\(\int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1256]

Optimal result

Integrand size = 29, antiderivative size = 254 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {x}{2 b^2}-\frac {3 \left (a^2-b^2\right ) x}{b^4}-\frac {2 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a b^4 d}+\frac {4 \left (2 a^6-3 a^4 b^2+b^6\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^4 \sqrt {a^2-b^2} d}+\frac {2 b \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {2 a \cos (c+d x)}{b^3 d}-\frac {\cot (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a^2 b^3 d (a+b \sin (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 3.85 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2 \left (-6 a^2+5 b^2\right ) (c+d x)}{b^4}+\frac {8 \left (a^2-b^2\right )^{3/2} \left (3 a^2+2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^4}-\frac {8 a \cos (c+d x)}{b^3}-\frac {2 \cot \left (\frac {1}{2} (c+d x)\right )}{a^2}+\frac {8 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}-\frac {8 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}-\frac {4 \left (a^2-b^2\right )^2 \cos (c+d x)}{a^2 b^3 (a+b \sin (c+d x))}+\frac {\sin (2 (c+d x))}{b^2}+\frac {2 \tan \left (\frac {1}{2} (c+d x)\right )}{a^2}}{4 d} \] Input:

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]
 

Output:

((2*(-6*a^2 + 5*b^2)*(c + d*x))/b^4 + (8*(a^2 - b^2)^(3/2)*(3*a^2 + 2*b^2) 
*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^4) - (8*a*Cos[c 
+ d*x])/b^3 - (2*Cot[(c + d*x)/2])/a^2 + (8*b*Log[Cos[(c + d*x)/2]])/a^3 - 
 (8*b*Log[Sin[(c + d*x)/2]])/a^3 - (4*(a^2 - b^2)^2*Cos[c + d*x])/(a^2*b^3 
*(a + b*Sin[c + d*x])) + Sin[2*(c + d*x)]/b^2 + (2*Tan[(c + d*x)/2])/a^2)/ 
(4*d)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3376, 2009}

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.

\(\displaystyle \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3376

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

\(\Big \downarrow \) 2009

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

Input:

Int[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3376

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) 
 + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(d*sin[ 
e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x] /; Fr 
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && ( 
LtQ[m, -1] || (EqQ[m, -1] && GtQ[p, 0]))
 

Maple [A] (verified)

Time = 11.43 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}+\frac {\frac {2 \left (-b^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-b a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 \left (3 a^{6}-4 a^{4} b^{2}-a^{2} b^{4}+2 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a^{3} b^{4}}-\frac {2 \left (\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+2 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+2 a b}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (6 a^{2}-5 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}-\frac {1}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}}{d}\)	\(304\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}+\frac {\frac {2 \left (-b^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-b a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 \left (3 a^{6}-4 a^{4} b^{2}-a^{2} b^{4}+2 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a^{3} b^{4}}-\frac {2 \left (\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+2 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+2 a b}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (6 a^{2}-5 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}-\frac {1}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}}{d}\)	\(304\)
risch	\(-\frac {3 x \,a^{2}}{b^{4}}+\frac {5 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 \left (i a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 i b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+a^{5} {\mathrm e}^{3 i \left (d x +c \right )}-2 a^{3} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}-i a^{4} b +2 i a^{2} b^{3}-2 i b^{5}-{\mathrm e}^{i \left (d x +c \right )} a^{5}+2 a^{3} b^{2} {\mathrm e}^{i \left (d x +c \right )}-3 a \,b^{4} {\mathrm e}^{i \left (d x +c \right )}\right )}{a^{2} d \,b^{4} \left (2 a \,{\mathrm e}^{3 i \left (d x +c \right )}-i b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}+2 i b \,{\mathrm e}^{2 i \left (d x +c \right )}-i b \right )}-\frac {3 \sqrt {-a^{2}+b^{2}}\, a \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}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2} a}+\frac {2 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{3}}+\frac {3 \sqrt {-a^{2}+b^{2}}\, a \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}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2} a}-\frac {2 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{3}}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\)	\(672\)

Input:

int(cos(d*x+c)^4*cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 901, normalized size of antiderivative = 3.55 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^4*cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

[-1/2*(3*a^4*b^2*cos(d*x + c)^3 + (6*a^5*b - 5*a^3*b^3)*d*x*cos(d*x + c)^2 
 - (6*a^5*b - 5*a^3*b^3)*d*x - (3*a^4*b - a^2*b^3 - 2*b^5 - (3*a^4*b - a^2 
*b^3 - 2*b^5)*cos(d*x + c)^2 + (3*a^5 - a^3*b^2 - 2*a*b^4)*sin(d*x + c))*s 
qrt(-a^2 + b^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*a^4*b^2 + 
2*a^2*b^4)*cos(d*x + c) - 2*(b^6*cos(d*x + c)^2 - a*b^5*sin(d*x + c) - b^6 
)*log(1/2*cos(d*x + c) + 1/2) + 2*(b^6*cos(d*x + c)^2 - a*b^5*sin(d*x + c) 
 - b^6)*log(-1/2*cos(d*x + c) + 1/2) - (a^3*b^3*cos(d*x + c)^3 + (6*a^6 - 
5*a^4*b^2)*d*x + (6*a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c))*sin(d*x + c 
))/(a^3*b^5*d*cos(d*x + c)^2 - a^4*b^4*d*sin(d*x + c) - a^3*b^5*d), -1/2*( 
3*a^4*b^2*cos(d*x + c)^3 + (6*a^5*b - 5*a^3*b^3)*d*x*cos(d*x + c)^2 - (6*a 
^5*b - 5*a^3*b^3)*d*x - 2*(3*a^4*b - a^2*b^3 - 2*b^5 - (3*a^4*b - a^2*b^3 
- 2*b^5)*cos(d*x + c)^2 + (3*a^5 - a^3*b^2 - 2*a*b^4)*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))) - ( 
3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c) - 2*(b^6*cos(d*x + c)^2 - a*b^5*sin(d* 
x + c) - b^6)*log(1/2*cos(d*x + c) + 1/2) + 2*(b^6*cos(d*x + c)^2 - a*b^5* 
sin(d*x + c) - b^6)*log(-1/2*cos(d*x + c) + 1/2) - (a^3*b^3*cos(d*x + c)^3 
 + (6*a^6 - 5*a^4*b^2)*d*x + (6*a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c)) 
*sin(d*x + c))/(a^3*b^5*d*cos(d*x + c)^2 - a^4*b^4*d*sin(d*x + c) - a^3...
 

Sympy [F]

\[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cos ^{4}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(cos(d*x+c)**4*cot(d*x+c)**2/(a+b*sin(d*x+c))**2,x)
 

Output:

Integral(cos(c + d*x)**4*cot(c + d*x)**2/(a + b*sin(c + d*x))**2, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^4*cot(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 = 0.24 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.51 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} + \frac {3 \, {\left (6 \, a^{2} - 5 \, b^{2}\right )} {\left (d x + c\right )}}{b^{4}} + \frac {6 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}} - \frac {12 \, {\left (3 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} + 2 \, b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3} b^{4}} - \frac {4 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 21 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 14 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b^{3}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{3} b^{3}}}{6 \, d} \] Input:

integrate(cos(d*x+c)^4*cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

-1/6*(12*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 3*tan(1/2*d*x + 1/2*c)/a^2 
 + 3*(6*a^2 - 5*b^2)*(d*x + c)/b^4 + 6*(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)/((tan(1/2*d*x + 1/2*c) 
^2 + 1)^2*b^3) - 12*(3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*(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)*a^3*b^4) - (4*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 12*a^ 
4*b*tan(1/2*d*x + 1/2*c)^2 + 21*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 - 4*b^5*tan 
(1/2*d*x + 1/2*c)^2 - 12*a^5*tan(1/2*d*x + 1/2*c) + 24*a^3*b^2*tan(1/2*d*x 
 + 1/2*c) - 14*a*b^4*tan(1/2*d*x + 1/2*c) - 3*a^2*b^3)/((a*tan(1/2*d*x + 1 
/2*c)^3 + 2*b*tan(1/2*d*x + 1/2*c)^2 + a*tan(1/2*d*x + 1/2*c))*a^3*b^3))/d
 

Mupad [B] (verification not implemented)

Time = 25.46 (sec) , antiderivative size = 5214, normalized size of antiderivative = 20.53 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)^4*cot(c + d*x)^2)/(a + b*sin(c + d*x))^2,x)
 

Output:

tan(c/2 + (d*x)/2)/(2*a^2*d) - (a + (2*tan(c/2 + (d*x)/2)*(6*a^4 + 3*b^4 - 
 4*a^2*b^2))/b^3 + (2*tan(c/2 + (d*x)/2)^5*(6*a^4 + 3*b^4 - 2*a^2*b^2))/b^ 
3 + (4*tan(c/2 + (d*x)/2)^3*(6*a^4 + 3*b^4 - 5*a^2*b^2))/b^3 + (tan(c/2 + 
(d*x)/2)^6*(6*a^4 + 4*b^4 - 7*a^2*b^2))/(a*b^2) + (tan(c/2 + (d*x)/2)^2*(1 
8*a^4 + 4*b^4 - 5*a^2*b^2))/(a*b^2) + (tan(c/2 + (d*x)/2)^4*(24*a^4 + 8*b^ 
4 - 13*a^2*b^2))/(a*b^2))/(d*(6*a^3*tan(c/2 + (d*x)/2)^3 + 6*a^3*tan(c/2 + 
 (d*x)/2)^5 + 2*a^3*tan(c/2 + (d*x)/2)^7 + 2*a^3*tan(c/2 + (d*x)/2) + 4*a^ 
2*b*tan(c/2 + (d*x)/2)^2 + 8*a^2*b*tan(c/2 + (d*x)/2)^4 + 4*a^2*b*tan(c/2 
+ (d*x)/2)^6)) + (atan((((a^2*6i - b^2*5i)*((8*(378*a^15 - 40*a^3*b^12 + 4 
88*a^5*b^10 - 1158*a^7*b^8 + 541*a^9*b^6 + 1008*a^11*b^4 - 1215*a^13*b^2)) 
/(a^6*b^8) - ((a^2*6i - b^2*5i)*(((a^2*6i - b^2*5i)*((((8*(16*a^8*b^13 - 1 
2*a^10*b^11))/(a^6*b^8) + (8*tan(c/2 + (d*x)/2)*(64*a^7*b^18 - 68*a^9*b^16 
 + 8*a^11*b^14))/(a^6*b^12))*(a^2*6i - b^2*5i))/(2*b^4) - (8*(64*a^5*b^14 
- 48*a^7*b^12 - 50*a^9*b^10 + 42*a^11*b^8))/(a^6*b^8) + (8*tan(c/2 + (d*x) 
/2)*(136*a^6*b^17 - 128*a^4*b^19 + 96*a^8*b^15 - 160*a^10*b^13 + 48*a^12*b 
^11))/(a^6*b^12)))/(2*b^4) - (8*(48*a^4*b^13 - 64*a^2*b^15 + 100*a^6*b^11 
- 184*a^8*b^9 + 315*a^10*b^7 - 324*a^12*b^5 + 108*a^14*b^3))/(a^6*b^8) + ( 
8*tan(c/2 + (d*x)/2)*(16*a^3*b^18 + 240*a^5*b^16 + 20*a^7*b^14 - 1696*a^9* 
b^12 + 2369*a^11*b^10 - 1020*a^13*b^8 + 72*a^15*b^6))/(a^6*b^12)))/(2*b^4) 
 + (8*tan(c/2 + (d*x)/2)*(32*b^19 - 32*a^2*b^17 + 680*a^4*b^15 - 2560*a...
 

Reduce [F]

\[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cos \left (d x +c \right )^{4} \cot \left (d x +c \right )^{2}}{\left (\sin \left (d x +c \right ) b +a \right )^{2}}d x \] Input:

int(cos(d*x+c)^4*cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x)
 

Output:

int(cos(d*x+c)^4*cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x)