\(\int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx\) [37]

Optimal result

Integrand size = 21, antiderivative size = 102 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {4 a^4 \csc (c+d x)}{d}-\frac {a^4 \csc ^2(c+d x)}{2 d}+\frac {5 a^4 \log (\sin (c+d x))}{d}-\frac {5 a^4 \sin ^2(c+d x)}{2 d}-\frac {4 a^4 \sin ^3(c+d x)}{3 d}-\frac {a^4 \sin ^4(c+d x)}{4 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {a^4 \left (48 \csc (c+d x)+6 \csc ^2(c+d x)-60 \log (\sin (c+d x))+30 \sin ^2(c+d x)+16 \sin ^3(c+d x)+3 \sin ^4(c+d x)\right )}{12 d} \] Input:

Integrate[Cot[c + d*x]^3*(a + a*Sin[c + d*x])^4,x]
 

Output:

-1/12*(a^4*(48*Csc[c + d*x] + 6*Csc[c + d*x]^2 - 60*Log[Sin[c + d*x]] + 30 
*Sin[c + d*x]^2 + 16*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4))/d
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 84, 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.

\(\displaystyle \int \cot ^3(c+d x) (a \sin (c+d x)+a)^4 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{\tan (c+d x)^3}dx\)

\(\Big \downarrow \) 3186

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

\(\Big \downarrow \) 84

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

\(\Big \downarrow \) 2009

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

Input:

Int[Cot[c + d*x]^3*(a + a*Sin[c + d*x])^4,x]
 

Output:

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

Defintions of rubi rules used

rule 84

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : 
> Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, 
 d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n + p + 2, 0 
] && GtQ[n + 2*p, 0])
 

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 = 5.48 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {-\frac {a^{4} \cos \left (d x +c \right )^{4}}{4}+\frac {4 a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+6 a^{4} \left (\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+4 a^{4} \left (-\frac {\cos \left (d x +c \right )^{4}}{\sin \left (d x +c \right )}-\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )\right )+a^{4} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(129\)
default	\(\frac {-\frac {a^{4} \cos \left (d x +c \right )^{4}}{4}+\frac {4 a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+6 a^{4} \left (\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+4 a^{4} \left (-\frac {\cos \left (d x +c \right )^{4}}{\sin \left (d x +c \right )}-\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )\right )+a^{4} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(129\)
risch	\(-5 i a^{4} x -\frac {i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{6 d}+\frac {11 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {11 a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}+\frac {i a^{4} {\mathrm e}^{-3 i \left (d x +c \right )}}{6 d}-\frac {10 i a^{4} c}{d}-\frac {2 i a^{4} \left (i {\mathrm e}^{2 i \left (d x +c \right )}+4 \,{\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {5 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {a^{4} \cos \left (4 d x +4 c \right )}{32 d}\)	\(219\)

Input:

int(cot(d*x+c)^3*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.28 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {24 \, a^{4} \cos \left (d x + c\right )^{6} - 312 \, a^{4} \cos \left (d x + c\right )^{4} + 423 \, a^{4} \cos \left (d x + c\right )^{2} - 183 \, a^{4} - 480 \, {\left (a^{4} \cos \left (d x + c\right )^{2} - a^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 128 \, {\left (a^{4} \cos \left (d x + c\right )^{4} - 2 \, a^{4} \cos \left (d x + c\right )^{2} + 4 \, a^{4}\right )} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:

integrate(cot(d*x+c)^3*(a+a*sin(d*x+c))^4,x, algorithm="fricas")
 

Output:

-1/96*(24*a^4*cos(d*x + c)^6 - 312*a^4*cos(d*x + c)^4 + 423*a^4*cos(d*x + 
c)^2 - 183*a^4 - 480*(a^4*cos(d*x + c)^2 - a^4)*log(1/2*sin(d*x + c)) - 12 
8*(a^4*cos(d*x + c)^4 - 2*a^4*cos(d*x + c)^2 + 4*a^4)*sin(d*x + c))/(d*cos 
(d*x + c)^2 - d)
 

Sympy [F]

\[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=a^{4} \left (\int 4 \sin {\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int \cot ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:

integrate(cot(d*x+c)**3*(a+a*sin(d*x+c))**4,x)
 

Output:

a**4*(Integral(4*sin(c + d*x)*cot(c + d*x)**3, x) + Integral(6*sin(c + d*x 
)**2*cot(c + d*x)**3, x) + Integral(4*sin(c + d*x)**3*cot(c + d*x)**3, x) 
+ Integral(sin(c + d*x)**4*cot(c + d*x)**3, x) + Integral(cot(c + d*x)**3, 
 x))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} \sin \left (d x + c\right )^{2} - 60 \, a^{4} \log \left (\sin \left (d x + c\right )\right ) + \frac {6 \, {\left (8 \, a^{4} \sin \left (d x + c\right ) + a^{4}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, d} \] Input:

integrate(cot(d*x+c)^3*(a+a*sin(d*x+c))^4,x, algorithm="maxima")
 

Output:

-1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 30*a^4*sin(d*x + c)^ 
2 - 60*a^4*log(sin(d*x + c)) + 6*(8*a^4*sin(d*x + c) + a^4)/sin(d*x + c)^2 
)/d
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} \sin \left (d x + c\right )^{2} - 60 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {6 \, {\left (8 \, a^{4} \sin \left (d x + c\right ) + a^{4}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, d} \] Input:

integrate(cot(d*x+c)^3*(a+a*sin(d*x+c))^4,x, algorithm="giac")
 

Output:

-1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 30*a^4*sin(d*x + c)^ 
2 - 60*a^4*log(abs(sin(d*x + c))) + 6*(8*a^4*sin(d*x + c) + a^4)/sin(d*x + 
 c)^2)/d
 

Mupad [B] (verification not implemented)

Time = 17.70 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.92 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {8\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {81\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+\frac {224\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+98\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {272\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+43\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^4}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d}-\frac {5\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:

int(cot(c + d*x)^3*(a + a*sin(c + d*x))^4,x)
 

Output:

(5*a^4*log(tan(c/2 + (d*x)/2)))/d - (a^4*tan(c/2 + (d*x)/2)^2)/(8*d) - (2* 
a^4*tan(c/2 + (d*x)/2)^2 + 32*a^4*tan(c/2 + (d*x)/2)^3 + 43*a^4*tan(c/2 + 
(d*x)/2)^4 + (272*a^4*tan(c/2 + (d*x)/2)^5)/3 + 98*a^4*tan(c/2 + (d*x)/2)^ 
6 + (224*a^4*tan(c/2 + (d*x)/2)^7)/3 + (81*a^4*tan(c/2 + (d*x)/2)^8)/2 + 8 
*a^4*tan(c/2 + (d*x)/2)^9 + a^4/2 + 8*a^4*tan(c/2 + (d*x)/2))/(d*(4*tan(c/ 
2 + (d*x)/2)^2 + 16*tan(c/2 + (d*x)/2)^4 + 24*tan(c/2 + (d*x)/2)^6 + 16*ta 
n(c/2 + (d*x)/2)^8 + 4*tan(c/2 + (d*x)/2)^10)) - (2*a^4*tan(c/2 + (d*x)/2) 
)/d - (5*a^4*log(tan(c/2 + (d*x)/2)^2 + 1))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.08 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^{4} \left (-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-3 \sin \left (d x +c \right )^{6}-16 \sin \left (d x +c \right )^{5}-30 \sin \left (d x +c \right )^{4}+9 \sin \left (d x +c \right )^{2}-48 \sin \left (d x +c \right )-6\right )}{12 \sin \left (d x +c \right )^{2} d} \] Input:

int(cot(d*x+c)^3*(a+a*sin(d*x+c))^4,x)
 

Output:

(a**4*( - 60*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 + 60*log(tan((c 
+ d*x)/2))*sin(c + d*x)**2 - 3*sin(c + d*x)**6 - 16*sin(c + d*x)**5 - 30*s 
in(c + d*x)**4 + 9*sin(c + d*x)**2 - 48*sin(c + d*x) - 6))/(12*sin(c + d*x 
)**2*d)