3*f*cos(f*x + e))
Sympy [F(-1)]
Timed out. \[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2)/(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. 854 vs. \(2 (189) = 378\).
Time = 0.33 (sec) , antiderivative size = 854, normalized size of antiderivative = 4.09
\[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display}
\]
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
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/(cos(f*x + e) + 1)^3 + 1065*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 1
220*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)*sin(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/(cos(f*x + e) + 1)^10 + 735
*c^(7/2)*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 147*c^(7/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12)*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/(co
s(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)))/f
Giac [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 774 vs. \(2 (189) = 378\).
Time = 0.56 (sec) , antiderivative size = 774, normalized size of antiderivative = 3.70
\[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display}
\]
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")
[Out]
-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*(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^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*p
i + 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))/(co
s(-1/4*pi + 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*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 +
15*A*c^3*(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 - 75*B*c^3*(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)/(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)^5))/f
Mupad [F(-1)]
Timed out. \[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3} \,d x
\]
[In]
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(7/2))/(a + a*sin(e + f*x))^3,x)
[Out]
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(7/2))/(a + a*sin(e + f*x))^3, x)