\(\int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{a+b \sin (c+d x)} \, dx\) [1322]

Optimal result

Integrand size = 29, antiderivative size = 197 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a x}{b^2}-\frac {2 \left (a^2-b^2\right )^{5/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 b^2 d}+\frac {b \text {arctanh}(\cos (c+d x))}{2 a^2 d}-\frac {b \left (3 a^2-b^2\right ) \text {arctanh}(\cos (c+d x))}{a^4 d}+\frac {\cos (c+d x)}{b d}-\frac {\cot (c+d x)}{a d}+\frac {\left (3 a^2-b^2\right ) \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {b \cot (c+d x) \csc (c+d x)}{2 a^2 d} \] Output:

a*x/b^2-2*(a^2-b^2)^(5/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2)) 
/a^4/b^2/d+1/2*b*arctanh(cos(d*x+c))/a^2/d-b*(3*a^2-b^2)*arctanh(cos(d*x+c 
))/a^4/d+cos(d*x+c)/b/d-cot(d*x+c)/a/d+(3*a^2-b^2)*cot(d*x+c)/a^3/d-1/3*co 
t(d*x+c)^3/a/d+1/2*b*cot(d*x+c)*csc(d*x+c)/a^2/d
 

Mathematica [A] (verified)

Time = 6.70 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.92 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a (c+d x)}{b^2 d}-\frac {2 \left (a^2-b^2\right )^{5/2} \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b \cos \left (\frac {1}{2} (c+d x)\right )+a \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^4 b^2 d}+\frac {\cos (c+d x)}{b d}+\frac {\left (7 a^2 \cos \left (\frac {1}{2} (c+d x)\right )-3 b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right ) \csc \left (\frac {1}{2} (c+d x)\right )}{6 a^3 d}+\frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 a d}+\frac {\left (-5 a^2 b+2 b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac {\left (5 a^2 b-2 b^3\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}-\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (-7 a^2 \sin \left (\frac {1}{2} (c+d x)\right )+3 b^2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^3 d}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{24 a d} \] Input:

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

Output:

(a*(c + d*x))/(b^2*d) - (2*(a^2 - b^2)^(5/2)*ArcTan[(Sec[(c + d*x)/2]*(b*C 
os[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]])/(a^4*b^2*d) + Cos 
[c + d*x]/(b*d) + ((7*a^2*Cos[(c + d*x)/2] - 3*b^2*Cos[(c + d*x)/2])*Csc[( 
c + d*x)/2])/(6*a^3*d) + (b*Csc[(c + d*x)/2]^2)/(8*a^2*d) - (Cot[(c + d*x) 
/2]*Csc[(c + d*x)/2]^2)/(24*a*d) + ((-5*a^2*b + 2*b^3)*Log[Cos[(c + d*x)/2 
]])/(2*a^4*d) + ((5*a^2*b - 2*b^3)*Log[Sin[(c + d*x)/2]])/(2*a^4*d) - (b*S 
ec[(c + d*x)/2]^2)/(8*a^2*d) + (Sec[(c + d*x)/2]*(-7*a^2*Sin[(c + d*x)/2] 
+ 3*b^2*Sin[(c + d*x)/2]))/(6*a^3*d) + (Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2 
])/(24*a*d)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 197, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^6}{\sin (c+d x)^4 (a+b \sin (c+d x))}dx\)

\(\Big \downarrow \) 3376

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \text {arctanh}(\cos (c+d x))}{2 a^2 d}+\frac {b \cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac {2 \left (a^2-b^2\right )^{5/2} \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 b^2 d}-\frac {b \left (3 a^2-b^2\right ) \text {arctanh}(\cos (c+d x))}{a^4 d}+\frac {\left (3 a^2-b^2\right ) \cot (c+d x)}{a^3 d}+\frac {a x}{b^2}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\cot (c+d x)}{a d}+\frac {\cos (c+d x)}{b d}\)

Input:

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

Output:

(a*x)/b^2 - (2*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 
- b^2]])/(a^4*b^2*d) + (b*ArcTanh[Cos[c + d*x]])/(2*a^2*d) - (b*(3*a^2 - b 
^2)*ArcTanh[Cos[c + d*x]])/(a^4*d) + Cos[c + d*x]/(b*d) - Cot[c + d*x]/(a* 
d) + ((3*a^2 - b^2)*Cot[c + d*x])/(a^3*d) - Cot[c + d*x]^3/(3*a*d) + (b*Co 
t[c + d*x]*Csc[c + d*x])/(2*a^2*d)
 

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 = 3.62 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.37

method	result	size
derivativedivides	\(\frac {\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}+4 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{3}}+\frac {\frac {2 b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}+\frac {\left (-16 a^{6}+48 a^{4} b^{2}-48 a^{2} b^{4}+16 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{8 a^{4} b^{2} \sqrt {a^{2}-b^{2}}}-\frac {1}{24 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-9 a^{2}+4 b^{2}}{8 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \left (5 a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{4}}}{d}\)	\(269\)
default	\(\frac {\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}+4 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{3}}+\frac {\frac {2 b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}+\frac {\left (-16 a^{6}+48 a^{4} b^{2}-48 a^{2} b^{4}+16 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{8 a^{4} b^{2} \sqrt {a^{2}-b^{2}}}-\frac {1}{24 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-9 a^{2}+4 b^{2}}{8 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \left (5 a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{4}}}{d}\)	\(269\)
risch	\(\frac {a x}{b^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}-\frac {-18 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+24 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-14 i a^{2}+6 i b^{2}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {5 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{4}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2}}-\frac {2 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{4}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2}}+\frac {2 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{2}}-\frac {\sqrt {-a^{2}+b^{2}}\, b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{4}}-\frac {5 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 a^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{4}}\)	\(556\)

Input:

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

Output:

1/d*(1/8/a^3*(1/3*a^2*tan(1/2*d*x+1/2*c)^3-a*b*tan(1/2*d*x+1/2*c)^2-9*tan( 
1/2*d*x+1/2*c)*a^2+4*b^2*tan(1/2*d*x+1/2*c))+2/b^2*(b/(1+tan(1/2*d*x+1/2*c 
)^2)+a*arctan(tan(1/2*d*x+1/2*c)))+1/8*(-16*a^6+48*a^4*b^2-48*a^2*b^4+16*b 
^6)/a^4/b^2/(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))-1/24/a/tan(1/2*d*x+1/2*c)^3-1/8*(-9*a^2+4*b^2)/a^3/tan(1/2*d*x+ 
1/2*c)+1/8/a^2*b/tan(1/2*d*x+1/2*c)^2+1/2/a^4*b*(5*a^2-2*b^2)*ln(tan(1/2*d 
*x+1/2*c)))
 

Fricas [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 801, normalized size of antiderivative = 4.07 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/12*(4*(7*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^3 - 6*(a^4 - 2*a^2*b^2 + b^4 - 
 (a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2)*sqrt(-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))*sin(d*x + c) + 3*(5*a^2*b^3 - 2*b^5 - (5*a^2*b^ 
3 - 2*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*(5 
*a^2*b^3 - 2*b^5 - (5*a^2*b^3 - 2*b^5)*cos(d*x + c)^2)*log(-1/2*cos(d*x + 
c) + 1/2)*sin(d*x + c) - 12*(2*a^3*b^2 - a*b^4)*cos(d*x + c) + 6*(2*a^5*d* 
x*cos(d*x + c)^2 + 2*a^4*b*cos(d*x + c)^3 - 2*a^5*d*x - (2*a^4*b + a^2*b^3 
)*cos(d*x + c))*sin(d*x + c))/((a^4*b^2*d*cos(d*x + c)^2 - a^4*b^2*d)*sin( 
d*x + c)), 1/12*(4*(7*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^3 - 12*(a^4 - 2*a^2* 
b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2)*sqrt(a^2 - b^2)*arctan 
(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*sin(d*x + c) + 3*(5 
*a^2*b^3 - 2*b^5 - (5*a^2*b^3 - 2*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c 
) + 1/2)*sin(d*x + c) - 3*(5*a^2*b^3 - 2*b^5 - (5*a^2*b^3 - 2*b^5)*cos(d*x 
 + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 12*(2*a^3*b^2 - a*b^4 
)*cos(d*x + c) + 6*(2*a^5*d*x*cos(d*x + c)^2 + 2*a^4*b*cos(d*x + c)^3 - 2* 
a^5*d*x - (2*a^4*b + a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^4*b^2*d*cos( 
d*x + c)^2 - a^4*b^2*d)*sin(d*x + c))]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(cos(d*x+c)^2*cot(d*x+c)^4/(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 = 0.88 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.61 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {24 \, {\left (d x + c\right )} a}{b^{2}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} + \frac {48}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b} + \frac {12 \, {\left (5 \, a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {48 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4} b^{2}} - \frac {110 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \] Input:

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

Output:

1/24*(24*(d*x + c)*a/b^2 + (a^2*tan(1/2*d*x + 1/2*c)^3 - 3*a*b*tan(1/2*d*x 
 + 1/2*c)^2 - 27*a^2*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c))/a 
^3 + 48/((tan(1/2*d*x + 1/2*c)^2 + 1)*b) + 12*(5*a^2*b - 2*b^3)*log(abs(ta 
n(1/2*d*x + 1/2*c)))/a^4 - 48*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(pi*floo 
r(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqr 
t(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^4*b^2) - (110*a^2*b*tan(1/2*d*x + 1/2*c) 
^3 - 44*b^3*tan(1/2*d*x + 1/2*c)^3 - 27*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a* 
b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/(a^4*tan( 
1/2*d*x + 1/2*c)^3))/d
 

Mupad [B] (verification not implemented)

Time = 22.23 (sec) , antiderivative size = 4712, normalized size of antiderivative = 23.92 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

cos(c + d*x)/(32*a*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) + (3*sin 
(c + d*x))/(32*b*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) - (7*cos(3 
*c + 3*d*x))/(96*a*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) + sin(2* 
c + 2*d*x)/(32*b*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) - sin(3*c 
+ 3*d*x)/(32*b*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) - sin(4*c + 
4*d*x)/(64*b*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) + (b^2*cos(3*c 
 + 3*d*x))/(32*a^3*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) - (b^2*c 
os(c + d*x))/(32*a^3*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) + (b*s 
in(2*c + 2*d*x))/(32*a^2*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) + 
(15*b*sin(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(64*a^2*d*( 
(3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) - (3*a*sin(c + d*x)*atan((2*a^ 
5*cos(c/2 + (d*x)/2) - 2*b^5*sin(c/2 + (d*x)/2) + 5*a^2*b^3*sin(c/2 + (d*x 
)/2))/(2*b^5*cos(c/2 + (d*x)/2) + 2*a^5*sin(c/2 + (d*x)/2) - 5*a^2*b^3*cos 
(c/2 + (d*x)/2))))/(16*b^2*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) 
- (5*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x))/(64*a^ 
2*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) - (3*b^3*sin(c + d*x)*log 
(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(32*a^4*d*((3*sin(c + d*x))/32 - 
sin(3*c + 3*d*x)/32)) + (a*sin(3*c + 3*d*x)*atan((2*a^5*cos(c/2 + (d*x)/2) 
 - 2*b^5*sin(c/2 + (d*x)/2) + 5*a^2*b^3*sin(c/2 + (d*x)/2))/(2*b^5*cos(c/2 
 + (d*x)/2) + 2*a^5*sin(c/2 + (d*x)/2) - 5*a^2*b^3*cos(c/2 + (d*x)/2)))...
 

Reduce [F]

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

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

Output:

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