\(\int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx\) [1108]

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

Optimal result

Integrand size = 27, antiderivative size = 178 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=2 a b x-\frac {3 \left (a^2-4 b^2\right ) \text {arctanh}(\cos (c+d x))}{8 d}-\frac {b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{24 a^2 d}+\frac {17 a b \cot (c+d x)}{12 d}+\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d} \] Output:

2*a*b*x-3/8*(a^2-4*b^2)*arctanh(cos(d*x+c))/d-1/24*b^2*(39*a^2+2*b^2)*cos( 
d*x+c)/a^2/d+17/12*a*b*cot(d*x+c)/d+5/8*cot(d*x+c)*csc(d*x+c)*(a+b*sin(d*x 
+c))^2/d+1/12*b*cot(d*x+c)*csc(d*x+c)^2*(a+b*sin(d*x+c))^3/a^2/d-1/4*cot(d 
*x+c)*csc(d*x+c)^3*(a+b*sin(d*x+c))^3/a/d
 

Mathematica [A] (verified)

Time = 4.16 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.52 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {384 a b c+384 a b d x-192 b^2 \cos (c+d x)+256 a b \cot \left (\frac {1}{2} (c+d x)\right )+30 a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )-24 b^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )-3 a^2 \csc ^4\left (\frac {1}{2} (c+d x)\right )-72 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+288 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+72 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-288 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-30 a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+24 b^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+3 a^2 \sec ^4\left (\frac {1}{2} (c+d x)\right )+128 a b \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-8 a b \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)-256 a b \tan \left (\frac {1}{2} (c+d x)\right )}{192 d} \] Input:

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

Output:

(384*a*b*c + 384*a*b*d*x - 192*b^2*Cos[c + d*x] + 256*a*b*Cot[(c + d*x)/2] 
 + 30*a^2*Csc[(c + d*x)/2]^2 - 24*b^2*Csc[(c + d*x)/2]^2 - 3*a^2*Csc[(c + 
d*x)/2]^4 - 72*a^2*Log[Cos[(c + d*x)/2]] + 288*b^2*Log[Cos[(c + d*x)/2]] + 
 72*a^2*Log[Sin[(c + d*x)/2]] - 288*b^2*Log[Sin[(c + d*x)/2]] - 30*a^2*Sec 
[(c + d*x)/2]^2 + 24*b^2*Sec[(c + d*x)/2]^2 + 3*a^2*Sec[(c + d*x)/2]^4 + 1 
28*a*b*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 8*a*b*Csc[(c + d*x)/2]^4*Sin[c 
+ d*x] - 256*a*b*Tan[(c + d*x)/2])/(192*d)
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {3042, 3372, 3042, 3526, 3042, 3510, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^4 (a+b \sin (c+d x))^2}{\sin (c+d x)^5}dx\)

\(\Big \downarrow \) 3372

\(\displaystyle -\frac {\int \csc ^3(c+d x) (a+b \sin (c+d x))^2 \left (15 a^2+2 b \sin (c+d x) a-\left (12 a^2+b^2\right ) \sin ^2(c+d x)\right )dx}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {(a+b \sin (c+d x))^2 \left (15 a^2+2 b \sin (c+d x) a-\left (12 a^2+b^2\right ) \sin (c+d x)^2\right )}{\sin (c+d x)^3}dx}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 3526

\(\displaystyle -\frac {\frac {1}{2} \int \csc ^2(c+d x) (a+b \sin (c+d x)) \left (34 b a^2-\left (9 a^2-2 b^2\right ) \sin (c+d x) a-b \left (39 a^2+2 b^2\right ) \sin ^2(c+d x)\right )dx-\frac {15 a^2 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{2} \int \frac {(a+b \sin (c+d x)) \left (34 b a^2-\left (9 a^2-2 b^2\right ) \sin (c+d x) a-b \left (39 a^2+2 b^2\right ) \sin (c+d x)^2\right )}{\sin (c+d x)^2}dx-\frac {15 a^2 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 3510

\(\displaystyle -\frac {\frac {1}{2} \left (-\int \csc (c+d x) \left (48 b \sin (c+d x) a^3+9 \left (a^2-4 b^2\right ) a^2+b^2 \left (39 a^2+2 b^2\right ) \sin ^2(c+d x)\right )dx-\frac {34 a^3 b \cot (c+d x)}{d}\right )-\frac {15 a^2 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{2} \left (-\int \frac {48 b \sin (c+d x) a^3+9 \left (a^2-4 b^2\right ) a^2+b^2 \left (39 a^2+2 b^2\right ) \sin (c+d x)^2}{\sin (c+d x)}dx-\frac {34 a^3 b \cot (c+d x)}{d}\right )-\frac {15 a^2 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {\frac {1}{2} \left (-\int 3 \csc (c+d x) \left (16 b \sin (c+d x) a^3+3 \left (a^2-4 b^2\right ) a^2\right )dx-\frac {34 a^3 b \cot (c+d x)}{d}+\frac {b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{d}\right )-\frac {15 a^2 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {1}{2} \left (-3 \int \csc (c+d x) \left (16 b \sin (c+d x) a^3+3 \left (a^2-4 b^2\right ) a^2\right )dx-\frac {34 a^3 b \cot (c+d x)}{d}+\frac {b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{d}\right )-\frac {15 a^2 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{2} \left (-3 \int \frac {16 b \sin (c+d x) a^3+3 \left (a^2-4 b^2\right ) a^2}{\sin (c+d x)}dx-\frac {34 a^3 b \cot (c+d x)}{d}+\frac {b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{d}\right )-\frac {15 a^2 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 3214

\(\displaystyle -\frac {\frac {1}{2} \left (-3 \left (3 a^2 \left (a^2-4 b^2\right ) \int \csc (c+d x)dx+16 a^3 b x\right )-\frac {34 a^3 b \cot (c+d x)}{d}+\frac {b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{d}\right )-\frac {15 a^2 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{2} \left (-3 \left (3 a^2 \left (a^2-4 b^2\right ) \int \csc (c+d x)dx+16 a^3 b x\right )-\frac {34 a^3 b \cot (c+d x)}{d}+\frac {b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{d}\right )-\frac {15 a^2 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{12 a^2}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\frac {1}{2} \left (-\frac {34 a^3 b \cot (c+d x)}{d}+\frac {b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{d}-3 \left (16 a^3 b x-\frac {3 a^2 \left (a^2-4 b^2\right ) \text {arctanh}(\cos (c+d x))}{d}\right )\right )-\frac {15 a^2 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}}{12 a^2}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\)

Input:

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

Output:

(b*Cot[c + d*x]*Csc[c + d*x]^2*(a + b*Sin[c + d*x])^3)/(12*a^2*d) - (Cot[c 
 + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(4*a*d) - ((-3*(16*a^3*b*x 
- (3*a^2*(a^2 - 4*b^2)*ArcTanh[Cos[c + d*x]])/d) + (b^2*(39*a^2 + 2*b^2)*C 
os[c + d*x])/d - (34*a^3*b*Cot[c + d*x])/d)/2 - (15*a^2*Cot[c + d*x]*Csc[c 
 + d*x]*(a + b*Sin[c + d*x])^2)/(2*d))/(12*a^2)
 

Defintions of rubi rules used

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 = 0.62 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.94

method result size
derivativedivides \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+2 a b \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+b^{2} \left (-\frac {\cos \left (d x +c \right )^{5}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{3}}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\) \(167\)
default \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+2 a b \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+b^{2} \left (-\frac {\cos \left (d x +c \right )^{5}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{3}}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\) \(167\)
risch \(2 a b x -\frac {{\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 d}-\frac {-96 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+15 a^{2} {\mathrm e}^{7 i \left (d x +c \right )}-12 b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+192 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+9 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-160 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+9 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+64 i a b +15 a^{2} {\mathrm e}^{i \left (d x +c \right )}-12 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{2 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{2 d}\) \(299\)

Input:

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

Output:

1/d*(a^2*(-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)))+2*a*b*(-1/3*co 
t(d*x+c)^3+cot(d*x+c)+d*x+c)+b^2*(-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 = 0.14 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.46 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {96 \, a b d x \cos \left (d x + c\right )^{4} - 48 \, b^{2} \cos \left (d x + c\right )^{5} - 192 \, a b d x \cos \left (d x + c\right )^{2} + 96 \, a b d x - 30 \, {\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 18 \, {\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right ) - 9 \, {\left ({\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} - 4 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 9 \, {\left ({\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} - 4 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 32 \, {\left (4 \, a b \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \] Input:

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

Output:

1/48*(96*a*b*d*x*cos(d*x + c)^4 - 48*b^2*cos(d*x + c)^5 - 192*a*b*d*x*cos( 
d*x + c)^2 + 96*a*b*d*x - 30*(a^2 - 4*b^2)*cos(d*x + c)^3 + 18*(a^2 - 4*b^ 
2)*cos(d*x + c) - 9*((a^2 - 4*b^2)*cos(d*x + c)^4 - 2*(a^2 - 4*b^2)*cos(d* 
x + c)^2 + a^2 - 4*b^2)*log(1/2*cos(d*x + c) + 1/2) + 9*((a^2 - 4*b^2)*cos 
(d*x + c)^4 - 2*(a^2 - 4*b^2)*cos(d*x + c)^2 + a^2 - 4*b^2)*log(-1/2*cos(d 
*x + c) + 1/2) - 32*(4*a*b*cos(d*x + c)^3 - 3*a*b*cos(d*x + c))*sin(d*x + 
c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.93 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {32 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a b - 3 \, a^{2} {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 12 \, b^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \] Input:

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

Output:

1/48*(32*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a*b - 3*a^2 
*(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)) + 12*b^2*(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 = 244, normalized size of antiderivative = 1.37 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 384 \, {\left (d x + c\right )} a b - 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, {\left (a^{2} - 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {384 \, b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - \frac {150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 600 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \] Input:

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

Output:

1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 + 16*a*b*tan(1/2*d*x + 1/2*c)^3 - 24*a 
^2*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*tan(1/2*d*x + 1/2*c)^2 + 384*(d*x + c)* 
a*b - 240*a*b*tan(1/2*d*x + 1/2*c) + 72*(a^2 - 4*b^2)*log(abs(tan(1/2*d*x 
+ 1/2*c))) - 384*b^2/(tan(1/2*d*x + 1/2*c)^2 + 1) - (150*a^2*tan(1/2*d*x + 
 1/2*c)^4 - 600*b^2*tan(1/2*d*x + 1/2*c)^4 - 240*a*b*tan(1/2*d*x + 1/2*c)^ 
3 - 24*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*tan(1/2*d*x + 1/2*c)^2 + 16*a*b 
*tan(1/2*d*x + 1/2*c) + 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d
 

Mupad [B] (verification not implemented)

Time = 35.28 (sec) , antiderivative size = 825, normalized size of antiderivative = 4.63 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Too large to display} \] Input:

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

Output:

-(3*a^2*cos(c/2 + (d*x)/2)^10 - 3*a^2*sin(c/2 + (d*x)/2)^10 + 21*a^2*cos(c 
/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 + 24*a^2*cos(c/2 + (d*x)/2)^4*sin(c/2 
 + (d*x)/2)^6 - 24*a^2*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 - 21*a^2* 
cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 - 24*b^2*cos(c/2 + (d*x)/2)^2*si 
n(c/2 + (d*x)/2)^8 - 24*b^2*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 40 
8*b^2*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 24*b^2*cos(c/2 + (d*x)/2 
)^8*sin(c/2 + (d*x)/2)^2 - 16*a*b*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^9 
+ 16*a*b*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2) - 72*a^2*log(sin(c/2 + (d 
*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 - 72* 
a^2*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)^4 + 288*b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/ 
2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 288*b^2*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)^4 + 224*a*b*cos(c/2 
 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^7 - 224*a*b*cos(c/2 + (d*x)/2)^7*sin(c/2 
+ (d*x)/2)^3 + 768*a*b*atan((3*a^2*sin(c/2 + (d*x)/2) - 12*b^2*sin(c/2 + ( 
d*x)/2) + 16*a*b*cos(c/2 + (d*x)/2))/(12*b^2*cos(c/2 + (d*x)/2) - 3*a^2*co 
s(c/2 + (d*x)/2) + 16*a*b*sin(c/2 + (d*x)/2)))*cos(c/2 + (d*x)/2)^4*sin(c/ 
2 + (d*x)/2)^6 + 768*a*b*atan((3*a^2*sin(c/2 + (d*x)/2) - 12*b^2*sin(c/2 + 
 (d*x)/2) + 16*a*b*cos(c/2 + (d*x)/2))/(12*b^2*cos(c/2 + (d*x)/2) - 3*a^2* 
cos(c/2 + (d*x)/2) + 16*a*b*sin(c/2 + (d*x)/2)))*cos(c/2 + (d*x)/2)^6*s...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.13 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{2}+64 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b +15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2}-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-16 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -6 \cos \left (d x +c \right ) a^{2}+9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a^{2}-36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} b^{2}-3 \sin \left (d x +c \right )^{4} a^{2}+48 \sin \left (d x +c \right )^{4} a b d x +27 \sin \left (d x +c \right )^{4} b^{2}}{24 \sin \left (d x +c \right )^{4} d} \] Input:

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

Output:

( - 24*cos(c + d*x)*sin(c + d*x)**4*b**2 + 64*cos(c + d*x)*sin(c + d*x)**3 
*a*b + 15*cos(c + d*x)*sin(c + d*x)**2*a**2 - 12*cos(c + d*x)*sin(c + d*x) 
**2*b**2 - 16*cos(c + d*x)*sin(c + d*x)*a*b - 6*cos(c + d*x)*a**2 + 9*log( 
tan((c + d*x)/2))*sin(c + d*x)**4*a**2 - 36*log(tan((c + d*x)/2))*sin(c + 
d*x)**4*b**2 - 3*sin(c + d*x)**4*a**2 + 48*sin(c + d*x)**4*a*b*d*x + 27*si 
n(c + d*x)**4*b**2)/(24*sin(c + d*x)**4*d)