\(\int \cot ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\) [527]

Optimal result

Integrand size = 21, antiderivative size = 131 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {5 a^3 \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {a^3 \csc ^3(c+d x)}{d}-\frac {a^3 \csc ^4(c+d x)}{4 d}-\frac {5 a^3 \log (\sin (c+d x))}{d}+\frac {a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin ^3(c+d x)}{3 d} \] Output:

5*a^3*csc(d*x+c)/d-1/2*a^3*csc(d*x+c)^2/d-a^3*csc(d*x+c)^3/d-1/4*a^3*csc(d 
*x+c)^4/d-5*a^3*ln(sin(d*x+c))/d+a^3*sin(d*x+c)/d+3/2*a^3*sin(d*x+c)^2/d+1 
/3*a^3*sin(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.66 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (60 \csc (c+d x)-6 \csc ^2(c+d x)-12 \csc ^3(c+d x)-3 \csc ^4(c+d x)-60 \log (\sin (c+d x))+12 \sin (c+d x)+18 \sin ^2(c+d x)+4 \sin ^3(c+d x)\right )}{12 d} \] Input:

Integrate[Cot[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]
 

Output:

(a^3*(60*Csc[c + d*x] - 6*Csc[c + d*x]^2 - 12*Csc[c + d*x]^3 - 3*Csc[c + d 
*x]^4 - 60*Log[Sin[c + d*x]] + 12*Sin[c + d*x] + 18*Sin[c + d*x]^2 + 4*Sin 
[c + d*x]^3))/(12*d)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 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.

\(\displaystyle \int \cot ^5(c+d x) (a \sin (c+d x)+a)^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^3}{\tan (c+d x)^5}dx\)

\(\Big \downarrow \) 3186

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{3} a^3 \sin ^3(c+d x)+\frac {3}{2} a^3 \sin ^2(c+d x)+a^3 \sin (c+d x)-\frac {1}{4} a^3 \csc ^4(c+d x)-a^3 \csc ^3(c+d x)-\frac {1}{2} a^3 \csc ^2(c+d x)+5 a^3 \csc (c+d x)-5 a^3 \log (a \sin (c+d x))}{d}\)

Input:

Int[Cot[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]
 

Output:

(5*a^3*Csc[c + d*x] - (a^3*Csc[c + d*x]^2)/2 - a^3*Csc[c + d*x]^3 - (a^3*C 
sc[c + d*x]^4)/4 - 5*a^3*Log[a*Sin[c + d*x]] + a^3*Sin[c + d*x] + (3*a^3*S 
in[c + d*x]^2)/2 + (a^3*Sin[c + d*x]^3)/3)/d
 

Defintions of rubi rules used

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 3186

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_.), x_Symbol] :> Simp[1/f   Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) 
^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E 
qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
 

Maple [A] (verified)

Time = 3.67 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.60

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )+3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )+3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{6}}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )+a^{3} \left (-\frac {\cot \left (d x +c \right )^{4}}{4}+\frac {\cot \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(210\)
default	\(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )+3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )+3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{6}}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )+a^{3} \left (-\frac {\cot \left (d x +c \right )^{4}}{4}+\frac {\cot \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(210\)
risch	\(5 i a^{3} x +\frac {i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {3 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {5 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {5 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {3 a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {10 i a^{3} c}{d}+\frac {2 i a^{3} \left (-i {\mathrm e}^{6 i \left (d x +c \right )}+5 \,{\mathrm e}^{7 i \left (d x +c \right )}+4 i {\mathrm e}^{4 i \left (d x +c \right )}-11 \,{\mathrm e}^{5 i \left (d x +c \right )}-i {\mathrm e}^{2 i \left (d x +c \right )}+11 \,{\mathrm e}^{3 i \left (d x +c \right )}-5 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {5 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(248\)

Input:

int(cot(d*x+c)^5*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.21 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {18 \, a^{3} \cos \left (d x + c\right )^{6} - 45 \, a^{3} \cos \left (d x + c\right )^{4} + 30 \, a^{3} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (a^{3} \cos \left (d x + c\right )^{6} - 6 \, a^{3} \cos \left (d x + c\right )^{4} + 24 \, a^{3} \cos \left (d x + c\right )^{2} - 16 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \] Input:

integrate(cot(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

-1/12*(18*a^3*cos(d*x + c)^6 - 45*a^3*cos(d*x + c)^4 + 30*a^3*cos(d*x + c) 
^2 + 60*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^2 + a^3)*log(1/2*sin(d*x 
+ c)) + 4*(a^3*cos(d*x + c)^6 - 6*a^3*cos(d*x + c)^4 + 24*a^3*cos(d*x + c) 
^2 - 16*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)
 

Sympy [F]

\[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=a^{3} \left (\int 3 \sin {\left (c + d x \right )} \cot ^{5}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \cot ^{5}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \cot ^{5}{\left (c + d x \right )}\, dx + \int \cot ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:

integrate(cot(d*x+c)**5*(a+a*sin(d*x+c))**3,x)
 

Output:

a**3*(Integral(3*sin(c + d*x)*cot(c + d*x)**5, x) + Integral(3*sin(c + d*x 
)**2*cot(c + d*x)**5, x) + Integral(sin(c + d*x)**3*cot(c + d*x)**5, x) + 
Integral(cot(c + d*x)**5, x))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.82 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {4 \, a^{3} \sin \left (d x + c\right )^{3} + 18 \, a^{3} \sin \left (d x + c\right )^{2} - 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 12 \, a^{3} \sin \left (d x + c\right ) + \frac {3 \, {\left (20 \, a^{3} \sin \left (d x + c\right )^{3} - 2 \, a^{3} \sin \left (d x + c\right )^{2} - 4 \, a^{3} \sin \left (d x + c\right ) - a^{3}\right )}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \] Input:

integrate(cot(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/12*(4*a^3*sin(d*x + c)^3 + 18*a^3*sin(d*x + c)^2 - 60*a^3*log(sin(d*x + 
c)) + 12*a^3*sin(d*x + c) + 3*(20*a^3*sin(d*x + c)^3 - 2*a^3*sin(d*x + c)^ 
2 - 4*a^3*sin(d*x + c) - a^3)/sin(d*x + c)^4)/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.83 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {4 \, a^{3} \sin \left (d x + c\right )^{3} + 18 \, a^{3} \sin \left (d x + c\right )^{2} - 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 12 \, a^{3} \sin \left (d x + c\right ) + \frac {3 \, {\left (20 \, a^{3} \sin \left (d x + c\right )^{3} - 2 \, a^{3} \sin \left (d x + c\right )^{2} - 4 \, a^{3} \sin \left (d x + c\right ) - a^{3}\right )}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \] Input:

integrate(cot(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

1/12*(4*a^3*sin(d*x + c)^3 + 18*a^3*sin(d*x + c)^2 - 60*a^3*log(abs(sin(d* 
x + c))) + 12*a^3*sin(d*x + c) + 3*(20*a^3*sin(d*x + c)^3 - 2*a^3*sin(d*x 
+ c)^2 - 4*a^3*sin(d*x + c) - a^3)/sin(d*x + c)^4)/d
 

Mupad [B] (verification not implemented)

Time = 34.17 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.46 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {66\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+93\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {620\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+\frac {347\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{4}+128\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {39\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^3}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {5\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {17\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {5\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:

int(cot(c + d*x)^5*(a + a*sin(c + d*x))^3,x)
 

Output:

(28*a^3*tan(c/2 + (d*x)/2)^3 - (15*a^3*tan(c/2 + (d*x)/2)^2)/4 - (39*a^3*t 
an(c/2 + (d*x)/2)^4)/4 + 128*a^3*tan(c/2 + (d*x)/2)^5 + (347*a^3*tan(c/2 + 
 (d*x)/2)^6)/4 + (620*a^3*tan(c/2 + (d*x)/2)^7)/3 + 93*a^3*tan(c/2 + (d*x) 
/2)^8 + 66*a^3*tan(c/2 + (d*x)/2)^9 - a^3/4 - 2*a^3*tan(c/2 + (d*x)/2))/(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)) - (a^3*tan(c/2 + (d*x)/2)^3)/(8*d) - (a^3 
*tan(c/2 + (d*x)/2)^4)/(64*d) - (3*a^3*tan(c/2 + (d*x)/2)^2)/(16*d) - (5*a 
^3*log(tan(c/2 + (d*x)/2)))/d + (17*a^3*tan(c/2 + (d*x)/2))/(8*d) + (5*a^3 
*log(tan(c/2 + (d*x)/2)^2 + 1))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.99 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \left (960 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4}-960 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}+64 \sin \left (d x +c \right )^{7}+288 \sin \left (d x +c \right )^{6}+192 \sin \left (d x +c \right )^{5}+183 \sin \left (d x +c \right )^{4}+960 \sin \left (d x +c \right )^{3}-96 \sin \left (d x +c \right )^{2}-192 \sin \left (d x +c \right )-48\right )}{192 \sin \left (d x +c \right )^{4} d} \] Input:

int(cot(d*x+c)^5*(a+a*sin(d*x+c))^3,x)
 

Output:

(a**3*(960*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4 - 960*log(tan((c + 
 d*x)/2))*sin(c + d*x)**4 + 64*sin(c + d*x)**7 + 288*sin(c + d*x)**6 + 192 
*sin(c + d*x)**5 + 183*sin(c + d*x)**4 + 960*sin(c + d*x)**3 - 96*sin(c + 
d*x)**2 - 192*sin(c + d*x) - 48))/(192*sin(c + d*x)**4*d)