\(\int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1119]

Optimal result

Integrand size = 29, antiderivative size = 227 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=3 a b^2 x-\frac {3 b \left (3 a^2-4 b^2\right ) \text {arctanh}(\cos (c+d x))}{8 d}-\frac {b^3 \left (83 a^2+2 b^2\right ) \cos (c+d x)}{40 a^2 d}-\frac {a \left (4 a^2-29 b^2\right ) \cot (c+d x)}{20 d}+\frac {27 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{40 d}+\frac {2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{5 d}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d} \] Output:

3*a*b^2*x-3/8*b*(3*a^2-4*b^2)*arctanh(cos(d*x+c))/d-1/40*b^3*(83*a^2+2*b^2 
)*cos(d*x+c)/a^2/d-1/20*a*(4*a^2-29*b^2)*cot(d*x+c)/d+27/40*b*cot(d*x+c)*c 
sc(d*x+c)*(a+b*sin(d*x+c))^2/d+2/5*cot(d*x+c)*csc(d*x+c)^2*(a+b*sin(d*x+c) 
)^3/d+1/20*b*cot(d*x+c)*csc(d*x+c)^3*(a+b*sin(d*x+c))^4/a^2/d-1/5*cot(d*x+ 
c)*csc(d*x+c)^4*(a+b*sin(d*x+c))^4/a/d
 

Mathematica [A] (verified)

Time = 3.29 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.78 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {960 a b^2 c+960 a b^2 d x-320 b^3 \cos (c+d x)-32 \left (a^3-20 a b^2\right ) \cot \left (\frac {1}{2} (c+d x)\right )+150 a^2 b \csc ^2\left (\frac {1}{2} (c+d x)\right )-40 b^3 \csc ^2\left (\frac {1}{2} (c+d x)\right )-15 a^2 b \csc ^4\left (\frac {1}{2} (c+d x)\right )-360 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+480 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+360 a^2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-480 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-150 a^2 b \sec ^2\left (\frac {1}{2} (c+d x)\right )+40 b^3 \sec ^2\left (\frac {1}{2} (c+d x)\right )+15 a^2 b \sec ^4\left (\frac {1}{2} (c+d x)\right )-112 a^3 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+320 a b^2 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+64 a^3 \csc ^5(c+d x) \sin ^6\left (\frac {1}{2} (c+d x)\right )+7 a^3 \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)-20 a b^2 \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)-a^3 \csc ^6\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)+32 a^3 \tan \left (\frac {1}{2} (c+d x)\right )-640 a b^2 \tan \left (\frac {1}{2} (c+d x)\right )}{320 d} \] Input:

Integrate[Cot[c + d*x]^4*Csc[c + d*x]^2*(a + b*Sin[c + d*x])^3,x]
 

Output:

(960*a*b^2*c + 960*a*b^2*d*x - 320*b^3*Cos[c + d*x] - 32*(a^3 - 20*a*b^2)* 
Cot[(c + d*x)/2] + 150*a^2*b*Csc[(c + d*x)/2]^2 - 40*b^3*Csc[(c + d*x)/2]^ 
2 - 15*a^2*b*Csc[(c + d*x)/2]^4 - 360*a^2*b*Log[Cos[(c + d*x)/2]] + 480*b^ 
3*Log[Cos[(c + d*x)/2]] + 360*a^2*b*Log[Sin[(c + d*x)/2]] - 480*b^3*Log[Si 
n[(c + d*x)/2]] - 150*a^2*b*Sec[(c + d*x)/2]^2 + 40*b^3*Sec[(c + d*x)/2]^2 
 + 15*a^2*b*Sec[(c + d*x)/2]^4 - 112*a^3*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 
 + 320*a*b^2*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 64*a^3*Csc[c + d*x]^5*Sin 
[(c + d*x)/2]^6 + 7*a^3*Csc[(c + d*x)/2]^4*Sin[c + d*x] - 20*a*b^2*Csc[(c 
+ d*x)/2]^4*Sin[c + d*x] - a^3*Csc[(c + d*x)/2]^6*Sin[c + d*x] + 32*a^3*Ta 
n[(c + d*x)/2] - 640*a*b^2*Tan[(c + d*x)/2])/(320*d)
 

Rubi [A] (verified)

Time = 1.80 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.07, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.621, Rules used = {3042, 3372, 3042, 3526, 27, 3042, 3526, 25, 3042, 3510, 25, 3042, 3502, 27, 3042, 3214, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^4 (a+b \sin (c+d x))^3}{\sin (c+d x)^6}dx\)

\(\Big \downarrow \) 3372

\(\displaystyle -\frac {\int \csc ^4(c+d x) (a+b \sin (c+d x))^3 \left (24 a^2+3 b \sin (c+d x) a-\left (20 a^2+b^2\right ) \sin ^2(c+d x)\right )dx}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {(a+b \sin (c+d x))^3 \left (24 a^2+3 b \sin (c+d x) a-\left (20 a^2+b^2\right ) \sin (c+d x)^2\right )}{\sin (c+d x)^4}dx}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3526

\(\displaystyle -\frac {\frac {1}{3} \int 3 \csc ^3(c+d x) (a+b \sin (c+d x))^2 \left (27 b a^2-2 \left (2 a^2-b^2\right ) \sin (c+d x) a-b \left (28 a^2+b^2\right ) \sin ^2(c+d x)\right )dx-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \csc ^3(c+d x) (a+b \sin (c+d x))^2 \left (27 b a^2-2 \left (2 a^2-b^2\right ) \sin (c+d x) a-b \left (28 a^2+b^2\right ) \sin ^2(c+d x)\right )dx-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {(a+b \sin (c+d x))^2 \left (27 b a^2-2 \left (2 a^2-b^2\right ) \sin (c+d x) a-b \left (28 a^2+b^2\right ) \sin (c+d x)^2\right )}{\sin (c+d x)^3}dx-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3526

\(\displaystyle -\frac {\frac {1}{2} \int -\csc ^2(c+d x) (a+b \sin (c+d x)) \left (2 \left (4 a^2-29 b^2\right ) a^2+b \left (37 a^2-2 b^2\right ) \sin (c+d x) a+b^2 \left (83 a^2+2 b^2\right ) \sin ^2(c+d x)\right )dx-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {-\frac {1}{2} \int \csc ^2(c+d x) (a+b \sin (c+d x)) \left (2 \left (4 a^2-29 b^2\right ) a^2+b \left (37 a^2-2 b^2\right ) \sin (c+d x) a+b^2 \left (83 a^2+2 b^2\right ) \sin ^2(c+d x)\right )dx-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {1}{2} \int \frac {(a+b \sin (c+d x)) \left (2 \left (4 a^2-29 b^2\right ) a^2+b \left (37 a^2-2 b^2\right ) \sin (c+d x) a+b^2 \left (83 a^2+2 b^2\right ) \sin (c+d x)^2\right )}{\sin (c+d x)^2}dx-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3510

\(\displaystyle -\frac {\frac {1}{2} \left (\int -\csc (c+d x) \left (120 b^2 \sin (c+d x) a^3+15 b \left (3 a^2-4 b^2\right ) a^2+b^3 \left (83 a^2+2 b^2\right ) \sin ^2(c+d x)\right )dx+\frac {2 a^3 \left (4 a^2-29 b^2\right ) \cot (c+d x)}{d}\right )-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {1}{2} \left (\frac {2 a^3 \left (4 a^2-29 b^2\right ) \cot (c+d x)}{d}-\int \csc (c+d x) \left (120 b^2 \sin (c+d x) a^3+15 b \left (3 a^2-4 b^2\right ) a^2+b^3 \left (83 a^2+2 b^2\right ) \sin ^2(c+d x)\right )dx\right )-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{2} \left (\frac {2 a^3 \left (4 a^2-29 b^2\right ) \cot (c+d x)}{d}-\int \frac {120 b^2 \sin (c+d x) a^3+15 b \left (3 a^2-4 b^2\right ) a^2+b^3 \left (83 a^2+2 b^2\right ) \sin (c+d x)^2}{\sin (c+d x)}dx\right )-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {\frac {1}{2} \left (-\int 15 \csc (c+d x) \left (8 b^2 \sin (c+d x) a^3+b \left (3 a^2-4 b^2\right ) a^2\right )dx+\frac {b^3 \left (83 a^2+2 b^2\right ) \cos (c+d x)}{d}+\frac {2 a^3 \left (4 a^2-29 b^2\right ) \cot (c+d x)}{d}\right )-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {1}{2} \left (-15 \int \csc (c+d x) \left (8 b^2 \sin (c+d x) a^3+b \left (3 a^2-4 b^2\right ) a^2\right )dx+\frac {b^3 \left (83 a^2+2 b^2\right ) \cos (c+d x)}{d}+\frac {2 a^3 \left (4 a^2-29 b^2\right ) \cot (c+d x)}{d}\right )-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{2} \left (-15 \int \frac {8 b^2 \sin (c+d x) a^3+b \left (3 a^2-4 b^2\right ) a^2}{\sin (c+d x)}dx+\frac {b^3 \left (83 a^2+2 b^2\right ) \cos (c+d x)}{d}+\frac {2 a^3 \left (4 a^2-29 b^2\right ) \cot (c+d x)}{d}\right )-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3214

\(\displaystyle -\frac {\frac {1}{2} \left (-15 \left (a^2 b \left (3 a^2-4 b^2\right ) \int \csc (c+d x)dx+8 a^3 b^2 x\right )+\frac {b^3 \left (83 a^2+2 b^2\right ) \cos (c+d x)}{d}+\frac {2 a^3 \left (4 a^2-29 b^2\right ) \cot (c+d x)}{d}\right )-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{2} \left (-15 \left (a^2 b \left (3 a^2-4 b^2\right ) \int \csc (c+d x)dx+8 a^3 b^2 x\right )+\frac {b^3 \left (83 a^2+2 b^2\right ) \cos (c+d x)}{d}+\frac {2 a^3 \left (4 a^2-29 b^2\right ) \cot (c+d x)}{d}\right )-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{20 a^2 d}-\frac {-\frac {8 a^2 \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{d}-\frac {27 a^2 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}+\frac {1}{2} \left (\frac {b^3 \left (83 a^2+2 b^2\right ) \cos (c+d x)}{d}-15 \left (8 a^3 b^2 x-\frac {a^2 b \left (3 a^2-4 b^2\right ) \text {arctanh}(\cos (c+d x))}{d}\right )+\frac {2 a^3 \left (4 a^2-29 b^2\right ) \cot (c+d x)}{d}\right )}{20 a^2}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^4}{5 a d}\)

Input:

Int[Cot[c + d*x]^4*Csc[c + d*x]^2*(a + b*Sin[c + d*x])^3,x]
 

Output:

(b*Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^4)/(20*a^2*d) - (Cot[c 
 + d*x]*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^4)/(5*a*d) - ((-15*(8*a^3*b^2* 
x - (a^2*b*(3*a^2 - 4*b^2)*ArcTanh[Cos[c + d*x]])/d) + (b^3*(83*a^2 + 2*b^ 
2)*Cos[c + d*x])/d + (2*a^3*(4*a^2 - 29*b^2)*Cot[c + d*x])/d)/2 - (27*a^2* 
b*Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^2)/(2*d) - (8*a^2*Cot[c + 
 d*x]*Csc[c + d*x]^2*(a + b*Sin[c + d*x])^3)/d)/(20*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3042

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

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3372

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[Cos[e + f*x]*(a + b* 
Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 1)/(a*d*f*(n + 1))), x] + (-Si 
mp[b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x] 
)^(n + 2)/(a^2*d^2*f*(n + 1)*(n + 2))), x] - Simp[1/(a^2*d^2*(n + 1)*(n + 2 
))   Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) 
 - b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) 
- b^2*(m + n + 2)*(m + n + 4))*Sin[e + f*x]^2, x], x], x]) /; FreeQ[{a, b, 
d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n]) 
 &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3510

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 

rule 3526

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 
 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* 
d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 
1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x 
] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f 
*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d 
, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

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 = 1.55 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {-\frac {a^{3} \cos \left (d x +c \right )^{5}}{5 \sin \left (d x +c \right )^{5}}+3 a^{2} b \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+b^{3} \left (-\frac {\cos \left (d x +c \right )^{5}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{3}}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(192\)
default	\(\frac {-\frac {a^{3} \cos \left (d x +c \right )^{5}}{5 \sin \left (d x +c \right )^{5}}+3 a^{2} b \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+b^{3} \left (-\frac {\cos \left (d x +c \right )^{5}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{3}}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(192\)
risch	\(3 a \,b^{2} x -\frac {b^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {-8 i a^{3}+160 i a \,b^{2}-75 a^{2} b \,{\mathrm e}^{9 i \left (d x +c \right )}+20 b^{3} {\mathrm e}^{9 i \left (d x +c \right )}+880 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+30 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-40 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-720 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-40 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+240 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-30 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+40 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-560 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-80 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+75 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-20 b^{3} {\mathrm e}^{i \left (d x +c \right )}}{20 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {9 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}-\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}-\frac {9 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}+\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}\)	\(367\)

Input:

int(cot(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.47 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {560 \, a b^{2} \cos \left (d x + c\right )^{3} + 16 \, {\left (a^{3} - 20 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 240 \, a b^{2} \cos \left (d x + c\right ) + 15 \, {\left ({\left (3 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b - 4 \, b^{3} - 2 \, {\left (3 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left ({\left (3 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b - 4 \, b^{3} - 2 \, {\left (3 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 10 \, {\left (24 \, a b^{2} d x \cos \left (d x + c\right )^{4} - 8 \, b^{3} \cos \left (d x + c\right )^{5} - 48 \, a b^{2} d x \cos \left (d x + c\right )^{2} + 24 \, a b^{2} d x - 5 \, {\left (3 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:

integrate(cot(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="frica 
s")
 

Output:

-1/80*(560*a*b^2*cos(d*x + c)^3 + 16*(a^3 - 20*a*b^2)*cos(d*x + c)^5 - 240 
*a*b^2*cos(d*x + c) + 15*((3*a^2*b - 4*b^3)*cos(d*x + c)^4 + 3*a^2*b - 4*b 
^3 - 2*(3*a^2*b - 4*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d 
*x + c) - 15*((3*a^2*b - 4*b^3)*cos(d*x + c)^4 + 3*a^2*b - 4*b^3 - 2*(3*a^ 
2*b - 4*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 1 
0*(24*a*b^2*d*x*cos(d*x + c)^4 - 8*b^3*cos(d*x + c)^5 - 48*a*b^2*d*x*cos(d 
*x + c)^2 + 24*a*b^2*d*x - 5*(3*a^2*b - 4*b^3)*cos(d*x + c)^3 + 3*(3*a^2*b 
 - 4*b^3)*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c 
)^2 + d)*sin(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**4*csc(d*x+c)**2*(a+b*sin(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.80 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {16 \, a^{3} \cot \left (d x + c\right )^{5} - 80 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a b^{2} + 15 \, a^{2} b {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 20 \, b^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{80 \, d} \] Input:

integrate(cot(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="maxim 
a")
 

Output:

-1/80*(16*a^3*cot(d*x + c)^5 - 80*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/ta 
n(d*x + c)^3)*a*b^2 + 15*a^2*b*(2*(5*cos(d*x + c)^3 - 3*cos(d*x + c))/(cos 
(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(cos( 
d*x + c) - 1)) - 20*b^3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - 4*cos(d*x + 
 c) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)))/d
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.57 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 40 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 960 \, {\left (d x + c\right )} a b^{2} + 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 600 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {640 \, b^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 120 \, {\left (3 \, a^{2} b - 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {822 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1096 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 600 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 40 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{320 \, d} \] Input:

integrate(cot(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="giac" 
)
 

Output:

1/320*(2*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 10 
*a^3*tan(1/2*d*x + 1/2*c)^3 + 40*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 120*a^2*b* 
tan(1/2*d*x + 1/2*c)^2 + 40*b^3*tan(1/2*d*x + 1/2*c)^2 + 960*(d*x + c)*a*b 
^2 + 20*a^3*tan(1/2*d*x + 1/2*c) - 600*a*b^2*tan(1/2*d*x + 1/2*c) - 640*b^ 
3/(tan(1/2*d*x + 1/2*c)^2 + 1) + 120*(3*a^2*b - 4*b^3)*log(abs(tan(1/2*d*x 
 + 1/2*c))) - (822*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 1096*b^3*tan(1/2*d*x + 1 
/2*c)^5 + 20*a^3*tan(1/2*d*x + 1/2*c)^4 - 600*a*b^2*tan(1/2*d*x + 1/2*c)^4 
 - 120*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 40*b^3*tan(1/2*d*x + 1/2*c)^3 - 10*a 
^3*tan(1/2*d*x + 1/2*c)^2 + 40*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 15*a^2*b*tan 
(1/2*d*x + 1/2*c) + 2*a^3)/tan(1/2*d*x + 1/2*c)^5)/d
 

Mupad [B] (verification not implemented)

Time = 36.72 (sec) , antiderivative size = 1007, normalized size of antiderivative = 4.44 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Too large to display} \] Input:

int((cot(c + d*x)^4*(a + b*sin(c + d*x))^3)/sin(c + d*x)^2,x)
 

Output:

-(2*a^3*cos(c/2 + (d*x)/2)^12 - 2*a^3*sin(c/2 + (d*x)/2)^12 + 8*a^3*cos(c/ 
2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^10 - 10*a^3*cos(c/2 + (d*x)/2)^4*sin(c/2 
 + (d*x)/2)^8 + 10*a^3*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4 - 8*a^3*c 
os(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^2 - 40*b^3*cos(c/2 + (d*x)/2)^3*si 
n(c/2 + (d*x)/2)^9 - 40*b^3*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^7 + 68 
0*b^3*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^5 + 40*b^3*cos(c/2 + (d*x)/2 
)^9*sin(c/2 + (d*x)/2)^3 + 480*b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/ 
2))*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^7 + 480*b^3*log(sin(c/2 + (d*x 
)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^5 - 15*a^ 
2*b*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^11 + 15*a^2*b*cos(c/2 + (d*x)/2) 
^11*sin(c/2 + (d*x)/2) - 40*a*b^2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^ 
10 + 560*a*b^2*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^8 - 560*a*b^2*cos(c 
/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4 + 40*a*b^2*cos(c/2 + (d*x)/2)^10*sin( 
c/2 + (d*x)/2)^2 + 105*a^2*b*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^9 + 1 
20*a^2*b*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^7 - 120*a^2*b*cos(c/2 + ( 
d*x)/2)^7*sin(c/2 + (d*x)/2)^5 - 105*a^2*b*cos(c/2 + (d*x)/2)^9*sin(c/2 + 
(d*x)/2)^3 + 1920*a*b^2*atan((3*a^2*sin(c/2 + (d*x)/2) - 4*b^2*sin(c/2 + ( 
d*x)/2) + 8*a*b*cos(c/2 + (d*x)/2))/(4*b^2*cos(c/2 + (d*x)/2) - 3*a^2*cos( 
c/2 + (d*x)/2) + 8*a*b*sin(c/2 + (d*x)/2)))*cos(c/2 + (d*x)/2)^5*sin(c/2 + 
 (d*x)/2)^7 + 1920*a*b^2*atan((3*a^2*sin(c/2 + (d*x)/2) - 4*b^2*sin(c/2...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.19 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b^{3}-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{3}+160 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}+75 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b -20 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{3}+16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -8 \cos \left (d x +c \right ) a^{3}+45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} a^{2} b -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} b^{3}-15 \sin \left (d x +c \right )^{5} a^{2} b +120 \sin \left (d x +c \right )^{5} a \,b^{2} d x +45 \sin \left (d x +c \right )^{5} b^{3}}{40 \sin \left (d x +c \right )^{5} d} \] Input:

int(cot(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x)
 

Output:

( - 40*cos(c + d*x)*sin(c + d*x)**5*b**3 - 8*cos(c + d*x)*sin(c + d*x)**4* 
a**3 + 160*cos(c + d*x)*sin(c + d*x)**4*a*b**2 + 75*cos(c + d*x)*sin(c + d 
*x)**3*a**2*b - 20*cos(c + d*x)*sin(c + d*x)**3*b**3 + 16*cos(c + d*x)*sin 
(c + d*x)**2*a**3 - 40*cos(c + d*x)*sin(c + d*x)**2*a*b**2 - 30*cos(c + d* 
x)*sin(c + d*x)*a**2*b - 8*cos(c + d*x)*a**3 + 45*log(tan((c + d*x)/2))*si 
n(c + d*x)**5*a**2*b - 60*log(tan((c + d*x)/2))*sin(c + d*x)**5*b**3 - 15* 
sin(c + d*x)**5*a**2*b + 120*sin(c + d*x)**5*a*b**2*d*x + 45*sin(c + d*x)* 
*5*b**3)/(40*sin(c + d*x)**5*d)