\(\int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [316]

Optimal result

Integrand size = 29, antiderivative size = 76 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 x}{a^3}+\frac {3 \cos (c+d x)}{a^3 d}-\frac {\cos ^3(c+d x)}{3 d (a+a \sin (c+d x))^3}+\frac {2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2} \] Output:

3*x/a^3+3*cos(d*x+c)/a^3/d-1/3*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^3+2*cos(d*x 
+c)^3/a/d/(a+a*sin(d*x+c))^2
 

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {9 c+9 d x+3 \cos (c+d x)-\frac {2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) (11+13 \sin (c+d x))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}}{3 a^3 d} \] Input:

Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
 

Output:

(9*c + 9*d*x + 3*Cos[c + d*x] - 2/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 
- (2*Sin[(c + d*x)/2]*(11 + 13*Sin[c + d*x]))/(Cos[(c + d*x)/2] + Sin[(c + 
 d*x)/2])^3)/(3*a^3*d)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 3350, 3042, 3159, 3042, 3161, 24}

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.

\(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^2(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^2}{(a \sin (c+d x)+a)^3}dx\)

\(\Big \downarrow \) 3350

\(\displaystyle -\int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^3}dx-\frac {\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^3}dx-\frac {\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3159

\(\displaystyle \frac {3 \int \frac {\cos ^2(c+d x)}{\sin (c+d x) a+a}dx}{a^2}+\frac {2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}-\frac {\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {\cos (c+d x)^2}{\sin (c+d x) a+a}dx}{a^2}+\frac {2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}-\frac {\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3161

\(\displaystyle \frac {3 \left (\frac {\int 1dx}{a}+\frac {\cos (c+d x)}{a d}\right )}{a^2}+\frac {2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}-\frac {\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {3 \left (\frac {\cos (c+d x)}{a d}+\frac {x}{a}\right )}{a^2}+\frac {2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}-\frac {\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3}\)

Input:

Int[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
 

Output:

(3*(x/a + Cos[c + d*x]/(a*d)))/a^2 - Cos[c + d*x]^3/(3*d*(a + a*Sin[c + d* 
x])^3) + (2*Cos[c + d*x]^3)/(a*d*(a + a*Sin[c + d*x])^2)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 3159

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f 
*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 
)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; 
FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & 
& NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
 

rule 3161

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si 
mp[g^2/a   Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x 
] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
 

rule 3350

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((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/(a*f*g*m)), x] - Simp[1/g^2   Int[(g*Cos[e + 
 f*x])^(p + 2)*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, m, p 
}, x] && EqQ[a^2 - b^2, 0] && EqQ[m + p + 1, 0]
 

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {6}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {8}{4+4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}\)	\(84\)
default	\(\frac {-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {6}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {8}{4+4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}\)	\(84\)
risch	\(\frac {3 x}{a^{3}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {-\frac {26}{3}+16 i {\mathrm e}^{i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{3}}\)	\(89\)
parallelrisch	\(\frac {54 d x \cos \left (d x +c \right )+18 d x \cos \left (3 d x +3 c \right )+72 \cos \left (2 d x +2 c \right )+84 \cos \left (d x +c \right )+6 \sin \left (d x +c \right )-26 \sin \left (3 d x +3 c \right )+3 \cos \left (4 d x +4 c \right )+28 \cos \left (3 d x +3 c \right )+37}{6 d \,a^{3} \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(114\)
norman	\(\frac {\frac {28}{3 a d}+\frac {252 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a}+\frac {213 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{a}+\frac {153 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a}+\frac {3 x}{a}+\frac {602 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}+\frac {824 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d a}+\frac {388 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d a}+\frac {1252 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d a}+\frac {922 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}+\frac {1100 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d a}+\frac {122 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}+\frac {30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d a}+\frac {302 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d a}+\frac {506 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d a}+\frac {238 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d a}+\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d a}+\frac {153 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {213 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a}+\frac {252 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {90 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{a}+\frac {42 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{a}+\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{a}+\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}+\frac {42 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {90 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(493\)

Input:

int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

8/d/a^3*(-1/3/(tan(1/2*d*x+1/2*c)+1)^3+1/2/(tan(1/2*d*x+1/2*c)+1)^2+3/4/(t 
an(1/2*d*x+1/2*c)+1)+1/4/(1+tan(1/2*d*x+1/2*c)^2)+3/4*arctan(tan(1/2*d*x+1 
/2*c)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.89 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {{\left (9 \, d x - 16\right )} \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )^{3} - 18 \, d x - {\left (9 \, d x + 17\right )} \cos \left (d x + c\right ) - {\left (18 \, d x + {\left (9 \, d x + 19\right )} \cos \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) + 2}{3 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - 2 \, a^{3} d - {\left (a^{3} d \cos \left (d x + c\right ) + 2 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="frica 
s")
 

Output:

1/3*((9*d*x - 16)*cos(d*x + c)^2 + 3*cos(d*x + c)^3 - 18*d*x - (9*d*x + 17 
)*cos(d*x + c) - (18*d*x + (9*d*x + 19)*cos(d*x + c) + 3*cos(d*x + c)^2 + 
2)*sin(d*x + c) + 2)/(a^3*d*cos(d*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*d 
- (a^3*d*cos(d*x + c) + 2*a^3*d)*sin(d*x + c))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1246 vs. \(2 (68) = 136\).

Time = 15.02 (sec) , antiderivative size = 1246, normalized size of antiderivative = 16.39 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)**2*sin(d*x+c)**2/(a+a*sin(d*x+c))**3,x)
 

Output:

Piecewise((9*d*x*tan(c/2 + d*x/2)**5/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a** 
3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/ 
2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 27*d*x*tan(c/2 + d 
*x/2)**4/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12 
*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan 
(c/2 + d*x/2) + 3*a**3*d) + 36*d*x*tan(c/2 + d*x/2)**3/(3*a**3*d*tan(c/2 + 
 d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 
+ 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 
36*d*x*tan(c/2 + d*x/2)**2/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/ 
2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2) 
**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 27*d*x*tan(c/2 + d*x/2)/(3*a 
**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c 
/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) 
 + 3*a**3*d) + 9*d*x/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d* 
x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 
9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 18*tan(c/2 + d*x/2)**4/(3*a**3*d*t 
an(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d* 
x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a* 
*3*d) + 54*tan(c/2 + d*x/2)**3/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*ta 
n(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 +...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (74) = 148\).

Time = 0.11 (sec) , antiderivative size = 228, normalized size of antiderivative = 3.00 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\frac {\frac {33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {29 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {27 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 14}{a^{3} + \frac {3 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{3 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxim 
a")
 

Output:

2/3*((33*sin(d*x + c)/(cos(d*x + c) + 1) + 29*sin(d*x + c)^2/(cos(d*x + c) 
 + 1)^2 + 27*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 9*sin(d*x + c)^4/(cos(d 
*x + c) + 1)^4 + 14)/(a^3 + 3*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^3* 
sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^3*sin(d*x + c)^3/(cos(d*x + c) + 
 1)^3 + 3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(co 
s(d*x + c) + 1)^5) + 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {9 \, {\left (d x + c\right )}}{a^{3}} + \frac {6}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} + \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{3}}}{3 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac" 
)
 

Output:

1/3*(9*(d*x + c)/a^3 + 6/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) + 2*(9*tan(1/2 
*d*x + 1/2*c)^2 + 24*tan(1/2*d*x + 1/2*c) + 11)/(a^3*(tan(1/2*d*x + 1/2*c) 
 + 1)^3))/d
 

Mupad [B] (verification not implemented)

Time = 19.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3\,x}{a^3}+\frac {6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {58\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+22\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {28}{3}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^3\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:

int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + a*sin(c + d*x))^3,x)
 

Output:

(3*x)/a^3 + (22*tan(c/2 + (d*x)/2) + (58*tan(c/2 + (d*x)/2)^2)/3 + 18*tan( 
c/2 + (d*x)/2)^3 + 6*tan(c/2 + (d*x)/2)^4 + 28/3)/(a^3*d*(tan(c/2 + (d*x)/ 
2) + 1)^3*(tan(c/2 + (d*x)/2)^2 + 1))
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.22 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x -11 \cos \left (d x +c \right ) \sin \left (d x +c \right )+9 \cos \left (d x +c \right ) d x -6 \cos \left (d x +c \right )-3 \sin \left (d x +c \right )^{3}-9 \sin \left (d x +c \right )^{2} d x -24 \sin \left (d x +c \right )^{2}-18 \sin \left (d x +c \right ) d x -11 \sin \left (d x +c \right )-9 d x +6}{3 a^{3} d \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+\cos \left (d x +c \right )-\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )-1\right )} \] Input:

int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x)
 

Output:

( - 3*cos(c + d*x)*sin(c + d*x)**2 + 9*cos(c + d*x)*sin(c + d*x)*d*x - 11* 
cos(c + d*x)*sin(c + d*x) + 9*cos(c + d*x)*d*x - 6*cos(c + d*x) - 3*sin(c 
+ d*x)**3 - 9*sin(c + d*x)**2*d*x - 24*sin(c + d*x)**2 - 18*sin(c + d*x)*d 
*x - 11*sin(c + d*x) - 9*d*x + 6)/(3*a**3*d*(cos(c + d*x)*sin(c + d*x) + c 
os(c + d*x) - sin(c + d*x)**2 - 2*sin(c + d*x) - 1))