\(\int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx\) [159]

Optimal result

Integrand size = 40, antiderivative size = 196 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 f (c-c \sin (e+f x))^{13/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{11/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{160 c^2 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{960 c^3 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

1/12*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(13/2)+1/40*(A-3*B)*cos(f*x+e)*(a+a*sin(f*x+e)
)^(5/2)/c/f/(c-c*sin(f*x+e))^(11/2)+1/160*(A-3*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/c^2/f/(c-c*sin(f*x+e))^(9/
2)+1/960*(A-3*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/c^3/f/(c-c*sin(f*x+e))^(7/2)

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 196, 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 = {3051, 2822, 2821} \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {(A-3 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{960 c^3 f (c-c \sin (e+f x))^{7/2}}+\frac {(A-3 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{160 c^2 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-3 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{40 c f (c-c \sin (e+f x))^{11/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{12 f (c-c \sin (e+f x))^{13/2}} \]

[In]

Int[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(13/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(12*f*(c - c*Sin[e + f*x])^(13/2)) + ((A - 3*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(5/2))/(40*c*f*(c - c*Sin[e + f*x])^(11/2)) + ((A - 3*B)*Cos[e + f*x]*(a + a*Sin[e + f*x
])^(5/2))/(160*c^2*f*(c - c*Sin[e + f*x])^(9/2)) + ((A - 3*B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(960*c^
3*f*(c - c*Sin[e + f*x])^(7/2))

Rule 2821

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

Rule 2822

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*((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Dist[(m + n + 1)/(a*(2*m
+ 1)), 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, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 1])

Rule 3051

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] + Dist[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)), 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, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2*m + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 f (c-c \sin (e+f x))^{13/2}}+\frac {(A-3 B) \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx}{4 c} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 f (c-c \sin (e+f x))^{13/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{11/2}}+\frac {(A-3 B) \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{20 c^2} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 f (c-c \sin (e+f x))^{13/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{11/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{160 c^2 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-3 B) \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{160 c^3} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 f (c-c \sin (e+f x))^{13/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{11/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{160 c^2 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{960 c^3 f (c-c \sin (e+f x))^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 14.09 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.73 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (29 A+13 B-15 (A+B) \cos (2 (e+f x))+6 (6 A+7 B) \sin (e+f x)-10 B \sin (3 (e+f x)))}{120 c^6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^6 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(13/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(29*A + 13*B - 15*(A + B)*Cos[2*(e + f*x
)] + 6*(6*A + 7*B)*Sin[e + f*x] - 10*B*Sin[3*(e + f*x)]))/(120*c^6*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1
+ Sin[e + f*x])^6*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 4.18 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.11

method	result	size
default	\(\frac {a^{2} \tan \left (f x +e \right ) \left (7 A \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )-B \left (\sin ^{5}\left (f x +e \right )\right )-42 A \left (\cos ^{4}\left (f x +e \right )\right )+6 \left (\sin ^{4}\left (f x +e \right )\right ) B -119 A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-15 B \left (\sin ^{3}\left (f x +e \right )\right )+224 A \left (\cos ^{2}\left (f x +e \right )\right )+202 A \sin \left (f x +e \right )-30 B \sin \left (f x +e \right )-242 A \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{60 c^{6} f \left (\left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )-5 \left (\cos ^{4}\left (f x +e \right )\right )-12 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+20 \left (\cos ^{2}\left (f x +e \right )\right )+16 \sin \left (f x +e \right )-16\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(217\)
parts	\(-\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2} \left (7 \left (\cos ^{5}\left (f x +e \right )\right )+42 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-126 \left (\cos ^{3}\left (f x +e \right )\right )-224 \cos \left (f x +e \right ) \sin \left (f x +e \right )+321 \cos \left (f x +e \right )+242 \tan \left (f x +e \right )-202 \sec \left (f x +e \right )\right )}{60 f \left (\left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )-5 \left (\cos ^{4}\left (f x +e \right )\right )-12 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+20 \left (\cos ^{2}\left (f x +e \right )\right )+16 \sin \left (f x +e \right )-16\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{6}}+\frac {B \sec \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) \left (\cos ^{4}\left (f x +e \right )+6 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-17 \left (\cos ^{2}\left (f x +e \right )\right )-6 \sin \left (f x +e \right )+46\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}{60 f \left (\left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )-5 \left (\cos ^{4}\left (f x +e \right )\right )-12 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+20 \left (\cos ^{2}\left (f x +e \right )\right )+16 \sin \left (f x +e \right )-16\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{6}}\)	\(343\)

[In]

int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(13/2),x,method=_RETURNVERBOSE)

[Out]

1/60*a^2/c^6/f*tan(f*x+e)*(7*A*cos(f*x+e)^4*sin(f*x+e)-B*sin(f*x+e)^5-42*A*cos(f*x+e)^4+6*sin(f*x+e)^4*B-119*A
*sin(f*x+e)*cos(f*x+e)^2-15*B*sin(f*x+e)^3+224*A*cos(f*x+e)^2+202*A*sin(f*x+e)-30*B*sin(f*x+e)-242*A)*(a*(1+si
n(f*x+e)))^(1/2)/(cos(f*x+e)^4*sin(f*x+e)-5*cos(f*x+e)^4-12*cos(f*x+e)^2*sin(f*x+e)+20*cos(f*x+e)^2+16*sin(f*x
+e)-16)/(-c*(sin(f*x+e)-1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {{\left (15 \, {\left (A + B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (11 \, A + 7 \, B\right )} a^{2} + 2 \, {\left (10 \, B a^{2} \cos \left (f x + e\right )^{2} - {\left (9 \, A + 13 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{60 \, {\left (c^{7} f \cos \left (f x + e\right )^{7} - 18 \, c^{7} f \cos \left (f x + e\right )^{5} + 48 \, c^{7} f \cos \left (f x + e\right )^{3} - 32 \, c^{7} f \cos \left (f x + e\right ) + 2 \, {\left (3 \, c^{7} f \cos \left (f x + e\right )^{5} - 16 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(13/2),x, algorithm="fricas")

[Out]

1/60*(15*(A + B)*a^2*cos(f*x + e)^2 - 2*(11*A + 7*B)*a^2 + 2*(10*B*a^2*cos(f*x + e)^2 - (9*A + 13*B)*a^2)*sin(
f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^7*f*cos(f*x + e)^7 - 18*c^7*f*cos(f*x + e)^5 +
48*c^7*f*cos(f*x + e)^3 - 32*c^7*f*cos(f*x + e) + 2*(3*c^7*f*cos(f*x + e)^5 - 16*c^7*f*cos(f*x + e)^3 + 16*c^
7*f*cos(f*x + e))*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(13/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {13}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(13/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(5/2)/(-c*sin(f*x + e) + c)^(13/2), x)

Giac [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.37 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx=-\frac {{\left (40 \, B a^{2} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 15 \, A a^{2} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 45 \, B a^{2} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 6 \, A a^{2} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 18 \, B a^{2} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{960 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{6} c^{7} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(13/2),x, algorithm="giac")

[Out]

-1/960*(40*B*a^2*sqrt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 15*A*a^2*sqrt(
c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 45*B*a^2*sqrt(c)*cos(-1/4*pi + 1/2*f
*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 6*A*a^2*sqrt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-
1/4*pi + 1/2*f*x + 1/2*e)) + 18*B*a^2*sqrt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2
*e)) + A*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*B*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))
)*sqrt(a)/((cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^6*c^7*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))

Mupad [B] (verification not implemented)

Time = 21.99 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.82 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\left (A\,29{}\mathrm {i}+B\,13{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,16{}\mathrm {i}}{15\,c^7\,f}-\frac {a^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\left (A\,1{}\mathrm {i}+B\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,16{}\mathrm {i}}{c^7\,f}-\frac {32\,a^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\left (6\,A+7\,B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{5\,c^7\,f}+\frac {32\,B\,a^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^7\,f}\right )}{-858\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}+858\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )-130\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )+2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (7\,e+7\,f\,x\right )+1144\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )-416\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )+24\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )} \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2))/(c - c*sin(e + f*x))^(13/2),x)

[Out]

((c - c*sin(e + f*x))^(1/2)*((a^2*exp(e*7i + f*x*7i)*(A*29i + B*13i)*(a + a*sin(e + f*x))^(1/2)*16i)/(15*c^7*f
) - (a^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)*16i)/(c^7*f) - (32*a^2*e
xp(e*7i + f*x*7i)*sin(e + f*x)*(6*A + 7*B)*(a + a*sin(e + f*x))^(1/2))/(5*c^7*f) + (32*B*a^2*exp(e*7i + f*x*7i
)*sin(3*e + 3*f*x)*(a + a*sin(e + f*x))^(1/2))/(3*c^7*f)))/(858*exp(e*7i + f*x*7i)*cos(3*e + 3*f*x) - 858*cos(
e + f*x)*exp(e*7i + f*x*7i) - 130*exp(e*7i + f*x*7i)*cos(5*e + 5*f*x) + 2*exp(e*7i + f*x*7i)*cos(7*e + 7*f*x)
+ 1144*exp(e*7i + f*x*7i)*sin(2*e + 2*f*x) - 416*exp(e*7i + f*x*7i)*sin(4*e + 4*f*x) + 24*exp(e*7i + f*x*7i)*s
in(6*e + 6*f*x))