\(\int \cot ^4(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x)) \, dx\) [375]

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

Optimal result

Integrand size = 27, antiderivative size = 98 \[ \int \cot ^4(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{16 d}-\frac {a \cot ^5(c+d x)}{5 d}-\frac {a \cot (c+d x) \csc (c+d x)}{16 d}+\frac {a \cot (c+d x) \csc ^3(c+d x)}{8 d}-\frac {a \cot ^3(c+d x) \csc ^3(c+d x)}{6 d} \] Output:

-1/16*a*arctanh(cos(d*x+c))/d-1/5*a*cot(d*x+c)^5/d-1/16*a*cot(d*x+c)*csc(d 
*x+c)/d+1/8*a*cot(d*x+c)*csc(d*x+c)^3/d-1/6*a*cot(d*x+c)^3*csc(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.79 \[ \int \cot ^4(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cot ^5(c+d x)}{5 d}-\frac {a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {a \csc ^6\left (\frac {1}{2} (c+d x)\right )}{384 d}-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{16 d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{16 d}+\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {a \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {a \sec ^6\left (\frac {1}{2} (c+d x)\right )}{384 d} \] Input:

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

Output:

-1/5*(a*Cot[c + d*x]^5)/d - (a*Csc[(c + d*x)/2]^2)/(64*d) + (a*Csc[(c + d* 
x)/2]^4)/(64*d) - (a*Csc[(c + d*x)/2]^6)/(384*d) - (a*Log[Cos[(c + d*x)/2] 
])/(16*d) + (a*Log[Sin[(c + d*x)/2]])/(16*d) + (a*Sec[(c + d*x)/2]^2)/(64* 
d) - (a*Sec[(c + d*x)/2]^4)/(64*d) + (a*Sec[(c + d*x)/2]^6)/(384*d)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3317, 3042, 3087, 15, 3091, 3042, 3091, 3042, 4255, 3042, 4257}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^4 (a \sin (c+d x)+a)}{\sin (c+d x)^7}dx\)

\(\Big \downarrow \) 3317

\(\displaystyle a \int \cot ^4(c+d x) \csc ^3(c+d x)dx+a \int \cot ^4(c+d x) \csc ^2(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \sec \left (c+d x-\frac {\pi }{2}\right )^2 \tan \left (c+d x-\frac {\pi }{2}\right )^4dx+a \int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^4dx\)

\(\Big \downarrow \) 3087

\(\displaystyle \frac {a \int \cot ^4(c+d x)d(-\cot (c+d x))}{d}+a \int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^4dx\)

\(\Big \downarrow \) 15

\(\displaystyle a \int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^4dx-\frac {a \cot ^5(c+d x)}{5 d}\)

\(\Big \downarrow \) 3091

\(\displaystyle a \left (-\frac {1}{2} \int \cot ^2(c+d x) \csc ^3(c+d x)dx-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (-\frac {1}{2} \int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\)

\(\Big \downarrow \) 3091

\(\displaystyle a \left (\frac {1}{2} \left (\frac {1}{4} \int \csc ^3(c+d x)dx+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}\right )-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {1}{2} \left (\frac {1}{4} \int \csc (c+d x)^3dx+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}\right )-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\)

\(\Big \downarrow \) 4255

\(\displaystyle a \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}\right )-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}\right )-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle a \left (\frac {1}{2} \left (\frac {1}{4} \left (-\frac {\text {arctanh}(\cos (c+d x))}{2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}\right )-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\)

Input:

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

Output:

-1/5*(a*Cot[c + d*x]^5)/d + a*(-1/6*(Cot[c + d*x]^3*Csc[c + d*x]^3)/d + (( 
Cot[c + d*x]*Csc[c + d*x]^3)/(4*d) + (-1/2*ArcTanh[Cos[c + d*x]]/d - (Cot[ 
c + d*x]*Csc[c + d*x])/(2*d))/4)/2)
 

Defintions of rubi rules used

rule 15
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3087
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S 
ymbol] :> Simp[1/f   Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + 
f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n - 1) 
/2] && LtQ[0, n, m - 1])
 

rule 3091
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m 
 + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1))   Int[(a*Sec[e + f*x])^m*( 
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & 
& NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
 

rule 3317
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[(g*Co 
s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d   Int[(g*Cos[e + f*x])^ 
p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
 

rule 4255
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.20

method result size
derivativedivides \(\frac {-\frac {a \cos \left (d x +c \right )^{5}}{5 \sin \left (d x +c \right )^{5}}+a \left (-\frac {\cos \left (d x +c \right )^{5}}{6 \sin \left (d x +c \right )^{6}}-\frac {\cos \left (d x +c \right )^{5}}{24 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{48 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{48}+\frac {\cos \left (d x +c \right )}{16}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{16}\right )}{d}\) \(118\)
default \(\frac {-\frac {a \cos \left (d x +c \right )^{5}}{5 \sin \left (d x +c \right )^{5}}+a \left (-\frac {\cos \left (d x +c \right )^{5}}{6 \sin \left (d x +c \right )^{6}}-\frac {\cos \left (d x +c \right )^{5}}{24 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{48 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{48}+\frac {\cos \left (d x +c \right )}{16}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{16}\right )}{d}\) \(118\)
risch \(\frac {a \left (15 \,{\mathrm e}^{11 i \left (d x +c \right )}+235 \,{\mathrm e}^{9 i \left (d x +c \right )}-240 i {\mathrm e}^{10 i \left (d x +c \right )}+390 \,{\mathrm e}^{7 i \left (d x +c \right )}+240 i {\mathrm e}^{8 i \left (d x +c \right )}+390 \,{\mathrm e}^{5 i \left (d x +c \right )}-480 i {\mathrm e}^{6 i \left (d x +c \right )}+235 \,{\mathrm e}^{3 i \left (d x +c \right )}+480 i {\mathrm e}^{4 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}-48 i {\mathrm e}^{2 i \left (d x +c \right )}+48 i\right )}{120 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{6}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{16 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{16 d}\) \(186\)

Input:

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

Output:

1/d*(-1/5*a/sin(d*x+c)^5*cos(d*x+c)^5+a*(-1/6/sin(d*x+c)^6*cos(d*x+c)^5-1/ 
24/sin(d*x+c)^4*cos(d*x+c)^5+1/48/sin(d*x+c)^2*cos(d*x+c)^5+1/48*cos(d*x+c 
)^3+1/16*cos(d*x+c)+1/16*ln(csc(d*x+c)-cot(d*x+c))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (88) = 176\).

Time = 0.09 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.91 \[ \int \cot ^4(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {96 \, a \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) + 30 \, a \cos \left (d x + c\right )^{5} + 80 \, a \cos \left (d x + c\right )^{3} - 30 \, a \cos \left (d x + c\right ) - 15 \, {\left (a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} + 3 \, a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} + 3 \, a \cos \left (d x + c\right )^{2} - a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{480 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:

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

Output:

1/480*(96*a*cos(d*x + c)^5*sin(d*x + c) + 30*a*cos(d*x + c)^5 + 80*a*cos(d 
*x + c)^3 - 30*a*cos(d*x + c) - 15*(a*cos(d*x + c)^6 - 3*a*cos(d*x + c)^4 
+ 3*a*cos(d*x + c)^2 - a)*log(1/2*cos(d*x + c) + 1/2) + 15*(a*cos(d*x + c) 
^6 - 3*a*cos(d*x + c)^4 + 3*a*cos(d*x + c)^2 - a)*log(-1/2*cos(d*x + c) + 
1/2))/(d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)
 

Sympy [F]

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

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

Output:

a*(Integral(cot(c + d*x)**4*csc(c + d*x)**3, x) + Integral(sin(c + d*x)*co 
t(c + d*x)**4*csc(c + d*x)**3, x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.08 \[ \int \cot ^4(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {5 \, a {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{5} + 8 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - \frac {96 \, a}{\tan \left (d x + c\right )^{5}}}{480 \, d} \] Input:

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

Output:

1/480*(5*a*(2*(3*cos(d*x + c)^5 + 8*cos(d*x + c)^3 - 3*cos(d*x + c))/(cos( 
d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - 1) - 3*log(cos(d*x + c) 
 + 1) + 3*log(cos(d*x + c) - 1)) - 96*a/tan(d*x + c)^5)/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (88) = 176\).

Time = 0.19 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.05 \[ \int \cot ^4(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {5 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 60 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 120 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {294 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 120 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 60 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6}}}{1920 \, d} \] Input:

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

Output:

1/1920*(5*a*tan(1/2*d*x + 1/2*c)^6 + 12*a*tan(1/2*d*x + 1/2*c)^5 - 15*a*ta 
n(1/2*d*x + 1/2*c)^4 - 60*a*tan(1/2*d*x + 1/2*c)^3 - 15*a*tan(1/2*d*x + 1/ 
2*c)^2 + 120*a*log(abs(tan(1/2*d*x + 1/2*c))) + 120*a*tan(1/2*d*x + 1/2*c) 
 - (294*a*tan(1/2*d*x + 1/2*c)^6 + 120*a*tan(1/2*d*x + 1/2*c)^5 - 15*a*tan 
(1/2*d*x + 1/2*c)^4 - 60*a*tan(1/2*d*x + 1/2*c)^3 - 15*a*tan(1/2*d*x + 1/2 
*c)^2 + 12*a*tan(1/2*d*x + 1/2*c) + 5*a)/tan(1/2*d*x + 1/2*c)^6)/d
 

Mupad [B] (verification not implemented)

Time = 18.16 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.44 \[ \int \cot ^4(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\left (5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+12\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-60\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+60\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+120\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\right )}{1920\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6} \] Input:

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

Output:

(a*(5*sin(c/2 + (d*x)/2)^12 - 5*cos(c/2 + (d*x)/2)^12 + 12*cos(c/2 + (d*x) 
/2)*sin(c/2 + (d*x)/2)^11 - 12*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2) - 
15*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^10 - 60*cos(c/2 + (d*x)/2)^3*si 
n(c/2 + (d*x)/2)^9 - 15*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^8 + 120*co 
s(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^7 - 120*cos(c/2 + (d*x)/2)^7*sin(c/2 
 + (d*x)/2)^5 + 15*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4 + 60*cos(c/2 
+ (d*x)/2)^9*sin(c/2 + (d*x)/2)^3 + 15*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d* 
x)/2)^2 + 120*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2 
)^6*sin(c/2 + (d*x)/2)^6))/(1920*d*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2) 
^6)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.23 \[ \int \cot ^4(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-48 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+96 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+70 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-48 \cos \left (d x +c \right ) \sin \left (d x +c \right )-40 \cos \left (d x +c \right )+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{6}\right )}{240 \sin \left (d x +c \right )^{6} d} \] Input:

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

Output:

(a*( - 48*cos(c + d*x)*sin(c + d*x)**5 - 15*cos(c + d*x)*sin(c + d*x)**4 + 
 96*cos(c + d*x)*sin(c + d*x)**3 + 70*cos(c + d*x)*sin(c + d*x)**2 - 48*co 
s(c + d*x)*sin(c + d*x) - 40*cos(c + d*x) + 15*log(tan((c + d*x)/2))*sin(c 
 + d*x)**6))/(240*sin(c + d*x)**6*d)