\(\int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx\) [592]

Optimal result

Integrand size = 27, antiderivative size = 161 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {5 a^2 x}{8}-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cos ^7(c+d x)}{7 d}+\frac {5 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {5 a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}+\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{3 d} \] Output:

5/8*a^2*x-a^2*arctanh(cos(d*x+c))/d+a^2*cos(d*x+c)/d+1/3*a^2*cos(d*x+c)^3/ 
d+1/5*a^2*cos(d*x+c)^5/d-1/7*a^2*cos(d*x+c)^7/d+5/8*a^2*cos(d*x+c)*sin(d*x 
+c)/d+5/12*a^2*cos(d*x+c)^3*sin(d*x+c)/d+1/3*a^2*cos(d*x+c)^5*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 6.00 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.70 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (4200 c+4200 d x+8715 \cos (c+d x)+665 \cos (3 (c+d x))-21 \cos (5 (c+d x))-15 \cos (7 (c+d x))-6720 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+6720 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+3150 \sin (2 (c+d x))+630 \sin (4 (c+d x))+70 \sin (6 (c+d x))\right )}{6720 d} \] Input:

Integrate[Cos[c + d*x]^5*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]
 

Output:

(a^2*(4200*c + 4200*d*x + 8715*Cos[c + d*x] + 665*Cos[3*(c + d*x)] - 21*Co 
s[5*(c + d*x)] - 15*Cos[7*(c + d*x)] - 6720*Log[Cos[(c + d*x)/2]] + 6720*L 
og[Sin[(c + d*x)/2]] + 3150*Sin[2*(c + d*x)] + 630*Sin[4*(c + d*x)] + 70*S 
in[6*(c + d*x)]))/(6720*d)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 3352, 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.

\(\displaystyle \int \cos ^5(c+d x) \cot (c+d x) (a \sin (c+d x)+a)^2 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3352

\(\displaystyle \int \left (2 a^2 \cos ^6(c+d x)+a^2 \sin (c+d x) \cos ^6(c+d x)+a^2 \cos ^5(c+d x) \cot (c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {a^2 \cos ^7(c+d x)}{7 d}+\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{3 d}+\frac {5 a^2 \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac {5 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a^2 x}{8}\)

Input:

Int[Cos[c + d*x]^5*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]
 

Output:

(5*a^2*x)/8 - (a^2*ArcTanh[Cos[c + d*x]])/d + (a^2*Cos[c + d*x])/d + (a^2* 
Cos[c + d*x]^3)/(3*d) + (a^2*Cos[c + d*x]^5)/(5*d) - (a^2*Cos[c + d*x]^7)/ 
(7*d) + (5*a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (5*a^2*Cos[c + d*x]^3*Si 
n[c + d*x])/(12*d) + (a^2*Cos[c + d*x]^5*Sin[c + d*x])/(3*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 3352

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig 
[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F 
reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
 

Maple [A] (verified)

Time = 7.54 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{7}}{7}+2 a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+a^{2} \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(114\)
default	\(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{7}}{7}+2 a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+a^{2} \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(114\)
risch	\(\frac {5 a^{2} x}{8}+\frac {83 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{128 d}+\frac {83 a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{128 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{2} \cos \left (7 d x +7 c \right )}{448 d}+\frac {a^{2} \sin \left (6 d x +6 c \right )}{96 d}-\frac {a^{2} \cos \left (5 d x +5 c \right )}{320 d}+\frac {3 a^{2} \sin \left (4 d x +4 c \right )}{32 d}+\frac {19 a^{2} \cos \left (3 d x +3 c \right )}{192 d}+\frac {15 a^{2} \sin \left (2 d x +2 c \right )}{32 d}\)	\(183\)

Input:

int(cos(d*x+c)^5*cot(d*x+c)*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/7*a^2*cos(d*x+c)^7+2*a^2*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8* 
cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+a^2*(1/5*cos(d*x+c)^5+1/3*cos(d*x+ 
c)^3+cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {120 \, a^{2} \cos \left (d x + c\right )^{7} - 168 \, a^{2} \cos \left (d x + c\right )^{5} - 280 \, a^{2} \cos \left (d x + c\right )^{3} - 525 \, a^{2} d x - 840 \, a^{2} \cos \left (d x + c\right ) + 420 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 420 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 35 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{5} + 10 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="fricas" 
)
 

Output:

-1/840*(120*a^2*cos(d*x + c)^7 - 168*a^2*cos(d*x + c)^5 - 280*a^2*cos(d*x 
+ c)^3 - 525*a^2*d*x - 840*a^2*cos(d*x + c) + 420*a^2*log(1/2*cos(d*x + c) 
 + 1/2) - 420*a^2*log(-1/2*cos(d*x + c) + 1/2) - 35*(8*a^2*cos(d*x + c)^5 
+ 10*a^2*cos(d*x + c)^3 + 15*a^2*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [F]

\[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cos ^{5}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx\right ) \] Input:

integrate(cos(d*x+c)**5*cot(d*x+c)*(a+a*sin(d*x+c))**2,x)
 

Output:

a**2*(Integral(cos(c + d*x)**5*cot(c + d*x), x) + Integral(2*sin(c + d*x)* 
cos(c + d*x)**5*cot(c + d*x), x) + Integral(sin(c + d*x)**2*cos(c + d*x)** 
5*cot(c + d*x), x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.76 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {480 \, a^{2} \cos \left (d x + c\right )^{7} - 112 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{3360 \, d} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="maxima" 
)
 

Output:

-1/3360*(480*a^2*cos(d*x + c)^7 - 112*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^ 
3 + 30*cos(d*x + c) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1)) 
*a^2 + 35*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48* 
sin(2*d*x + 2*c))*a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.52 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {525 \, {\left (d x + c\right )} a^{2} + 840 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (1155 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 1680 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} + 980 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 10080 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 2975 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 16240 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 24640 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 2975 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 14448 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 980 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6496 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1155 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1168 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7}}}{840 \, d} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="giac")
 

Output:

1/840*(525*(d*x + c)*a^2 + 840*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(115 
5*a^2*tan(1/2*d*x + 1/2*c)^13 - 1680*a^2*tan(1/2*d*x + 1/2*c)^12 + 980*a^2 
*tan(1/2*d*x + 1/2*c)^11 - 10080*a^2*tan(1/2*d*x + 1/2*c)^10 + 2975*a^2*ta 
n(1/2*d*x + 1/2*c)^9 - 16240*a^2*tan(1/2*d*x + 1/2*c)^8 - 24640*a^2*tan(1/ 
2*d*x + 1/2*c)^6 - 2975*a^2*tan(1/2*d*x + 1/2*c)^5 - 14448*a^2*tan(1/2*d*x 
 + 1/2*c)^4 - 980*a^2*tan(1/2*d*x + 1/2*c)^3 - 6496*a^2*tan(1/2*d*x + 1/2* 
c)^2 - 1155*a^2*tan(1/2*d*x + 1/2*c) - 1168*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 
 1)^7)/d
 

Mupad [B] (verification not implemented)

Time = 34.98 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.39 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {5\,a^2\,\mathrm {atan}\left (\frac {25\,a^4}{16\,\left (\frac {5\,a^4}{2}-\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}+\frac {5\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {5\,a^4}{2}-\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d}+\frac {-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}+24\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-\frac {85\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}+\frac {116\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {176\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {85\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {172\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {232\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {11\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {292\,a^2}{105}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:

int(cos(c + d*x)^5*cot(c + d*x)*(a + a*sin(c + d*x))^2,x)
 

Output:

(a^2*log(tan(c/2 + (d*x)/2)))/d + (5*a^2*atan((25*a^4)/(16*((5*a^4)/2 - (2 
5*a^4*tan(c/2 + (d*x)/2))/16)) + (5*a^4*tan(c/2 + (d*x)/2))/(2*((5*a^4)/2 
- (25*a^4*tan(c/2 + (d*x)/2))/16))))/(4*d) + ((232*a^2*tan(c/2 + (d*x)/2)^ 
2)/15 + (7*a^2*tan(c/2 + (d*x)/2)^3)/3 + (172*a^2*tan(c/2 + (d*x)/2)^4)/5 
+ (85*a^2*tan(c/2 + (d*x)/2)^5)/12 + (176*a^2*tan(c/2 + (d*x)/2)^6)/3 + (1 
16*a^2*tan(c/2 + (d*x)/2)^8)/3 - (85*a^2*tan(c/2 + (d*x)/2)^9)/12 + 24*a^2 
*tan(c/2 + (d*x)/2)^10 - (7*a^2*tan(c/2 + (d*x)/2)^11)/3 + 4*a^2*tan(c/2 + 
 (d*x)/2)^12 - (11*a^2*tan(c/2 + (d*x)/2)^13)/4 + (292*a^2)/105 + (11*a^2* 
tan(c/2 + (d*x)/2))/4)/(d*(7*tan(c/2 + (d*x)/2)^2 + 21*tan(c/2 + (d*x)/2)^ 
4 + 35*tan(c/2 + (d*x)/2)^6 + 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2 + (d*x) 
/2)^10 + 7*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.81 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-192 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-910 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-256 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+1155 \cos \left (d x +c \right ) \sin \left (d x +c \right )+1168 \cos \left (d x +c \right )+840 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+525 c +525 d x -1168\right )}{840 d} \] Input:

int(cos(d*x+c)^5*cot(d*x+c)*(a+a*sin(d*x+c))^2,x)
 

Output:

(a**2*(120*cos(c + d*x)*sin(c + d*x)**6 + 280*cos(c + d*x)*sin(c + d*x)**5 
 - 192*cos(c + d*x)*sin(c + d*x)**4 - 910*cos(c + d*x)*sin(c + d*x)**3 - 2 
56*cos(c + d*x)*sin(c + d*x)**2 + 1155*cos(c + d*x)*sin(c + d*x) + 1168*co 
s(c + d*x) + 840*log(tan((c + d*x)/2)) + 525*c + 525*d*x - 1168))/(840*d)