\(\int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [1283]

Optimal result

Integrand size = 29, antiderivative size = 148 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac {2 a^2 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 d}+\frac {\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{3 b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {-12 a^3 c+6 a b^2 c-12 a^3 d x+6 a b^2 d x+24 a^2 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+3 b \left (-4 a^2+b^2\right ) \cos (c+d x)+b^3 \cos (3 (c+d x))+3 a b^2 \sin (2 (c+d x))}{12 b^4 d} \] Input:

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

Output:

-1/12*(-12*a^3*c + 6*a*b^2*c - 12*a^3*d*x + 6*a*b^2*d*x + 24*a^2*Sqrt[a^2 
- b^2]*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 3*b*(-4*a^2 + b^ 
2)*Cos[c + d*x] + b^3*Cos[3*(c + d*x)] + 3*a*b^2*Sin[2*(c + d*x)])/(b^4*d)
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.16, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.586, Rules used = {3042, 3368, 3042, 3529, 25, 3042, 3528, 25, 3042, 3502, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3368

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3529

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 3528

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (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)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 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])))
 

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 = 0.63 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {-\frac {2 a^{2} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{4}}+\frac {\frac {2 \left (\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}+\left (a^{2} b -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+a^{2} b -\frac {b^{3}}{3}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+a \left (2 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\)	\(184\)
default	\(\frac {-\frac {2 a^{2} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{4}}+\frac {\frac {2 \left (\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}+\left (a^{2} b -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+a^{2} b -\frac {b^{3}}{3}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+a \left (2 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\)	\(184\)
risch	\(\frac {a^{3} x}{b^{4}}-\frac {a x}{2 b^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{8 b d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{8 b d}-\frac {\sqrt {-a^{2}+b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}+\frac {\sqrt {-a^{2}+b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}-\frac {\cos \left (3 d x +3 c \right )}{12 b d}-\frac {a \sin \left (2 d x +2 c \right )}{4 b^{2} d}\)	\(232\)

Input:

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

Output:

1/d*(-2*a^2*(a^2-b^2)^(1/2)/b^4*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a 
^2-b^2)^(1/2))+2/b^4*((1/2*a*b^2*tan(1/2*d*x+1/2*c)^5+(a^2*b-b^3)*tan(1/2* 
d*x+1/2*c)^4+2*a^2*b*tan(1/2*d*x+1/2*c)^2-1/2*a*b^2*tan(1/2*d*x+1/2*c)+a^2 
*b-1/3*b^3)/(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*a*(2*a^2-b^2)*arctan(tan(1/2*d* 
x+1/2*c))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.13 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {2 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} b \cos \left (d x + c\right ) - 3 \, \sqrt {-a^{2} + b^{2}} a^{2} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 3 \, {\left (2 \, a^{3} - a b^{2}\right )} d x}{6 \, b^{4} d}, -\frac {2 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} b \cos \left (d x + c\right ) - 6 \, \sqrt {a^{2} - b^{2}} a^{2} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 3 \, {\left (2 \, a^{3} - a b^{2}\right )} d x}{6 \, b^{4} d}\right ] \] Input:

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

Output:

[-1/6*(2*b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)*sin(d*x + c) - 6*a^2*b* 
cos(d*x + c) - 3*sqrt(-a^2 + b^2)*a^2*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 
2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(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)) - 3*(2*a^3 - a*b^2)*d*x)/(b^4*d), -1/6*(2*b^3*cos(d*x + c)^3 + 3*a* 
b^2*cos(d*x + c)*sin(d*x + c) - 6*a^2*b*cos(d*x + c) - 6*sqrt(a^2 - b^2)*a 
^2*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 3*(2*a^3 
 - a*b^2)*d*x)/(b^4*d)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

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 = 0.20 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {3 \, {\left (2 \, a^{3} - a b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {12 \, {\left (a^{4} - a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} - 2 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \] Input:

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

Output:

1/6*(3*(2*a^3 - a*b^2)*(d*x + c)/b^4 - 12*(a^4 - a^2*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) + 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*tan 
(1/2*d*x + 1/2*c)^4 - 6*b^2*tan(1/2*d*x + 1/2*c)^4 + 12*a^2*tan(1/2*d*x + 
1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*a^2 - 2*b^2)/((tan(1/2*d*x + 1/2 
*c)^2 + 1)^3*b^3))/d
 

Mupad [B] (verification not implemented)

Time = 21.59 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.52 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2\,\cos \left (c+d\,x\right )}{b^3\,d}-\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4}}{b^2\,d}-\frac {\frac {\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{12}}{b\,d}+\frac {2\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^4\,d}+\frac {2\,a^2\,\mathrm {atanh}\left (\frac {-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{\sqrt {b^2-a^2}\,\left (a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )\,\sqrt {b^2-a^2}}{b^4\,d} \] Input:

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

Output:

(a^2*cos(c + d*x))/(b^3*d) - (a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2) 
) + (a*sin(2*c + 2*d*x))/4)/(b^2*d) - (cos(c + d*x)/4 + cos(3*c + 3*d*x)/1 
2)/(b*d) + (2*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^4*d) + ( 
2*a^2*atanh((2*b^2*sin(c/2 + (d*x)/2) - a^2*sin(c/2 + (d*x)/2) + a*b*cos(c 
/2 + (d*x)/2))/((b^2 - a^2)^(1/2)*(a*cos(c/2 + (d*x)/2) + 2*b*sin(c/2 + (d 
*x)/2))))*(b^2 - a^2)^(1/2))/(b^4*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{3}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2}+6 \cos \left (d x +c \right ) a^{2} b -2 \cos \left (d x +c \right ) b^{3}+6 a^{3} c +6 a^{3} d x +2 a^{2} b -3 a \,b^{2} c -3 a \,b^{2} d x +2 b^{3}}{6 b^{4} d} \] Input:

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

Output:

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