\(\int \cos ^2(c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx\) [666]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 118 \[ \int \cos ^2(c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 a \csc (c+d x)}{d}+\frac {3 a \csc ^2(c+d x)}{2 d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^4(c+d x)}{4 d}+\frac {3 a \log (\sin (c+d x))}{d}+\frac {3 a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin ^3(c+d x)}{3 d} \] Output:

3*a*csc(d*x+c)/d+3/2*a*csc(d*x+c)^2/d-1/3*a*csc(d*x+c)^3/d-1/4*a*csc(d*x+c 
)^4/d+3*a*ln(sin(d*x+c))/d+3*a*sin(d*x+c)/d-1/2*a*sin(d*x+c)^2/d-1/3*a*sin 
(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00 \[ \int \cos ^2(c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 a \csc (c+d x)}{d}+\frac {3 a \csc ^2(c+d x)}{2 d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^4(c+d x)}{4 d}+\frac {3 a \log (\sin (c+d x))}{d}+\frac {3 a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin ^3(c+d x)}{3 d} \] Input:

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

Output:

(3*a*Csc[c + d*x])/d + (3*a*Csc[c + d*x]^2)/(2*d) - (a*Csc[c + d*x]^3)/(3* 
d) - (a*Csc[c + d*x]^4)/(4*d) + (3*a*Log[Sin[c + d*x]])/d + (3*a*Sin[c + d 
*x])/d - (a*Sin[c + d*x]^2)/(2*d) - (a*Sin[c + d*x]^3)/(3*d)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 99, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^7 (a \sin (c+d x)+a)}{\sin (c+d x)^5}dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {\int \csc ^5(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\csc ^5(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^4}{a^5}d(a \sin (c+d x))}{a^2 d}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (a^2 \csc ^5(c+d x)+a^2 \csc ^4(c+d x)-3 a^2 \csc ^3(c+d x)-3 a^2 \csc ^2(c+d x)+3 a^2 \csc (c+d x)+3 a^2-a^2 \sin ^2(c+d x)-a^2 \sin (c+d x)\right )d(a \sin (c+d x))}{a^2 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {1}{3} a^3 \sin ^3(c+d x)-\frac {1}{2} a^3 \sin ^2(c+d x)+3 a^3 \sin (c+d x)-\frac {1}{4} a^3 \csc ^4(c+d x)-\frac {1}{3} a^3 \csc ^3(c+d x)+\frac {3}{2} a^3 \csc ^2(c+d x)+3 a^3 \csc (c+d x)+3 a^3 \log (a \sin (c+d x))}{a^2 d}\)

Input:

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

Output:

(3*a^3*Csc[c + d*x] + (3*a^3*Csc[c + d*x]^2)/2 - (a^3*Csc[c + d*x]^3)/3 - 
(a^3*Csc[c + d*x]^4)/4 + 3*a^3*Log[a*Sin[c + d*x]] + 3*a^3*Sin[c + d*x] - 
(a^3*Sin[c + d*x]^2)/2 - (a^3*Sin[c + d*x]^3)/3)/(a^2*d)
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 99
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

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 3315
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, 
 x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege 
rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
 
Maple [A] (verified)

Time = 3.64 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.36

method result size
derivativedivides \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{8}}{3 \sin \left (d x +c \right )^{3}}+\frac {5 \cos \left (d x +c \right )^{8}}{3 \sin \left (d x +c \right )}+\frac {5 \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{3}\right )+a \left (-\frac {\cos \left (d x +c \right )^{8}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{8}}{2 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{6}}{2}+\frac {3 \cos \left (d x +c \right )^{4}}{4}+\frac {3 \cos \left (d x +c \right )^{2}}{2}+3 \ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) \(161\)
default \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{8}}{3 \sin \left (d x +c \right )^{3}}+\frac {5 \cos \left (d x +c \right )^{8}}{3 \sin \left (d x +c \right )}+\frac {5 \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{3}\right )+a \left (-\frac {\cos \left (d x +c \right )^{8}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{8}}{2 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{6}}{2}+\frac {3 \cos \left (d x +c \right )^{4}}{4}+\frac {3 \cos \left (d x +c \right )^{2}}{2}+3 \ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) \(161\)
risch \(-3 i a x -\frac {i a \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {11 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {11 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {i a \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {6 i a c}{d}+\frac {2 i a \left (9 i {\mathrm e}^{6 i \left (d x +c \right )}+9 \,{\mathrm e}^{7 i \left (d x +c \right )}-12 i {\mathrm e}^{4 i \left (d x +c \right )}-23 \,{\mathrm e}^{5 i \left (d x +c \right )}+9 i {\mathrm e}^{2 i \left (d x +c \right )}+23 \,{\mathrm e}^{3 i \left (d x +c \right )}-9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) \(228\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.20 \[ \int \cos ^2(c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {6 \, a \cos \left (d x + c\right )^{6} - 15 \, a \cos \left (d x + c\right )^{4} - 6 \, a \cos \left (d x + c\right )^{2} + 36 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (a \cos \left (d x + c\right )^{6} + 6 \, a \cos \left (d x + c\right )^{4} - 24 \, a \cos \left (d x + c\right )^{2} + 16 \, a\right )} \sin \left (d x + c\right ) + 12 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \] Input:

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

Output:

1/12*(6*a*cos(d*x + c)^6 - 15*a*cos(d*x + c)^4 - 6*a*cos(d*x + c)^2 + 36*( 
a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(1/2*sin(d*x + c)) + 4*(a*co 
s(d*x + c)^6 + 6*a*cos(d*x + c)^4 - 24*a*cos(d*x + c)^2 + 16*a)*sin(d*x + 
c) + 12*a)/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {4 \, a \sin \left (d x + c\right )^{3} + 6 \, a \sin \left (d x + c\right )^{2} - 36 \, a \log \left (\sin \left (d x + c\right )\right ) - 36 \, a \sin \left (d x + c\right ) - \frac {36 \, a \sin \left (d x + c\right )^{3} + 18 \, a \sin \left (d x + c\right )^{2} - 4 \, a \sin \left (d x + c\right ) - 3 \, a}{\sin \left (d x + c\right )^{4}}}{12 \, d} \] Input:

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

Output:

-1/12*(4*a*sin(d*x + c)^3 + 6*a*sin(d*x + c)^2 - 36*a*log(sin(d*x + c)) - 
36*a*sin(d*x + c) - (36*a*sin(d*x + c)^3 + 18*a*sin(d*x + c)^2 - 4*a*sin(d 
*x + c) - 3*a)/sin(d*x + c)^4)/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79 \[ \int \cos ^2(c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {4 \, a \sin \left (d x + c\right )^{3} + 6 \, a \sin \left (d x + c\right )^{2} - 36 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 36 \, a \sin \left (d x + c\right ) - \frac {36 \, a \sin \left (d x + c\right )^{3} + 18 \, a \sin \left (d x + c\right )^{2} - 4 \, a \sin \left (d x + c\right ) - 3 \, a}{\sin \left (d x + c\right )^{4}}}{12 \, d} \] Input:

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

Output:

-1/12*(4*a*sin(d*x + c)^3 + 6*a*sin(d*x + c)^2 - 36*a*log(abs(sin(d*x + c) 
)) - 36*a*sin(d*x + c) - (36*a*sin(d*x + c)^3 + 18*a*sin(d*x + c)^2 - 4*a* 
sin(d*x + c) - 3*a)/sin(d*x + c)^4)/d
 

Mupad [B] (verification not implemented)

Time = 34.08 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.46 \[ \int \cos ^2(c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {11\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {118\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-27\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {644\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {69\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{4}+160\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {57\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {17\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {a}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {3\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}+\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {3\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \] Input:

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

Output:

(11*a*tan(c/2 + (d*x)/2))/(8*d) + ((17*a*tan(c/2 + (d*x)/2)^2)/4 - (2*a*ta 
n(c/2 + (d*x)/2))/3 - a/4 + 20*a*tan(c/2 + (d*x)/2)^3 + (57*a*tan(c/2 + (d 
*x)/2)^4)/4 + 160*a*tan(c/2 + (d*x)/2)^5 - (69*a*tan(c/2 + (d*x)/2)^6)/4 + 
 (644*a*tan(c/2 + (d*x)/2)^7)/3 - 27*a*tan(c/2 + (d*x)/2)^8 + 118*a*tan(c/ 
2 + (d*x)/2)^9)/(d*(16*tan(c/2 + (d*x)/2)^4 + 48*tan(c/2 + (d*x)/2)^6 + 48 
*tan(c/2 + (d*x)/2)^8 + 16*tan(c/2 + (d*x)/2)^10)) - (3*a*log(tan(c/2 + (d 
*x)/2)^2 + 1))/d + (5*a*tan(c/2 + (d*x)/2)^2)/(16*d) - (a*tan(c/2 + (d*x)/ 
2)^3)/(24*d) - (a*tan(c/2 + (d*x)/2)^4)/(64*d) + (3*a*log(tan(c/2 + (d*x)/ 
2)))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.08 \[ \int \cos ^2(c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-576 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4}+576 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}-64 \sin \left (d x +c \right )^{7}-96 \sin \left (d x +c \right )^{6}+576 \sin \left (d x +c \right )^{5}-297 \sin \left (d x +c \right )^{4}+576 \sin \left (d x +c \right )^{3}+288 \sin \left (d x +c \right )^{2}-64 \sin \left (d x +c \right )-48\right )}{192 \sin \left (d x +c \right )^{4} d} \] Input:

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

Output:

(a*( - 576*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4 + 576*log(tan((c + 
 d*x)/2))*sin(c + d*x)**4 - 64*sin(c + d*x)**7 - 96*sin(c + d*x)**6 + 576* 
sin(c + d*x)**5 - 297*sin(c + d*x)**4 + 576*sin(c + d*x)**3 + 288*sin(c + 
d*x)**2 - 64*sin(c + d*x) - 48))/(192*sin(c + d*x)**4*d)