\(\int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx\) [22]

Optimal result

Integrand size = 34, antiderivative size = 72 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {a B x}{c^2}-\frac {a (A+7 B) \cos (e+f x)}{3 c^2 f (1-\sin (e+f x))}+\frac {2 a (A+B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^2} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(160\) vs. \(2(72)=144\).

Time = 6.61 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.22 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=-\frac {a \left (-9 B f x \cos \left (\frac {f x}{2}\right )-6 (A+3 B) \cos \left (e+\frac {f x}{2}\right )+2 A \cos \left (e+\frac {3 f x}{2}\right )+14 B \cos \left (e+\frac {3 f x}{2}\right )+3 B f x \cos \left (2 e+\frac {3 f x}{2}\right )+24 B \sin \left (\frac {f x}{2}\right )+9 B f x \sin \left (e+\frac {f x}{2}\right )+3 B f x \sin \left (e+\frac {3 f x}{2}\right )\right )}{6 c^2 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \] Input:

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

Output:

-1/6*(a*(-9*B*f*x*Cos[(f*x)/2] - 6*(A + 3*B)*Cos[e + (f*x)/2] + 2*A*Cos[e 
+ (3*f*x)/2] + 14*B*Cos[e + (3*f*x)/2] + 3*B*f*x*Cos[2*e + (3*f*x)/2] + 24 
*B*Sin[(f*x)/2] + 9*B*f*x*Sin[e + (f*x)/2] + 3*B*f*x*Sin[e + (3*f*x)/2]))/ 
(c^2*f*(Cos[e/2] - Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 3446, 3042, 3336, 25, 3042, 3214, 3042, 3127}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a \sin (e+f x)+a) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2}dx\)

\(\Big \downarrow \) 3446

\(\displaystyle a c \int \frac {\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \int \frac {\cos (e+f x)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3}dx\)

\(\Big \downarrow \) 3336

\(\displaystyle a c \left (\frac {\int -\frac {(A+4 B) c+3 B \sin (e+f x) c}{c-c \sin (e+f x)}dx}{3 c^3}+\frac {2 (A+B) \cos (e+f x)}{3 c f (c-c \sin (e+f x))^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{3 c f (c-c \sin (e+f x))^2}-\frac {\int \frac {(A+4 B) c+3 B \sin (e+f x) c}{c-c \sin (e+f x)}dx}{3 c^3}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{3 c f (c-c \sin (e+f x))^2}-\frac {\int \frac {(A+4 B) c+3 B \sin (e+f x) c}{c-c \sin (e+f x)}dx}{3 c^3}\right )\)

\(\Big \downarrow \) 3214

\(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{3 c f (c-c \sin (e+f x))^2}-\frac {c (A+7 B) \int \frac {1}{c-c \sin (e+f x)}dx-3 B x}{3 c^3}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{3 c f (c-c \sin (e+f x))^2}-\frac {c (A+7 B) \int \frac {1}{c-c \sin (e+f x)}dx-3 B x}{3 c^3}\right )\)

\(\Big \downarrow \) 3127

\(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{3 c f (c-c \sin (e+f x))^2}-\frac {\frac {c (A+7 B) \cos (e+f x)}{f (c-c \sin (e+f x))}-3 B x}{3 c^3}\right )\)

Input:

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

Output:

a*c*((2*(A + B)*Cos[e + f*x])/(3*c*f*(c - c*Sin[e + f*x])^2) - (-3*B*x + ( 
(A + 7*B)*c*Cos[e + f*x])/(f*(c - c*Sin[e + f*x])))/(3*c^3))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3127

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 3214

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

rule 3336

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] + Simp[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 3446

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.71 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.11

method	result	size
risch	\(\frac {a B x}{c^{2}}-\frac {2 \left (3 A a \,{\mathrm e}^{2 i \left (f x +e \right )}-12 i B a \,{\mathrm e}^{i \left (f x +e \right )}+9 B a \,{\mathrm e}^{2 i \left (f x +e \right )}-A a -7 B a \right )}{3 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,c^{2}}\)	\(80\)
derivativedivides	\(\frac {2 a \left (-\frac {A -B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {4 A +4 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{2}}\)	\(87\)
default	\(\frac {2 a \left (-\frac {A -B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {4 A +4 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{2}}\)	\(87\)
parallelrisch	\(-\frac {2 a \left (-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} x f B}{2}+\left (\frac {3}{2} f x B +A -B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+B \left (-\frac {3 f x}{2}+4\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {f x B}{2}+\frac {A}{3}-\frac {5 B}{3}\right )}{f \,c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\)	\(90\)
norman	\(\frac {\frac {a x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c}-\frac {2 A a -10 B a}{3 c f}-\frac {a x B}{c}-\frac {16 B a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f c}-\frac {8 B a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f c}-\frac {8 B a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c f}-\frac {\left (2 A a -2 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c f}-\frac {\left (10 A a -26 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 c f}-\frac {\left (14 A a -22 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3 c f}+\frac {3 a x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}-\frac {5 a x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c}+\frac {7 a x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c}-\frac {7 a x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c}+\frac {5 a x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c}-\frac {3 a x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\)	\(334\)

Input:

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

Output:

a*B*x/c^2-2/3*(3*A*a*exp(2*I*(f*x+e))-12*I*B*a*exp(I*(f*x+e))+9*B*a*exp(2* 
I*(f*x+e))-A*a-7*B*a)/(exp(I*(f*x+e))-I)^3/f/c^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (67) = 134\).

Time = 0.08 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.25 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=-\frac {6 \, B a f x - {\left (3 \, B a f x + {\left (A + 7 \, B\right )} a\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (A + B\right )} a + {\left (3 \, B a f x + {\left (A - 5 \, B\right )} a\right )} \cos \left (f x + e\right ) - {\left (6 \, B a f x - 2 \, {\left (A + B\right )} a + {\left (3 \, B a f x - {\left (A + 7 \, B\right )} a\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

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

Output:

-1/3*(6*B*a*f*x - (3*B*a*f*x + (A + 7*B)*a)*cos(f*x + e)^2 + 2*(A + B)*a + 
 (3*B*a*f*x + (A - 5*B)*a)*cos(f*x + e) - (6*B*a*f*x - 2*(A + B)*a + (3*B* 
a*f*x - (A + 7*B)*a)*cos(f*x + e))*sin(f*x + e))/(c^2*f*cos(f*x + e)^2 - c 
^2*f*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^2*f)*sin(f*x + e))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 700 vs. \(2 (65) = 130\).

Time = 2.23 (sec) , antiderivative size = 700, normalized size of antiderivative = 9.72 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

Piecewise((-6*A*a*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c* 
*2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 2*A*a/( 
3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan 
(e/2 + f*x/2) - 3*c**2*f) + 3*B*a*f*x*tan(e/2 + f*x/2)**3/(3*c**2*f*tan(e/ 
2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 
 3*c**2*f) - 9*B*a*f*x*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**3 - 
 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 9* 
B*a*f*x*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 
+ f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 3*B*a*f*x/(3*c**2*f* 
tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f* 
x/2) - 3*c**2*f) + 6*B*a*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**3 
 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 
24*B*a*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + 
 f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 10*B*a/(3*c**2*f*tan( 
e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) 
 - 3*c**2*f), Ne(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + a)/(-c*sin(e) + c)* 
*2, True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (67) = 134\).

Time = 0.13 (sec) , antiderivative size = 456, normalized size of antiderivative = 6.33 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {2 \, {\left (B a {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 4}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{2}}\right )} - \frac {A a {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {A a {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {B a {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \] Input:

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

Output:

2/3*(B*a*((9*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + 
 e) + 1)^2 - 4)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f 
*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) 
+ 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) - A*a*(3*sin(f*x + e)/(co 
s(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c^2 - 3*c^2* 
sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^ 
2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + A*a*(3*sin(f*x + e)/(cos(f* 
x + e) + 1) - 1)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin( 
f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) 
 + B*a*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 1)/(c^2 - 3*c^2*sin(f*x + e)/( 
cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f* 
x + e)^3/(cos(f*x + e) + 1)^3))/f
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.21 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {\frac {3 \, {\left (f x + e\right )} B a}{c^{2}} - \frac {2 \, {\left (3 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A a - 5 \, B a\right )}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}}{3 \, f} \] Input:

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

Output:

1/3*(3*(f*x + e)*B*a/c^2 - 2*(3*A*a*tan(1/2*f*x + 1/2*e)^2 - 3*B*a*tan(1/2 
*f*x + 1/2*e)^2 + 12*B*a*tan(1/2*f*x + 1/2*e) + A*a - 5*B*a)/(c^2*(tan(1/2 
*f*x + 1/2*e) - 1)^3))/f
 

Mupad [B] (verification not implemented)

Time = 34.88 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.83 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {B\,a\,x}{c^2}-\frac {\left (\frac {a\,\left (6\,A-6\,B+9\,B\,\left (e+f\,x\right )\right )}{3}-3\,B\,a\,\left (e+f\,x\right )\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\left (\frac {a\,\left (24\,B-9\,B\,\left (e+f\,x\right )\right )}{3}+3\,B\,a\,\left (e+f\,x\right )\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+\frac {a\,\left (2\,A-10\,B+3\,B\,\left (e+f\,x\right )\right )}{3}-B\,a\,\left (e+f\,x\right )}{c^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^3} \] Input:

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

Output:

(B*a*x)/c^2 - ((a*(2*A - 10*B + 3*B*(e + f*x)))/3 + tan(e/2 + (f*x)/2)^2*( 
(a*(6*A - 6*B + 9*B*(e + f*x)))/3 - 3*B*a*(e + f*x)) + tan(e/2 + (f*x)/2)* 
((a*(24*B - 9*B*(e + f*x)))/3 + 3*B*a*(e + f*x)) - B*a*(e + f*x))/(c^2*f*( 
tan(e/2 + (f*x)/2) - 1)^3)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.15 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {a \left (-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a +3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b f x +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b -9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b f x -6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b f x -18 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b -3 b f x +8 b \right )}{3 c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )} \] Input:

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

Output:

(a*( - 2*tan((e + f*x)/2)**3*a + 3*tan((e + f*x)/2)**3*b*f*x + 2*tan((e + 
f*x)/2)**3*b - 9*tan((e + f*x)/2)**2*b*f*x - 6*tan((e + f*x)/2)*a + 9*tan( 
(e + f*x)/2)*b*f*x - 18*tan((e + f*x)/2)*b - 3*b*f*x + 8*b))/(3*c**2*f*(ta 
n((e + f*x)/2)**3 - 3*tan((e + f*x)/2)**2 + 3*tan((e + f*x)/2) - 1))