\(\int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx\) [23]

Optimal result

Integrand size = 34, antiderivative size = 104 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}-\frac {a (A+11 B) c \cos (e+f x)}{15 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac {a (A-4 B) \cos (e+f x)}{15 f \left (c^3-c^3 \sin (e+f x)\right )} \]

[Out]

2/5*a*(A+B)*cos(f*x+e)/f/(c-c*sin(f*x+e))^3-1/15*a*(A+11*B)*c*cos(f*x+e)/f/(c^2-c^2*sin(f*x+e))^2-1/15*a*(A-4*
B)*cos(f*x+e)/f/(c^3-c^3*sin(f*x+e))

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3046, 2936, 2829, 2727} \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {a (A-4 B) \cos (e+f x)}{15 f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {a c (A+11 B) \cos (e+f x)}{15 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3} \]

[In]

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

[Out]

(2*a*(A + B)*Cos[e + f*x])/(5*f*(c - c*Sin[e + f*x])^3) - (a*(A + 11*B)*c*Cos[e + f*x])/(15*f*(c^2 - c^2*Sin[e
+ f*x])^2) - (a*(A - 4*B)*Cos[e + f*x])/(15*f*(c^3 - c^3*Sin[e + f*x]))

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2829

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2936

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[2*(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(2*m + 3))), x] + Dist[
1/(b^3*(2*m + 3)), Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d*(2*m + 3)*Sin[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]

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}& = (a c) \int \frac {\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx \\ & = \frac {2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}+\frac {a \int \frac {-A c-6 B c-5 B c \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx}{5 c^2} \\ & = \frac {2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}-\frac {a (A+11 B) \cos (e+f x)}{15 c f (c-c \sin (e+f x))^2}-\frac {(a (A-4 B)) \int \frac {1}{c-c \sin (e+f x)} \, dx}{15 c^2} \\ & = \frac {2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}-\frac {a (A+11 B) \cos (e+f x)}{15 c f (c-c \sin (e+f x))^2}-\frac {a (A-4 B) \cos (e+f x)}{15 f \left (c^3-c^3 \sin (e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.41 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {a \left (15 (A-B) \cos \left (e+\frac {f x}{2}\right )-5 (A-B) \cos \left (e+\frac {3 f x}{2}\right )+5 A \sin \left (\frac {f x}{2}\right )+25 B \sin \left (\frac {f x}{2}\right )+15 B \sin \left (2 e+\frac {3 f x}{2}\right )+A \sin \left (2 e+\frac {5 f x}{2}\right )-4 B \sin \left (2 e+\frac {5 f x}{2}\right )\right )}{30 c^3 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

[In]

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

[Out]

(a*(15*(A - B)*Cos[e + (f*x)/2] - 5*(A - B)*Cos[e + (3*f*x)/2] + 5*A*Sin[(f*x)/2] + 25*B*Sin[(f*x)/2] + 15*B*S
in[2*e + (3*f*x)/2] + A*Sin[2*e + (5*f*x)/2] - 4*B*Sin[2*e + (5*f*x)/2]))/(30*c^3*f*(Cos[e/2] - Sin[e/2])*(Cos
[(e + f*x)/2] - Sin[(e + f*x)/2])^5)

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(-\frac {2 \left (A \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-A +B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {\left (5 A +B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+\frac {\left (-A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3}+\frac {4 A}{15}-\frac {B}{15}\right ) a}{f \,c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\)	\(94\)
derivativedivides	\(\frac {2 a \left (-\frac {14 A +10 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {8 A +8 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 B +6 A}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {16 A +16 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f \,c^{3}}\)	\(115\)
default	\(\frac {2 a \left (-\frac {14 A +10 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {8 A +8 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 B +6 A}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {16 A +16 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f \,c^{3}}\)	\(115\)
risch	\(\frac {-\frac {10 B a \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}-2 i B a \,{\mathrm e}^{3 i \left (f x +e \right )}+\frac {2 i B a \,{\mathrm e}^{i \left (f x +e \right )}}{3}+2 B a \,{\mathrm e}^{4 i \left (f x +e \right )}-\frac {2 i A a \,{\mathrm e}^{i \left (f x +e \right )}}{3}-\frac {2 A a \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}-\frac {2 a A}{15}+\frac {8 B a}{15}+2 i A a \,{\mathrm e}^{3 i \left (f x +e \right )}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} f \,c^{3}}\)	\(127\)
norman	\(\frac {-\frac {8 a A -2 B a}{15 c f}-\frac {2 a A \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {10 \left (a A -B a \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}+\frac {2 \left (a A -B a \right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {\left (2 a A -2 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 c f}-\frac {2 \left (11 a A +B a \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f}-\frac {2 \left (11 a A +B a \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}+\frac {2 \left (7 a A -7 B a \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {2 \left (23 a A +3 B a \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\)	\(262\)

[In]

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

[Out]

-2*(A*tan(1/2*f*x+1/2*e)^4+(-A+B)*tan(1/2*f*x+1/2*e)^3+1/3*(5*A+B)*tan(1/2*f*x+1/2*e)^2+1/3*(-A+B)*tan(1/2*f*x
+1/2*e)+4/15*A-1/15*B)*a/f/c^3/(tan(1/2*f*x+1/2*e)-1)^5

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.76 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {{\left (A - 4 \, B\right )} a \cos \left (f x + e\right )^{3} - {\left (2 \, A + 7 \, B\right )} a \cos \left (f x + e\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (f x + e\right ) + 6 \, {\left (A + B\right )} a + {\left ({\left (A - 4 \, B\right )} a \cos \left (f x + e\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (f x + e\right ) + 6 \, {\left (A + B\right )} a\right )} \sin \left (f x + e\right )}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

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

[Out]

-1/15*((A - 4*B)*a*cos(f*x + e)^3 - (2*A + 7*B)*a*cos(f*x + e)^2 + 3*(A + B)*a*cos(f*x + e) + 6*(A + B)*a + ((
A - 4*B)*a*cos(f*x + e)^2 + 3*(A + B)*a*cos(f*x + e) + 6*(A + B)*a)*sin(f*x + e))/(c^3*f*cos(f*x + e)^3 + 3*c^
3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*
sin(f*x + e))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1035 vs. \(2 (92) = 184\).

Time = 4.34 (sec) , antiderivative size = 1035, normalized size of antiderivative = 9.95 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-30*A*a*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c*
*3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 30*A*a*t
an(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2
)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 50*A*a*tan(e/2 + f*x/2)**2/(
15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*ta
n(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 10*A*a*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2 + f*x
/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*
c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 8*A*a/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 +
150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 30
*B*a*tan(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 +
f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 10*B*a*tan(e/2 + f*x/2
)**2/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**
3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 10*B*a*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2
+ f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2
+ 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 2*B*a/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)
**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f
), Ne(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + a)/(-c*sin(e) + c)**3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 737 vs. \(2 (101) = 202\).

Time = 0.23 (sec) , antiderivative size = 737, normalized size of antiderivative = 7.09 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-2/15*(A*a*(20*sin(f*x + e)/(cos(f*x + e) + 1) - 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 7)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*A*a*(5*sin(f*x + e)/(cos(f*x + e) + 1)
- 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 1)/(c^3 - 5*c^3*sin(f*x + e)
/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*B*a*(5*sin(f*x + e)
/(cos(f*x + e) + 1) - 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 1)/(c^3
- 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(
cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 2
*B*a*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)/(c^3 - 5*c^3*sin(f*x + e
)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.26 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 25 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, A a - B a\right )}}{15 \, c^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}} \]

[In]

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

[Out]

-2/15*(15*A*a*tan(1/2*f*x + 1/2*e)^4 - 15*A*a*tan(1/2*f*x + 1/2*e)^3 + 15*B*a*tan(1/2*f*x + 1/2*e)^3 + 25*A*a*
tan(1/2*f*x + 1/2*e)^2 + 5*B*a*tan(1/2*f*x + 1/2*e)^2 - 5*A*a*tan(1/2*f*x + 1/2*e) + 5*B*a*tan(1/2*f*x + 1/2*e
) + 4*A*a - B*a)/(c^3*f*(tan(1/2*f*x + 1/2*e) - 1)^5)

Mupad [B] (verification not implemented)

Time = 12.46 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.65 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {11\,A\,a\,\cos \left (e+f\,x\right )}{2}-\frac {B\,a}{4}-\frac {41\,A\,a}{4}+\frac {B\,a\,\cos \left (e+f\,x\right )}{2}+5\,A\,a\,\sin \left (e+f\,x\right )-5\,B\,a\,\sin \left (e+f\,x\right )+\frac {3\,A\,a\,\cos \left (2\,e+2\,f\,x\right )}{4}+\frac {3\,B\,a\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {5\,A\,a\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {5\,B\,a\,\sin \left (2\,e+2\,f\,x\right )}{4}\right )}{15\,c^3\,f\,\left (\frac {5\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}-\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{4}-\frac {5\,\sqrt {2}\,\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}+\frac {\sqrt {2}\,\cos \left (\frac {5\,e}{2}+\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{4}\right )} \]

[In]

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

[Out]

(2*cos(e/2 + (f*x)/2)*((11*A*a*cos(e + f*x))/2 - (B*a)/4 - (41*A*a)/4 + (B*a*cos(e + f*x))/2 + 5*A*a*sin(e + f
*x) - 5*B*a*sin(e + f*x) + (3*A*a*cos(2*e + 2*f*x))/4 + (3*B*a*cos(2*e + 2*f*x))/4 - (5*A*a*sin(2*e + 2*f*x))/
4 + (5*B*a*sin(2*e + 2*f*x))/4))/(15*c^3*f*((5*2^(1/2)*cos((3*e)/2 - pi/4 + (3*f*x)/2))/4 - (5*2^(1/2)*cos(e/2
+ pi/4 + (f*x)/2))/2 + (2^(1/2)*cos((5*e)/2 + pi/4 + (5*f*x)/2))/4))