\(\int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1075]

Optimal result

Integrand size = 29, antiderivative size = 153 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\left (6 a^2-b^2\right ) x}{2 b^4}+\frac {2 a \left (3 a^2-2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2} d}-\frac {3 a \cos (c+d x)}{b^3 d}+\frac {3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac {\cos (c+d x) \sin ^2(c+d x)}{b d (a+b \sin (c+d x))} \] Output:

-1/2*(6*a^2-b^2)*x/b^4+2*a*(3*a^2-2*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/( 
a^2-b^2)^(1/2))/b^4/(a^2-b^2)^(1/2)/d-3*a*cos(d*x+c)/b^3/d+3/2*cos(d*x+c)* 
sin(d*x+c)/b^2/d-cos(d*x+c)*sin(d*x+c)^2/b/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \left (-6 a^2+b^2\right ) (c+d x)+\frac {8 a \left (3 a^2-2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-8 a b \cos (c+d x)-\frac {4 a^2 b \cos (c+d x)}{a+b \sin (c+d x)}+b^2 \sin (2 (c+d x))}{4 b^4 d} \] Input:

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

Output:

(2*(-6*a^2 + b^2)*(c + d*x) + (8*a*(3*a^2 - 2*b^2)*ArcTan[(b + a*Tan[(c + 
d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 8*a*b*Cos[c + d*x] - (4*a^2*b 
*Cos[c + d*x])/(a + b*Sin[c + d*x]) + b^2*Sin[2*(c + d*x)])/(4*b^4*d)
 

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.39, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.552, Rules used = {3042, 3368, 3042, 3527, 25, 3042, 3529, 25, 3042, 3502, 3042, 3214, 3042, 3139, 1083, 217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3368

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3529

\(\displaystyle \frac {\frac {\int -\frac {-6 a \left (a^2-b^2\right ) \sin ^2(c+d x)-b \left (a^2-b^2\right ) \sin (c+d x)+3 a \left (a^2-b^2\right )}{a+b \sin (c+d x)}dx}{2 b}+\frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\int \frac {-6 a \left (a^2-b^2\right ) \sin ^2(c+d x)-b \left (a^2-b^2\right ) \sin (c+d x)+3 a \left (a^2-b^2\right )}{a+b \sin (c+d x)}dx}{2 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\int \frac {-6 a \left (a^2-b^2\right ) \sin (c+d x)^2-b \left (a^2-b^2\right ) \sin (c+d x)+3 a \left (a^2-b^2\right )}{a+b \sin (c+d x)}dx}{2 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {\int \frac {3 a b \left (a^2-b^2\right )+\left (6 a^2-b^2\right ) \sin (c+d x) \left (a^2-b^2\right )}{a+b \sin (c+d x)}dx}{b}+\frac {6 a \left (a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {\int \frac {3 a b \left (a^2-b^2\right )+\left (6 a^2-b^2\right ) \sin (c+d x) \left (a^2-b^2\right )}{a+b \sin (c+d x)}dx}{b}+\frac {6 a \left (a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (6 a^2-b^2\right )}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{b}+\frac {6 a \left (a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (6 a^2-b^2\right )}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{b}+\frac {6 a \left (a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (6 a^2-b^2\right )}{b}-\frac {4 a \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}+\frac {6 a \left (a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {\frac {8 a \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {x \left (a^2-b^2\right ) \left (6 a^2-b^2\right )}{b}}{b}+\frac {6 a \left (a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (6 a^2-b^2\right )}{b}-\frac {4 a \left (3 a^2-2 b^2\right ) \sqrt {a^2-b^2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d}}{b}+\frac {6 a \left (a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

Input:

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

Output:

-((Cos[c + d*x]*Sin[c + d*x]^2)/(b*d*(a + b*Sin[c + d*x]))) + (-1/2*((((a^ 
2 - b^2)*(6*a^2 - b^2)*x)/b - (4*a*(3*a^2 - 2*b^2)*Sqrt[a^2 - b^2]*ArcTan[ 
(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(b*d))/b + (6*a*(a^2 - 
b^2)*Cos[c + d*x])/(b*d))/b + (3*(a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x])/(2 
*b*d))/(b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

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

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[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 3368

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a 
 + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n 
}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3527

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

rule 3529

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] : 
> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 
1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2))   Int[(a + b*Sin[e + f*x 
])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*( 
n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* 
(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 
0])))
 

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.37

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+2 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+2 a b}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (6 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}+\frac {2 a \left (\frac {-b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a b}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (3 a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{b^{4}}}{d}\)	\(210\)
default	\(\frac {-\frac {2 \left (\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+2 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+2 a b}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (6 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}+\frac {2 a \left (\frac {-b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a b}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (3 a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{b^{4}}}{d}\)	\(210\)
risch	\(-\frac {3 x \,a^{2}}{b^{4}}+\frac {x}{2 b^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{b^{3} d}-\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{b^{3} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} d}+\frac {2 i a^{2} \left (i b +a \,{\mathrm e}^{i \left (d x +c \right )}\right )}{b^{4} d \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\)	\(433\)

Input:

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

Output:

1/d*(-2/b^4*((1/2*tan(1/2*d*x+1/2*c)^3*b^2+2*a*b*tan(1/2*d*x+1/2*c)^2-1/2* 
b^2*tan(1/2*d*x+1/2*c)+2*a*b)/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(6*a^2-b^2)*a 
rctan(tan(1/2*d*x+1/2*c)))+2*a/b^4*((-b^2*tan(1/2*d*x+1/2*c)-a*b)/(tan(1/2 
*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)+(3*a^2-2*b^2)/(a^2-b^2)^(1/2)*ar 
ctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 568, normalized size of antiderivative = 3.71 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [-\frac {{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, a^{5} - 7 \, a^{3} b^{2} + a b^{4}\right )} d x - {\left (3 \, a^{4} - 2 \, a^{2} b^{2} + {\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + {\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + {\left ({\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} d x + 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{5} - b^{7}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{4} - a b^{6}\right )} d\right )}}, -\frac {{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, a^{5} - 7 \, a^{3} b^{2} + a b^{4}\right )} d x + 2 \, {\left (3 \, a^{4} - 2 \, a^{2} b^{2} + {\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + {\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + {\left ({\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} d x + 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{5} - b^{7}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{4} - a b^{6}\right )} d\right )}}\right ] \] Input:

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

Output:

[-1/2*((a^2*b^3 - b^5)*cos(d*x + c)^3 + (6*a^5 - 7*a^3*b^2 + a*b^4)*d*x - 
(3*a^4 - 2*a^2*b^2 + (3*a^3*b - 2*a*b^3)*sin(d*x + c))*sqrt(-a^2 + b^2)*lo 
g(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*c 
os(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x 
+ c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + (6*a^4*b - 7*a^2*b^3 + b^5)*co 
s(d*x + c) + ((6*a^4*b - 7*a^2*b^3 + b^5)*d*x + 3*(a^3*b^2 - a*b^4)*cos(d* 
x + c))*sin(d*x + c))/((a^2*b^5 - b^7)*d*sin(d*x + c) + (a^3*b^4 - a*b^6)* 
d), -1/2*((a^2*b^3 - b^5)*cos(d*x + c)^3 + (6*a^5 - 7*a^3*b^2 + a*b^4)*d*x 
 + 2*(3*a^4 - 2*a^2*b^2 + (3*a^3*b - 2*a*b^3)*sin(d*x + c))*sqrt(a^2 - b^2 
)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + (6*a^4*b 
- 7*a^2*b^3 + b^5)*cos(d*x + c) + ((6*a^4*b - 7*a^2*b^3 + b^5)*d*x + 3*(a^ 
3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^5 - b^7)*d*sin(d*x + c) 
 + (a^3*b^4 - a*b^6)*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {{\left (6 \, a^{2} - b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {4 \, {\left (3 \, a^{3} - 2 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {4 \, {\left (a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} b^{3}} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, d} \] Input:

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

Output:

-1/2*((6*a^2 - b^2)*(d*x + c)/b^4 - 4*(3*a^3 - 2*a*b^2)*(pi*floor(1/2*(d*x 
 + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^ 
2)))/(sqrt(a^2 - b^2)*b^4) + 4*(a*b*tan(1/2*d*x + 1/2*c) + a^2)/((a*tan(1/ 
2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*b^3) + 2*(b*tan(1/2*d*x + 
 1/2*c)^3 + 4*a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c) + 4*a)/((t 
an(1/2*d*x + 1/2*c)^2 + 1)^2*b^3))/d
 

Mupad [B] (verification not implemented)

Time = 36.73 (sec) , antiderivative size = 479, normalized size of antiderivative = 3.13 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {6\,a^2}{b^3}+\frac {9\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^2+b^2\right )}{b^3}+\frac {12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{b^2}+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{b^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (6\,a^2-b^2\right )}{b^3}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sqrt {b^2-a^2}\right )\,\left (3\,a^3\,\sqrt {b^2-a^2}-2\,a\,b^2\,\sqrt {b^2-a^2}\right )}{b^4\,d\,\left (a^2-b^2\right )}-\frac {a\,\ln \left (b+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (3\,a^2-2\,b^2\right )}{d\,\left (b^6-a^2\,b^4\right )}+\frac {\mathrm {atan}\left (\frac {24\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{24\,a^3-8\,a\,b^2+\frac {144\,a^5}{b^2}}+\frac {144\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{144\,a^5+24\,a^3\,b^2-8\,a\,b^4}-\frac {8\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {24\,a^3}{b^2}-8\,a+\frac {144\,a^5}{b^4}}\right )\,\left (a^2\,6{}\mathrm {i}-b^2\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b^4\,d} \] Input:

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

Output:

(atan((24*a^3*tan(c/2 + (d*x)/2))/(24*a^3 - 8*a*b^2 + (144*a^5)/b^2) + (14 
4*a^5*tan(c/2 + (d*x)/2))/(144*a^5 - 8*a*b^4 + 24*a^3*b^2) - (8*a*tan(c/2 
+ (d*x)/2))/((24*a^3)/b^2 - 8*a + (144*a^5)/b^4))*(a^2*6i - b^2*1i)*1i)/(b 
^4*d) - ((6*a^2)/b^3 + (9*a*tan(c/2 + (d*x)/2))/b^2 + (2*tan(c/2 + (d*x)/2 
)^4*(3*a^2 + b^2))/b^3 + (12*a*tan(c/2 + (d*x)/2)^3)/b^2 + (3*a*tan(c/2 + 
(d*x)/2)^5)/b^2 + (2*tan(c/2 + (d*x)/2)^2*(6*a^2 - b^2))/b^3)/(d*(a + 2*b* 
tan(c/2 + (d*x)/2) + 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2)^4 + 
 a*tan(c/2 + (d*x)/2)^6 + 4*b*tan(c/2 + (d*x)/2)^3 + 2*b*tan(c/2 + (d*x)/2 
)^5)) - (log(b + a*tan(c/2 + (d*x)/2) - (b^2 - a^2)^(1/2))*(3*a^3*(b^2 - a 
^2)^(1/2) - 2*a*b^2*(b^2 - a^2)^(1/2)))/(b^4*d*(a^2 - b^2)) - (a*log(b + a 
*tan(c/2 + (d*x)/2) + (b^2 - a^2)^(1/2))*(-(a + b)*(a - b))^(1/2)*(3*a^2 - 
 2*b^2))/(d*(b^6 - a^2*b^4))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.88 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) a^{3} b -8 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) a \,b^{3}+12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4}-8 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2} b^{2}+\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b^{3}-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{5}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{4}-6 \cos \left (d x +c \right ) a^{4} b +6 \cos \left (d x +c \right ) a^{2} b^{3}-6 \sin \left (d x +c \right ) a^{4} b d x -3 \sin \left (d x +c \right ) a^{3} b^{2}+7 \sin \left (d x +c \right ) a^{2} b^{3} d x +3 \sin \left (d x +c \right ) a \,b^{4}-\sin \left (d x +c \right ) b^{5} d x -6 a^{5} d x -3 a^{4} b +7 a^{3} b^{2} d x +3 a^{2} b^{3}-a \,b^{4} d x}{2 b^{4} d \left (\sin \left (d x +c \right ) a^{2} b -\sin \left (d x +c \right ) b^{3}+a^{3}-a \,b^{2}\right )} \] Input:

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

Output:

(12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin 
(c + d*x)*a**3*b - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt( 
a**2 - b**2))*sin(c + d*x)*a*b**3 + 12*sqrt(a**2 - b**2)*atan((tan((c + d* 
x)/2)*a + b)/sqrt(a**2 - b**2))*a**4 - 8*sqrt(a**2 - b**2)*atan((tan((c + 
d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**2*b**2 + cos(c + d*x)*sin(c + d*x)**2 
*a**2*b**3 - cos(c + d*x)*sin(c + d*x)**2*b**5 - 3*cos(c + d*x)*sin(c + d* 
x)*a**3*b**2 + 3*cos(c + d*x)*sin(c + d*x)*a*b**4 - 6*cos(c + d*x)*a**4*b 
+ 6*cos(c + d*x)*a**2*b**3 - 6*sin(c + d*x)*a**4*b*d*x - 3*sin(c + d*x)*a* 
*3*b**2 + 7*sin(c + d*x)*a**2*b**3*d*x + 3*sin(c + d*x)*a*b**4 - sin(c + d 
*x)*b**5*d*x - 6*a**5*d*x - 3*a**4*b + 7*a**3*b**2*d*x + 3*a**2*b**3 - a*b 
**4*d*x)/(2*b**4*d*(sin(c + d*x)*a**2*b - sin(c + d*x)*b**3 + a**3 - a*b** 
2))