\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx\) [64]

Optimal result

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(156\) vs. \(2(72)=144\).

Time = 6.52 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.17 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\frac {c \left (-9 B f x \cos \left (\frac {f x}{2}\right )-6 (A-3 B) \cos \left (e+\frac {f x}{2}\right )+2 A \cos \left (e+\frac {3 f x}{2}\right )-14 B \cos \left (e+\frac {3 f x}{2}\right )+3 B f x \cos \left (2 e+\frac {3 f x}{2}\right )+24 B \sin \left (\frac {f x}{2}\right )-9 B f x \sin \left (e+\frac {f x}{2}\right )-3 B f x \sin \left (e+\frac {3 f x}{2}\right )\right )}{6 a^2 f \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \] Input:

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

Output:

(c*(-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*Si 
n[(f*x)/2] - 9*B*f*x*Sin[e + (f*x)/2] - 3*B*f*x*Sin[e + (3*f*x)/2]))/(6*a^ 
2*f*(Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3446, 3042, 3336, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3336

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3127

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

Input:

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

Output:

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

Defintions of rubi rules used

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 = 0.47 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.12

method	result	size
risch	\(-\frac {B c x}{a^{2}}+\frac {2 A c \,{\mathrm e}^{2 i \left (f x +e \right )}-8 i B c \,{\mathrm e}^{i \left (f x +e \right )}-6 B c \,{\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 A c}{3}+\frac {14 B c}{3}}{f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3}}\)	\(81\)
derivativedivides	\(\frac {2 c \left (-\frac {-4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {A +B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {4 A -4 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,a^{2}}\)	\(86\)
default	\(\frac {2 c \left (-\frac {-4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {A +B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {4 A -4 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,a^{2}}\)	\(86\)
parallelrisch	\(-\frac {2 \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} x f B}{2}+\left (\frac {3}{2} f x B +A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+B \left (\frac {3 f x}{2}+4\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {f x B}{2}+\frac {A}{3}+\frac {5 B}{3}\right ) c}{f \,a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\)	\(88\)
norman	\(\frac {-\frac {2 A c +10 B c}{3 a f}-\frac {B c x}{a}-\frac {16 B c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a f}-\frac {8 B c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}-\frac {8 B c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {\left (14 A c +22 B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3 a f}-\frac {\left (10 A c +26 B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 a f}-\frac {\left (2 A c +2 B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a f}-\frac {3 B c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}-\frac {5 B c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a}-\frac {7 B c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a}-\frac {7 B c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}-\frac {5 B c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a}-\frac {3 B c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a}-\frac {B c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2} a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\)	\(335\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (68) = 136\).

Time = 0.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.31 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\frac {6 \, B c f x - {\left (3 \, B c f x + {\left (A - 7 \, B\right )} c\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (A - B\right )} c + {\left (3 \, B c f x + {\left (A + 5 \, B\right )} c\right )} \cos \left (f x + e\right ) + {\left (6 \, B c f x - 2 \, {\left (A - B\right )} c + {\left (3 \, B c f x - {\left (A - 7 \, B\right )} c\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 702 vs. \(2 (70) = 140\).

Time = 2.16 (sec) , antiderivative size = 702, normalized size of antiderivative = 9.75 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (68) = 136\).

Time = 0.12 (sec) , antiderivative size = 452, normalized size of antiderivative = 6.28 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (B c {\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}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{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 )}{a^{2}}\right )} + \frac {A c {\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 )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} - \frac {A c {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {B c {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=-\frac {\frac {3 \, {\left (f x + e\right )} B c}{a^{2}} + \frac {2 \, {\left (3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A c + 5 \, B c\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 37.11 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.85 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=-\frac {B\,c\,x}{a^2}-\frac {\left (\frac {c\,\left (6\,A+6\,B+9\,B\,\left (e+f\,x\right )\right )}{3}-3\,B\,c\,\left (e+f\,x\right )\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\left (\frac {c\,\left (24\,B+9\,B\,\left (e+f\,x\right )\right )}{3}-3\,B\,c\,\left (e+f\,x\right )\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+\frac {c\,\left (2\,A+10\,B+3\,B\,\left (e+f\,x\right )\right )}{3}-B\,c\,\left (e+f\,x\right )}{a^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^3} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.15 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\frac {c \left (2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a -3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b f x +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b -9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b f x +6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a -9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b f x -18 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b -3 b f x -8 b \right )}{3 a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \] Input:

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

Output:

(c*(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*a**2*f*(tan(( 
e + f*x)/2)**3 + 3*tan((e + f*x)/2)**2 + 3*tan((e + f*x)/2) + 1))