\(\int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1109]

Optimal result

Integrand size = 29, antiderivative size = 209 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=b^2 x-\frac {3 a b \text {arctanh}(\cos (c+d x))}{4 d}-\frac {\left (3 a^4-14 a^2 b^2+b^4\right ) \cot (c+d x)}{15 a^2 d}+\frac {b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{60 a d}+\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{30 a^2 d}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d} \] Output:

b^2*x-3/4*a*b*arctanh(cos(d*x+c))/d-1/15*(3*a^4-14*a^2*b^2+b^4)*cot(d*x+c) 
/a^2/d+1/60*b*(27*a^2-2*b^2)*cot(d*x+c)*csc(d*x+c)/a/d+1/30*(12*a^2-b^2)*c 
ot(d*x+c)*csc(d*x+c)^2*(a+b*sin(d*x+c))^2/a^2/d+1/10*b*cot(d*x+c)*csc(d*x+ 
c)^3*(a+b*sin(d*x+c))^3/a^2/d-1/5*cot(d*x+c)*csc(d*x+c)^4*(a+b*sin(d*x+c)) 
^3/a/d
 

Mathematica [A] (verified)

Time = 2.81 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.36 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {960 b^2 c+960 b^2 d x+\left (-96 a^2+640 b^2\right ) \cot \left (\frac {1}{2} (c+d x)\right )+300 a b \csc ^2\left (\frac {1}{2} (c+d x)\right )-720 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+720 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-300 a b \sec ^2\left (\frac {1}{2} (c+d x)\right )+30 a b \sec ^4\left (\frac {1}{2} (c+d x)\right )-336 a^2 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+320 b^2 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+192 a^2 \csc ^5(c+d x) \sin ^6\left (\frac {1}{2} (c+d x)\right )-3 a^2 \csc ^6\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)+\csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (-30 a b+\left (21 a^2-20 b^2\right ) \sin (c+d x)\right )+96 a^2 \tan \left (\frac {1}{2} (c+d x)\right )-640 b^2 \tan \left (\frac {1}{2} (c+d x)\right )}{960 d} \] Input:

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

Output:

(960*b^2*c + 960*b^2*d*x + (-96*a^2 + 640*b^2)*Cot[(c + d*x)/2] + 300*a*b* 
Csc[(c + d*x)/2]^2 - 720*a*b*Log[Cos[(c + d*x)/2]] + 720*a*b*Log[Sin[(c + 
d*x)/2]] - 300*a*b*Sec[(c + d*x)/2]^2 + 30*a*b*Sec[(c + d*x)/2]^4 - 336*a^ 
2*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 320*b^2*Csc[c + d*x]^3*Sin[(c + d*x) 
/2]^4 + 192*a^2*Csc[c + d*x]^5*Sin[(c + d*x)/2]^6 - 3*a^2*Csc[(c + d*x)/2] 
^6*Sin[c + d*x] + Csc[(c + d*x)/2]^4*(-30*a*b + (21*a^2 - 20*b^2)*Sin[c + 
d*x]) + 96*a^2*Tan[(c + d*x)/2] - 640*b^2*Tan[(c + d*x)/2])/(960*d)
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.483, Rules used = {3042, 3372, 27, 3042, 3526, 3042, 3510, 3042, 3500, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^4 (a+b \sin (c+d x))^2}{\sin (c+d x)^6}dx\)

\(\Big \downarrow \) 3372

\(\displaystyle -\frac {\int 2 \csc ^4(c+d x) (a+b \sin (c+d x))^2 \left (-10 \sin ^2(c+d x) a^2+12 a^2+b \sin (c+d x) a-b^2\right )dx}{20 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \csc ^4(c+d x) (a+b \sin (c+d x))^2 \left (-10 \sin ^2(c+d x) a^2+12 a^2+b \sin (c+d x) a-b^2\right )dx}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {(a+b \sin (c+d x))^2 \left (-10 \sin (c+d x)^2 a^2+12 a^2+b \sin (c+d x) a-b^2\right )}{\sin (c+d x)^4}dx}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 3526

\(\displaystyle -\frac {\frac {1}{3} \int \csc ^3(c+d x) (a+b \sin (c+d x)) \left (-30 a^2 b \sin ^2(c+d x)-a \left (6 a^2-b^2\right ) \sin (c+d x)+b \left (27 a^2-2 b^2\right )\right )dx-\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{3} \int \frac {(a+b \sin (c+d x)) \left (-30 a^2 b \sin (c+d x)^2-a \left (6 a^2-b^2\right ) \sin (c+d x)+b \left (27 a^2-2 b^2\right )\right )}{\sin (c+d x)^3}dx-\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 3510

\(\displaystyle -\frac {\frac {1}{3} \left (-\frac {1}{2} \int \csc ^2(c+d x) \left (45 b \sin (c+d x) a^3+60 b^2 \sin ^2(c+d x) a^2+4 \left (3 a^4-14 b^2 a^2+b^4\right )\right )dx-\frac {a b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{3} \left (-\frac {1}{2} \int \frac {45 b \sin (c+d x) a^3+60 b^2 \sin (c+d x)^2 a^2+4 \left (3 a^4-14 b^2 a^2+b^4\right )}{\sin (c+d x)^2}dx-\frac {a b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 3500

\(\displaystyle -\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {4 \left (3 a^4-14 a^2 b^2+b^4\right ) \cot (c+d x)}{d}-\int 15 \csc (c+d x) \left (3 b a^3+4 b^2 \sin (c+d x) a^2\right )dx\right )-\frac {a b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {4 \left (3 a^4-14 a^2 b^2+b^4\right ) \cot (c+d x)}{d}-15 \int \csc (c+d x) \left (3 b a^3+4 b^2 \sin (c+d x) a^2\right )dx\right )-\frac {a b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {4 \left (3 a^4-14 a^2 b^2+b^4\right ) \cot (c+d x)}{d}-15 \int \frac {3 b a^3+4 b^2 \sin (c+d x) a^2}{\sin (c+d x)}dx\right )-\frac {a b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 3214

\(\displaystyle -\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {4 \left (3 a^4-14 a^2 b^2+b^4\right ) \cot (c+d x)}{d}-15 \left (3 a^3 b \int \csc (c+d x)dx+4 a^2 b^2 x\right )\right )-\frac {a b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {4 \left (3 a^4-14 a^2 b^2+b^4\right ) \cot (c+d x)}{d}-15 \left (3 a^3 b \int \csc (c+d x)dx+4 a^2 b^2 x\right )\right )-\frac {a b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}}{10 a^2}+\frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {4 \left (3 a^4-14 a^2 b^2+b^4\right ) \cot (c+d x)}{d}-15 \left (4 a^2 b^2 x-\frac {3 a^3 b \text {arctanh}(\cos (c+d x))}{d}\right )\right )-\frac {a b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}}{10 a^2}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\)

Input:

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

Output:

(b*Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(10*a^2*d) - (Cot[c 
 + d*x]*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^3)/(5*a*d) - (((-15*(4*a^2*b^2 
*x - (3*a^3*b*ArcTanh[Cos[c + d*x]])/d) + (4*(3*a^4 - 14*a^2*b^2 + b^4)*Co 
t[c + d*x])/d)/2 - (a*b*(27*a^2 - 2*b^2)*Cot[c + d*x]*Csc[c + d*x])/(2*d)) 
/3 - ((12*a^2 - b^2)*Cot[c + d*x]*Csc[c + d*x]^2*(a + b*Sin[c + d*x])^2)/( 
3*d))/(10*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 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 4257

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.62

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{5}}{5 \sin \left (d x +c \right )^{5}}+2 a b \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+b^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(129\)
default	\(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{5}}{5 \sin \left (d x +c \right )^{5}}+2 a b \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+b^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(129\)
risch	\(b^{2} x -\frac {60 i a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-120 i b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+75 a b \,{\mathrm e}^{9 i \left (d x +c \right )}+360 i b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-30 a b \,{\mathrm e}^{7 i \left (d x +c \right )}+120 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-440 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+280 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+30 a b \,{\mathrm e}^{3 i \left (d x +c \right )}+12 i a^{2}-80 i b^{2}-75 a b \,{\mathrm e}^{i \left (d x +c \right )}}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d}\)	\(218\)

Input:

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

Output:

1/d*(-1/5*a^2/sin(d*x+c)^5*cos(d*x+c)^5+2*a*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(c 
sc(d*x+c)-cot(d*x+c)))+b^2*(-1/3*cot(d*x+c)^3+cot(d*x+c)+d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.15 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {8 \, {\left (3 \, a^{2} - 20 \, b^{2}\right )} \cos \left (d x + c\right )^{5} + 280 \, b^{2} \cos \left (d x + c\right )^{3} - 120 \, b^{2} \cos \left (d x + c\right ) + 45 \, {\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 45 \, {\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (4 \, b^{2} d x \cos \left (d x + c\right )^{4} - 8 \, b^{2} d x \cos \left (d x + c\right )^{2} - 5 \, a b \cos \left (d x + c\right )^{3} + 4 \, b^{2} d x + 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:

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

Output:

-1/120*(8*(3*a^2 - 20*b^2)*cos(d*x + c)^5 + 280*b^2*cos(d*x + c)^3 - 120*b 
^2*cos(d*x + c) + 45*(a*b*cos(d*x + c)^4 - 2*a*b*cos(d*x + c)^2 + a*b)*log 
(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 45*(a*b*cos(d*x + c)^4 - 2*a*b*cos 
(d*x + c)^2 + a*b)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 30*(4*b^2*d 
*x*cos(d*x + c)^4 - 8*b^2*d*x*cos(d*x + c)^2 - 5*a*b*cos(d*x + c)^3 + 4*b^ 
2*d*x + 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)*sin(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.59 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {24 \, a^{2} \cot \left (d x + c\right )^{5} - 40 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} b^{2} + 15 \, a b {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \] Input:

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

Output:

-1/120*(24*a^2*cot(d*x + c)^5 - 40*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/t 
an(d*x + c)^3)*b^2 + 15*a*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)))/d
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.26 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 480 \, {\left (d x + c\right )} b^{2} + 360 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 300 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {822 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 300 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \] Input:

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

Output:

1/480*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 15*a*b*tan(1/2*d*x + 1/2*c)^4 - 15*a 
^2*tan(1/2*d*x + 1/2*c)^3 + 20*b^2*tan(1/2*d*x + 1/2*c)^3 - 120*a*b*tan(1/ 
2*d*x + 1/2*c)^2 + 480*(d*x + c)*b^2 + 360*a*b*log(abs(tan(1/2*d*x + 1/2*c 
))) + 30*a^2*tan(1/2*d*x + 1/2*c) - 300*b^2*tan(1/2*d*x + 1/2*c) - (822*a* 
b*tan(1/2*d*x + 1/2*c)^5 + 30*a^2*tan(1/2*d*x + 1/2*c)^4 - 300*b^2*tan(1/2 
*d*x + 1/2*c)^4 - 120*a*b*tan(1/2*d*x + 1/2*c)^3 - 15*a^2*tan(1/2*d*x + 1/ 
2*c)^2 + 20*b^2*tan(1/2*d*x + 1/2*c)^2 + 15*a*b*tan(1/2*d*x + 1/2*c) + 3*a 
^2)/tan(1/2*d*x + 1/2*c)^5)/d
 

Mupad [B] (verification not implemented)

Time = 34.52 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.66 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32\,d}-\frac {a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {b^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a^2\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}+\frac {5\,b^2\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {5\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {2\,b^2\,\mathrm {atan}\left (\frac {4\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+3\,a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a\,b\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {a\,b\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}+\frac {3\,a\,b\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d} \] Input:

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

Output:

(a^2*cot(c/2 + (d*x)/2)^3)/(32*d) - (a^2*cot(c/2 + (d*x)/2)^5)/(160*d) - ( 
b^2*cot(c/2 + (d*x)/2)^3)/(24*d) - (a^2*tan(c/2 + (d*x)/2)^3)/(32*d) + (a^ 
2*tan(c/2 + (d*x)/2)^5)/(160*d) + (b^2*tan(c/2 + (d*x)/2)^3)/(24*d) - (a^2 
*cot(c/2 + (d*x)/2))/(16*d) + (5*b^2*cot(c/2 + (d*x)/2))/(8*d) + (a^2*tan( 
c/2 + (d*x)/2))/(16*d) - (5*b^2*tan(c/2 + (d*x)/2))/(8*d) + (2*b^2*atan((4 
*b*cos(c/2 + (d*x)/2) + 3*a*sin(c/2 + (d*x)/2))/(3*a*cos(c/2 + (d*x)/2) - 
4*b*sin(c/2 + (d*x)/2))))/d + (a*b*cot(c/2 + (d*x)/2)^2)/(4*d) - (a*b*cot( 
c/2 + (d*x)/2)^4)/(32*d) - (a*b*tan(c/2 + (d*x)/2)^2)/(4*d) + (a*b*tan(c/2 
 + (d*x)/2)^4)/(32*d) + (3*a*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))) 
/(4*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.82 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{2}+80 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{2}+75 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b +24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2}-20 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -12 \cos \left (d x +c \right ) a^{2}+45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} a b +60 \sin \left (d x +c \right )^{5} b^{2} d x}{60 \sin \left (d x +c \right )^{5} d} \] Input:

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

Output:

( - 12*cos(c + d*x)*sin(c + d*x)**4*a**2 + 80*cos(c + d*x)*sin(c + d*x)**4 
*b**2 + 75*cos(c + d*x)*sin(c + d*x)**3*a*b + 24*cos(c + d*x)*sin(c + d*x) 
**2*a**2 - 20*cos(c + d*x)*sin(c + d*x)**2*b**2 - 30*cos(c + d*x)*sin(c + 
d*x)*a*b - 12*cos(c + d*x)*a**2 + 45*log(tan((c + d*x)/2))*sin(c + d*x)**5 
*a*b + 60*sin(c + d*x)**5*b**2*d*x)/(60*sin(c + d*x)**5*d)