\(\int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [629]

Optimal result

Integrand size = 29, antiderivative size = 91 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 x}{2 a}+\frac {\text {arctanh}(\cos (c+d x))}{a d}-\frac {\cos (c+d x)}{a d}-\frac {\cos ^3(c+d x)}{3 a d}-\frac {\cot (c+d x)}{a d}-\frac {\cos (c+d x) \sin (c+d x)}{2 a d} \] Output:

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

Mathematica [A] (verified)

Time = 2.19 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.34 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (27 \cos (c+d x)+6 \left (6 c+6 d x+5 \cos (c+d x)-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin (c+d x)+\cos (3 (c+d x)) (-3+2 \sin (c+d x))\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{48 a d (1+\sin (c+d x))} \] Input:

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

Output:

-1/48*((1 + Cot[(c + d*x)/2])^2*(27*Cos[c + d*x] + 6*(6*c + 6*d*x + 5*Cos[ 
c + d*x] - 4*Log[Cos[(c + d*x)/2]] + 4*Log[Sin[(c + d*x)/2]])*Sin[c + d*x] 
 + Cos[3*(c + d*x)]*(-3 + 2*Sin[c + d*x]))*Tan[(c + d*x)/2])/(a*d*(1 + Sin 
[c + d*x]))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {3042, 3318, 3042, 25, 3071, 252, 262, 216, 3072, 254, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3071

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 3072

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

\(\Big \downarrow \) 254

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\text {arctanh}(\cos (c+d x))-\frac {1}{3} \cos ^3(c+d x)-\cos (c+d x)}{a d}-\frac {\frac {3}{2} (\cot (c+d x)-\arctan (\cot (c+d x)))-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}}{a d}\)

Input:

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

Output:

(ArcTanh[Cos[c + d*x]] - Cos[c + d*x] - Cos[c + d*x]^3/3)/(a*d) - ((3*(-Ar 
cTan[Cot[c + d*x]] + Cot[c + d*x]))/2 - Cot[c + d*x]^3/(2*(1 + Cot[c + d*x 
]^2)))/(a*d)
 

Defintions of rubi rules used

rule 25

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

rule 216

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

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 254

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, 
 a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

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 3071

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S 
ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f)   Subst[I 
nt[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff)], 
x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
 

rule 3072

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f   Subst[Int[ 
(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x 
]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
 

rule 3318

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

Maple [A] (verified)

Time = 1.70 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {2}{3}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}\)	\(121\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {2}{3}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}\)	\(121\)
risch	\(-\frac {3 x}{2 a}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d a}-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d a}-\frac {2 i}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}-\frac {\cos \left (3 d x +3 c \right )}{12 a d}\)	\(156\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \, \cos \left (d x + c\right )^{3} - {\left (2 \, \cos \left (d x + c\right )^{3} + 9 \, d x + 6 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 3 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 9 \, \cos \left (d x + c\right )}{6 \, a d \sin \left (d x + c\right )} \] Input:

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

Output:

1/6*(3*cos(d*x + c)^3 - (2*cos(d*x + c)^3 + 9*d*x + 6*cos(d*x + c))*sin(d* 
x + c) + 3*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*log(-1/2*cos(d*x + 
 c) + 1/2)*sin(d*x + c) - 9*cos(d*x + c))/(a*d*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos ^{4}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

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

Output:

Integral(cos(c + d*x)**4*cot(c + d*x)**2/(sin(c + d*x) + 1), x)/a
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (85) = 170\).

Time = 0.11 (sec) , antiderivative size = 277, normalized size of antiderivative = 3.04 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {\frac {16 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {24 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {24 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 3}{\frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac {18 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, d} \] Input:

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

Output:

-1/6*((16*sin(d*x + c)/(cos(d*x + c) + 1) + 15*sin(d*x + c)^2/(cos(d*x + c 
) + 1)^2 + 24*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 9*sin(d*x + c)^4/(cos( 
d*x + c) + 1)^4 + 24*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3*sin(d*x + c)^ 
6/(cos(d*x + c) + 1)^6 + 3)/(a*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a*sin(d 
*x + c)^3/(cos(d*x + c) + 1)^3 + 3*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 
 a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) + 18*arctan(sin(d*x + c)/(cos(d*x 
+ c) + 1))/a + 6*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - 3*sin(d*x + c)/( 
a*(cos(d*x + c) + 1)))/d
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.62 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {9 \, {\left (d x + c\right )}}{a} + \frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {3 \, {\left (2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \] Input:

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

Output:

-1/6*(9*(d*x + c)/a + 6*log(abs(tan(1/2*d*x + 1/2*c)))/a - 3*tan(1/2*d*x + 
 1/2*c)/a - 3*(2*tan(1/2*d*x + 1/2*c) - 1)/(a*tan(1/2*d*x + 1/2*c)) - 2*(3 
*tan(1/2*d*x + 1/2*c)^5 - 12*tan(1/2*d*x + 1/2*c)^4 - 12*tan(1/2*d*x + 1/2 
*c)^2 - 3*tan(1/2*d*x + 1/2*c) - 8)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a))/d
 

Mupad [B] (verification not implemented)

Time = 32.95 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.52 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3\,\mathrm {atan}\left (\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-6}+\frac {9}{9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-6}\right )}{a\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1}{d\,\left (2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d} \] Input:

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

Output:

(3*atan((6*tan(c/2 + (d*x)/2))/(9*tan(c/2 + (d*x)/2) - 6) + 9/(9*tan(c/2 + 
 (d*x)/2) - 6)))/(a*d) - log(tan(c/2 + (d*x)/2))/(a*d) - ((16*tan(c/2 + (d 
*x)/2))/3 + 5*tan(c/2 + (d*x)/2)^2 + 8*tan(c/2 + (d*x)/2)^3 + 3*tan(c/2 + 
(d*x)/2)^4 + 8*tan(c/2 + (d*x)/2)^5 - tan(c/2 + (d*x)/2)^6 + 1)/(d*(2*a*ta 
n(c/2 + (d*x)/2) + 6*a*tan(c/2 + (d*x)/2)^3 + 6*a*tan(c/2 + (d*x)/2)^5 + 2 
*a*tan(c/2 + (d*x)/2)^7)) + tan(c/2 + (d*x)/2)/(2*a*d)
 

Reduce [F]

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

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

Output:

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