\(\int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1072]

Optimal result

Integrand size = 29, antiderivative size = 183 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {b \left (3 a^2+4 b^2\right ) \text {arctanh}(\cos (c+d x))}{8 d}+\frac {a \left (2 a^2+15 b^2\right ) \cot (c+d x)}{15 d}+\frac {3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}+\frac {a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d} \] Output:

1/8*b*(3*a^2+4*b^2)*arctanh(cos(d*x+c))/d+1/15*a*(2*a^2+15*b^2)*cot(d*x+c) 
/d+3/40*b*(5*a^2-2*b^2)*cot(d*x+c)*csc(d*x+c)/d+1/30*a*(2*a^2-3*b^2)*cot(d 
*x+c)*csc(d*x+c)^2/d-3/20*b*cot(d*x+c)*csc(d*x+c)^3*(a+b*sin(d*x+c))^2/d-1 
/5*cot(d*x+c)*csc(d*x+c)^4*(a+b*sin(d*x+c))^3/d
 

Mathematica [A] (verified)

Time = 2.73 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.88 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {32 \left (2 a^3+15 a b^2\right ) \cot \left (\frac {1}{2} (c+d x)\right )+30 \left (3 a^2 b-4 b^3\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )+360 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+480 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-360 a^2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-480 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-90 a^2 b \sec ^2\left (\frac {1}{2} (c+d x)\right )+120 b^3 \sec ^2\left (\frac {1}{2} (c+d x)\right )+45 a^2 b \sec ^4\left (\frac {1}{2} (c+d x)\right )-16 a^3 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+960 a b^2 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-3 a^3 \csc ^6\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)+a \csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (-45 a b+\left (a^2-60 b^2\right ) \sin (c+d x)\right )-64 a^3 \tan \left (\frac {1}{2} (c+d x)\right )-480 a b^2 \tan \left (\frac {1}{2} (c+d x)\right )+6 a^3 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{960 d} \] Input:

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

Output:

(32*(2*a^3 + 15*a*b^2)*Cot[(c + d*x)/2] + 30*(3*a^2*b - 4*b^3)*Csc[(c + d* 
x)/2]^2 + 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[Sin[(c + d*x)/2]] - 90*a^2 
*b*Sec[(c + d*x)/2]^2 + 120*b^3*Sec[(c + d*x)/2]^2 + 45*a^2*b*Sec[(c + d*x 
)/2]^4 - 16*a^3*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 960*a*b^2*Csc[c + d*x] 
^3*Sin[(c + d*x)/2]^4 - 3*a^3*Csc[(c + d*x)/2]^6*Sin[c + d*x] + a*Csc[(c + 
 d*x)/2]^4*(-45*a*b + (a^2 - 60*b^2)*Sin[c + d*x]) - 64*a^3*Tan[(c + d*x)/ 
2] - 480*a*b^2*Tan[(c + d*x)/2] + 6*a^3*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2 
])/(960*d)
 

Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.09, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.621, Rules used = {3042, 3368, 3042, 3527, 3042, 3526, 25, 3042, 3510, 25, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^2 (a+b \sin (c+d x))^3}{\sin (c+d x)^6}dx\)

\(\Big \downarrow \) 3368

\(\displaystyle \int \left (1-\sin ^2(c+d x)\right ) \csc ^6(c+d x) (a+b \sin (c+d x))^3dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int -\csc ^4(c+d x) (a+b \sin (c+d x)) \left (13 b^2 \sin ^2(c+d x)+11 a b \sin (c+d x)+2 \left (2 a^2-3 b^2\right )\right )dx-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} \left (-\frac {1}{4} \int \csc ^4(c+d x) (a+b \sin (c+d x)) \left (13 b^2 \sin ^2(c+d x)+11 a b \sin (c+d x)+2 \left (2 a^2-3 b^2\right )\right )dx-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (-\frac {1}{4} \int \frac {(a+b \sin (c+d x)) \left (13 b^2 \sin (c+d x)^2+11 a b \sin (c+d x)+2 \left (2 a^2-3 b^2\right )\right )}{\sin (c+d x)^4}dx-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3510

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int -\csc ^3(c+d x) \left (39 \sin ^2(c+d x) b^3+9 \left (5 a^2-2 b^2\right ) b+4 a \left (2 a^2+15 b^2\right ) \sin (c+d x)\right )dx+\frac {2 a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{3 d}\right )-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{3 d}-\frac {1}{3} \int \csc ^3(c+d x) \left (39 \sin ^2(c+d x) b^3+9 \left (5 a^2-2 b^2\right ) b+4 a \left (2 a^2+15 b^2\right ) \sin (c+d x)\right )dx\right )-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{3 d}-\frac {1}{3} \int \frac {39 \sin (c+d x)^2 b^3+9 \left (5 a^2-2 b^2\right ) b+4 a \left (2 a^2+15 b^2\right ) \sin (c+d x)}{\sin (c+d x)^3}dx\right )-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {9 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}-\frac {1}{2} \int \csc ^2(c+d x) \left (8 a \left (2 a^2+15 b^2\right )+15 b \left (3 a^2+4 b^2\right ) \sin (c+d x)\right )dx\right )+\frac {2 a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{3 d}\right )-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {9 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}-\frac {1}{2} \int \frac {8 a \left (2 a^2+15 b^2\right )+15 b \left (3 a^2+4 b^2\right ) \sin (c+d x)}{\sin (c+d x)^2}dx\right )+\frac {2 a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{3 d}\right )-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (-8 a \left (2 a^2+15 b^2\right ) \int \csc ^2(c+d x)dx-15 b \left (3 a^2+4 b^2\right ) \int \csc (c+d x)dx\right )+\frac {9 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{3 d}\right )-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (-15 b \left (3 a^2+4 b^2\right ) \int \csc (c+d x)dx-8 a \left (2 a^2+15 b^2\right ) \int \csc (c+d x)^2dx\right )+\frac {9 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{3 d}\right )-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {8 a \left (2 a^2+15 b^2\right ) \int 1d\cot (c+d x)}{d}-15 b \left (3 a^2+4 b^2\right ) \int \csc (c+d x)dx\right )+\frac {9 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{3 d}\right )-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {8 a \left (2 a^2+15 b^2\right ) \cot (c+d x)}{d}-15 b \left (3 a^2+4 b^2\right ) \int \csc (c+d x)dx\right )+\frac {9 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{3 d}\right )-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {15 b \left (3 a^2+4 b^2\right ) \text {arctanh}(\cos (c+d x))}{d}+\frac {8 a \left (2 a^2+15 b^2\right ) \cot (c+d x)}{d}\right )+\frac {9 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{3 d}\right )-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\right )-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\)

Input:

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

Output:

-1/5*(Cot[c + d*x]*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^3)/d + (((2*a*(2*a^ 
2 - 3*b^2)*Cot[c + d*x]*Csc[c + d*x]^2)/(3*d) + (((15*b*(3*a^2 + 4*b^2)*Ar 
cTanh[Cos[c + d*x]])/d + (8*a*(2*a^2 + 15*b^2)*Cot[c + d*x])/d)/2 + (9*b*( 
5*a^2 - 2*b^2)*Cot[c + d*x]*Csc[c + d*x])/(2*d))/3)/4 - (3*b*Cot[c + d*x]* 
Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2)/(4*d))/5
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25

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

rule 3042

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

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3368

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

rule 3500

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

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 3527

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b 
*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( 
A*d^2*(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, 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 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

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 = 0.50 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{3}}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \cos \left (d x +c \right )^{3}}{15 \sin \left (d x +c \right )^{3}}\right )+3 a^{2} b \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )-\frac {a \,b^{2} \cos \left (d x +c \right )^{3}}{\sin \left (d x +c \right )^{3}}+b^{3} \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(187\)
default	\(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{3}}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \cos \left (d x +c \right )^{3}}{15 \sin \left (d x +c \right )^{3}}\right )+3 a^{2} b \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )-\frac {a \,b^{2} \cos \left (d x +c \right )^{3}}{\sin \left (d x +c \right )^{3}}+b^{3} \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(187\)
risch	\(\frac {16 i a^{3}-45 a^{2} b \,{\mathrm e}^{9 i \left (d x +c \right )}+60 b^{3} {\mathrm e}^{9 i \left (d x +c \right )}+120 i a \,b^{2}-720 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-270 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-120 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-80 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-80 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-240 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+360 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+270 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+120 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+480 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-240 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+45 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-60 b^{3} {\mathrm e}^{i \left (d x +c \right )}}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}\)	\(341\)

Input:

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

Output:

1/d*(a^3*(-1/5/sin(d*x+c)^5*cos(d*x+c)^3-2/15/sin(d*x+c)^3*cos(d*x+c)^3)+3 
*a^2*b*(-1/4/sin(d*x+c)^4*cos(d*x+c)^3-1/8/sin(d*x+c)^2*cos(d*x+c)^3-1/8*c 
os(d*x+c)-1/8*ln(csc(d*x+c)-cot(d*x+c)))-a*b^2/sin(d*x+c)^3*cos(d*x+c)^3+b 
^3*(-1/2/sin(d*x+c)^2*cos(d*x+c)^3-1/2*cos(d*x+c)-1/2*ln(csc(d*x+c)-cot(d* 
x+c))))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.50 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {16 \, {\left (2 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 80 \, {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left ({\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b + 4 \, b^{3} - 2 \, {\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left ({\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b + 4 \, b^{3} - 2 \, {\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left ({\left (3 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:

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

Output:

1/240*(16*(2*a^3 + 15*a*b^2)*cos(d*x + c)^5 - 80*(a^3 + 3*a*b^2)*cos(d*x + 
 c)^3 + 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)*c 
os(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 30*((3*a^2*b - 
4*b^3)*cos(d*x + c)^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. \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.86 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {45 \, a^{2} b {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 60 \, b^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {240 \, a b^{2}}{\tan \left (d x + c\right )^{3}} + \frac {16 \, {\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{3}}{\tan \left (d x + c\right )^{5}}}{240 \, d} \] Input:

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

Output:

-1/240*(45*a^2*b*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*co 
s(d*x + c)^2 + 1) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 60*b^ 
3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) + 1) - log(cos(d 
*x + c) - 1)) + 240*a*b^2/tan(d*x + c)^3 + 16*(5*tan(d*x + c)^2 + 3)*a^3/t 
an(d*x + c)^5)/d
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.58 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 45 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, {\left (3 \, a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {822 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1096 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \] Input:

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

Output:

1/960*(6*a^3*tan(1/2*d*x + 1/2*c)^5 + 45*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 10 
*a^3*tan(1/2*d*x + 1/2*c)^3 + 120*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*b^3*t 
an(1/2*d*x + 1/2*c)^2 - 60*a^3*tan(1/2*d*x + 1/2*c) - 360*a*b^2*tan(1/2*d* 
x + 1/2*c) - 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 + 60*a^3*tan 
(1/2*d*x + 1/2*c)^4 + 360*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 120*b^3*tan(1/2*d 
*x + 1/2*c)^3 - 10*a^3*tan(1/2*d*x + 1/2*c)^2 - 120*a*b^2*tan(1/2*d*x + 1/ 
2*c)^2 - 45*a^2*b*tan(1/2*d*x + 1/2*c) - 6*a^3)/tan(1/2*d*x + 1/2*c)^5)/d
 

Mupad [B] (verification not implemented)

Time = 33.81 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.32 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a^3}{96}+\frac {a\,b^2}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^2\,b}{8}+\frac {b^3}{2}\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (4\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3}{3}+4\,a\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^3+12\,a\,b^2\right )+\frac {a^3}{5}+\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{32\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^3}{16}+\frac {3\,a\,b^2}{8}\right )}{d}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d} \] Input:

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

Output:

(a^3*tan(c/2 + (d*x)/2)^5)/(160*d) + (b^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (t 
an(c/2 + (d*x)/2)^3*((a*b^2)/8 + a^3/96))/d - (log(tan(c/2 + (d*x)/2))*((3 
*a^2*b)/8 + b^3/2))/d - (cot(c/2 + (d*x)/2)^5*(4*b^3*tan(c/2 + (d*x)/2)^3 
+ tan(c/2 + (d*x)/2)^2*(4*a*b^2 + a^3/3) - tan(c/2 + (d*x)/2)^4*(12*a*b^2 
+ 2*a^3) + a^3/5 + (3*a^2*b*tan(c/2 + (d*x)/2))/2))/(32*d) - (tan(c/2 + (d 
*x)/2)*((3*a*b^2)/8 + a^3/16))/d + (3*a^2*b*tan(c/2 + (d*x)/2)^4)/(64*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.13 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{3}+120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}+45 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b -60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{3}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-90 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -24 \cos \left (d x +c \right ) a^{3}-45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} a^{2} b -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} b^{3}}{120 \sin \left (d x +c \right )^{5} d} \] Input:

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

Output:

(16*cos(c + d*x)*sin(c + d*x)**4*a**3 + 120*cos(c + d*x)*sin(c + d*x)**4*a 
*b**2 + 45*cos(c + d*x)*sin(c + d*x)**3*a**2*b - 60*cos(c + d*x)*sin(c + d 
*x)**3*b**3 + 8*cos(c + d*x)*sin(c + d*x)**2*a**3 - 120*cos(c + d*x)*sin(c 
 + d*x)**2*a*b**2 - 90*cos(c + d*x)*sin(c + d*x)*a**2*b - 24*cos(c + d*x)* 
a**3 - 45*log(tan((c + d*x)/2))*sin(c + d*x)**5*a**2*b - 60*log(tan((c + d 
*x)/2))*sin(c + d*x)**5*b**3)/(120*sin(c + d*x)**5*d)