\(\int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx\) [382]

Optimal result

Integrand size = 27, antiderivative size = 119 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 a^2 x}{4}-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 5.88 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.81 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (270 \cos (c+d x)+5 \cos (3 (c+d x))-3 \cos (5 (c+d x))+15 \left (4 \left (3 c+3 d x-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+8 \sin (2 (c+d x))+\sin (4 (c+d x))\right )\right )}{240 d} \] Input:

Integrate[Cos[c + d*x]^3*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]
 

Output:

(a^2*(270*Cos[c + d*x] + 5*Cos[3*(c + d*x)] - 3*Cos[5*(c + d*x)] + 15*(4*( 
3*c + 3*d*x - 4*Log[Cos[(c + d*x)/2]] + 4*Log[Sin[(c + d*x)/2]]) + 8*Sin[2 
*(c + d*x)] + Sin[4*(c + d*x)])))/(240*d)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 119, 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.

\(\displaystyle \int \cos ^3(c+d x) \cot (c+d x) (a \sin (c+d x)+a)^2 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3352

\(\displaystyle \int \left (2 a^2 \cos ^4(c+d x)+a^2 \sin (c+d x) \cos ^4(c+d x)+a^2 \cos ^3(c+d x) \cot (c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac {3 a^2 x}{4}\)

Input:

Int[Cos[c + d*x]^3*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]
 

Output:

(3*a^2*x)/4 - (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) + (3*a^2*Cos[c + d*x]*S 
in[c + d*x])/(4*d) + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(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 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 = 2.34 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{5}}{5}+2 a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} \left (\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(94\)
default	\(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{5}}{5}+2 a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} \left (\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(94\)
risch	\(\frac {3 a^{2} x}{4}+\frac {9 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {9 a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{16 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {a^{2} \cos \left (5 d x +5 c \right )}{80 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{16 d}+\frac {a^{2} \cos \left (3 d x +3 c \right )}{48 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{2 d}\)	\(149\)

Input:

int(cos(d*x+c)^3*cot(d*x+c)*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/5*a^2*cos(d*x+c)^5+2*a^2*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d* 
x+c)+3/8*d*x+3/8*c)+a^2*(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 = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.97 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {12 \, a^{2} \cos \left (d x + c\right )^{5} - 20 \, a^{2} \cos \left (d x + c\right )^{3} - 45 \, a^{2} d x - 60 \, a^{2} \cos \left (d x + c\right ) + 30 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 30 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \] Input:

integrate(cos(d*x+c)^3*cot(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="fricas" 
)
 

Output:

-1/60*(12*a^2*cos(d*x + c)^5 - 20*a^2*cos(d*x + c)^3 - 45*a^2*d*x - 60*a^2 
*cos(d*x + c) + 30*a^2*log(1/2*cos(d*x + c) + 1/2) - 30*a^2*log(-1/2*cos(d 
*x + c) + 1/2) - 15*(2*a^2*cos(d*x + c)^3 + 3*a^2*cos(d*x + c))*sin(d*x + 
c))/d
 

Sympy [F]

\[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cos ^{3}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx\right ) \] Input:

integrate(cos(d*x+c)**3*cot(d*x+c)*(a+a*sin(d*x+c))**2,x)
 

Output:

a**2*(Integral(cos(c + d*x)**3*cot(c + d*x), x) + Integral(2*sin(c + d*x)* 
cos(c + d*x)**3*cot(c + d*x), x) + Integral(sin(c + d*x)**2*cos(c + d*x)** 
3*cot(c + d*x), x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {48 \, a^{2} \cos \left (d x + c\right )^{5} - 40 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{240 \, d} \] Input:

integrate(cos(d*x+c)^3*cot(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="maxima" 
)
 

Output:

-1/240*(48*a^2*cos(d*x + c)^5 - 40*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3* 
log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1))*a^2 - 15*(12*d*x + 12*c + 
 sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.52 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {45 \, {\left (d x + c\right )} a^{2} + 60 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 320 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 280 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 68 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{60 \, d} \] Input:

integrate(cos(d*x+c)^3*cot(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="giac")
 

Output:

1/60*(45*(d*x + c)*a^2 + 60*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(75*a^2 
*tan(1/2*d*x + 1/2*c)^9 - 60*a^2*tan(1/2*d*x + 1/2*c)^8 + 30*a^2*tan(1/2*d 
*x + 1/2*c)^7 - 360*a^2*tan(1/2*d*x + 1/2*c)^6 - 320*a^2*tan(1/2*d*x + 1/2 
*c)^4 - 30*a^2*tan(1/2*d*x + 1/2*c)^3 - 280*a^2*tan(1/2*d*x + 1/2*c)^2 - 7 
5*a^2*tan(1/2*d*x + 1/2*c) - 68*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
 

Mupad [B] (verification not implemented)

Time = 18.66 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.46 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {3\,a^2\,\mathrm {atan}\left (\frac {9\,a^4}{4\,\left (3\,a^4-\frac {9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )}+\frac {3\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,a^4-\frac {9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}\right )}{2\,d}+\frac {-\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {32\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {28\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {34\,a^2}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:

int(cos(c + d*x)^3*cot(c + d*x)*(a + a*sin(c + d*x))^2,x)
 

Output:

(a^2*log(tan(c/2 + (d*x)/2)))/d + (3*a^2*atan((9*a^4)/(4*(3*a^4 - (9*a^4*t 
an(c/2 + (d*x)/2))/4)) + (3*a^4*tan(c/2 + (d*x)/2))/(3*a^4 - (9*a^4*tan(c/ 
2 + (d*x)/2))/4)))/(2*d) + ((28*a^2*tan(c/2 + (d*x)/2)^2)/3 + a^2*tan(c/2 
+ (d*x)/2)^3 + (32*a^2*tan(c/2 + (d*x)/2)^4)/3 + 12*a^2*tan(c/2 + (d*x)/2) 
^6 - a^2*tan(c/2 + (d*x)/2)^7 + 2*a^2*tan(c/2 + (d*x)/2)^8 - (5*a^2*tan(c/ 
2 + (d*x)/2)^9)/2 + (34*a^2)/15 + (5*a^2*tan(c/2 + (d*x)/2))/2)/(d*(5*tan( 
c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*t 
an(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+75 \cos \left (d x +c \right ) \sin \left (d x +c \right )+68 \cos \left (d x +c \right )+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+45 c +45 d x -68\right )}{60 d} \] Input:

int(cos(d*x+c)^3*cot(d*x+c)*(a+a*sin(d*x+c))^2,x)
 

Output:

(a**2*( - 12*cos(c + d*x)*sin(c + d*x)**4 - 30*cos(c + d*x)*sin(c + d*x)** 
3 + 4*cos(c + d*x)*sin(c + d*x)**2 + 75*cos(c + d*x)*sin(c + d*x) + 68*cos 
(c + d*x) + 60*log(tan((c + d*x)/2)) + 45*c + 45*d*x - 68))/(60*d)