\(\int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1063]

Optimal result

Integrand size = 29, antiderivative size = 148 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a b \text {arctanh}(\cos (c+d x))}{4 d}+\frac {\left (2 a^2+5 b^2\right ) \cot (c+d x)}{15 d}+\frac {a b \cot (c+d x) \csc (c+d x)}{4 d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac {a b \cot (c+d x) \csc ^3(c+d x)}{10 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2}{5 d} \] Output:

1/4*a*b*arctanh(cos(d*x+c))/d+1/15*(2*a^2+5*b^2)*cot(d*x+c)/d+1/4*a*b*cot( 
d*x+c)*csc(d*x+c)/d+1/15*(a^2-2*b^2)*cot(d*x+c)*csc(d*x+c)^2/d-1/10*a*b*co 
t(d*x+c)*csc(d*x+c)^3/d-1/5*cot(d*x+c)*csc(d*x+c)^4*(a+b*sin(d*x+c))^2/d
 

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.59 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\csc ^5(c+d x) \left (-40 \left (4 a^2+b^2\right ) \cos (c+d x)+20 \left (-2 a^2+b^2\right ) \cos (3 (c+d x))+8 a^2 \cos (5 (c+d x))+20 b^2 \cos (5 (c+d x))+150 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-150 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-180 a b \sin (2 (c+d x))-75 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+75 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-30 a b \sin (4 (c+d x))+15 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-15 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{960 d} \] Input:

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

Output:

(Csc[c + d*x]^5*(-40*(4*a^2 + b^2)*Cos[c + d*x] + 20*(-2*a^2 + b^2)*Cos[3* 
(c + d*x)] + 8*a^2*Cos[5*(c + d*x)] + 20*b^2*Cos[5*(c + d*x)] + 150*a*b*Lo 
g[Cos[(c + d*x)/2]]*Sin[c + d*x] - 150*a*b*Log[Sin[(c + d*x)/2]]*Sin[c + d 
*x] - 180*a*b*Sin[2*(c + d*x)] - 75*a*b*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d 
*x)] + 75*a*b*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 30*a*b*Sin[4*(c + d 
*x)] + 15*a*b*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] - 15*a*b*Log[Sin[(c + 
 d*x)/2]]*Sin[5*(c + d*x)]))/(960*d)
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.09, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.586, Rules used = {3042, 3368, 3042, 3527, 3042, 3510, 27, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4255, 3042, 4257}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3368

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3510

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 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 4255

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

rule 4257

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \cos \left (d x +c \right )^{3}}{15 \sin \left (d x +c \right )^{3}}\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )-\frac {b^{2} \cos \left (d x +c \right )^{3}}{3 \sin \left (d x +c \right )^{3}}}{d}\)	\(135\)
default	\(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \cos \left (d x +c \right )^{3}}{15 \sin \left (d x +c \right )^{3}}\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )-\frac {b^{2} \cos \left (d x +c \right )^{3}}{3 \sin \left (d x +c \right )^{3}}}{d}\)	\(135\)
risch	\(-\frac {-60 i b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+15 a b \,{\mathrm e}^{9 i \left (d x +c \right )}+120 i a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+120 i b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+90 a b \,{\mathrm e}^{7 i \left (d x +c \right )}+40 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-80 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+40 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+40 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-90 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-8 i a^{2}-20 i b^{2}-15 a b \,{\mathrm e}^{i \left (d x +c \right )}}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d}+\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d}\)	\(228\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.32 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {8 \, {\left (2 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 40 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:

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

Output:

1/120*(8*(2*a^2 + 5*b^2)*cos(d*x + c)^5 - 40*(a^2 + b^2)*cos(d*x + c)^3 + 
15*(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) - 15*(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*(a*b*cos(d*x + c)^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)*s 
in(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.73 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {15 \, a b {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {40 \, b^{2}}{\tan \left (d x + c\right )^{3}} + \frac {8 \, {\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{2}}{\tan \left (d x + c\right )^{5}}}{120 \, d} \] Input:

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

Output:

-1/120*(15*a*b*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos( 
d*x + c)^2 + 1) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) + 40*b^2/ 
tan(d*x + c)^3 + 8*(5*tan(d*x + c)^2 + 3)*a^2/tan(d*x + c)^5)/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.50 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {274 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 20 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \] Input:

integrate(cot(d*x+c)^2*csc(d*x+c)^4*(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 + 5*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*log(abs 
(tan(1/2*d*x + 1/2*c))) - 30*a^2*tan(1/2*d*x + 1/2*c) - 60*b^2*tan(1/2*d*x 
 + 1/2*c) + (274*a*b*tan(1/2*d*x + 1/2*c)^5 + 30*a^2*tan(1/2*d*x + 1/2*c)^ 
4 + 60*b^2*tan(1/2*d*x + 1/2*c)^4 - 5*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 = 33.17 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.26 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{3}+\frac {4\,b^2}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2+4\,b^2\right )+\frac {a^2}{5}+a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{32\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2}{16}+\frac {b^2}{8}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a^2}{96}+\frac {b^2}{24}\right )}{d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}-\frac {a\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \] Input:

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

Output:

(a^2*tan(c/2 + (d*x)/2)^5)/(160*d) - (cot(c/2 + (d*x)/2)^5*(tan(c/2 + (d*x 
)/2)^2*(a^2/3 + (4*b^2)/3) - tan(c/2 + (d*x)/2)^4*(2*a^2 + 4*b^2) + a^2/5 
+ a*b*tan(c/2 + (d*x)/2)))/(32*d) - (tan(c/2 + (d*x)/2)*(a^2/16 + b^2/8))/ 
d + (tan(c/2 + (d*x)/2)^3*(a^2/96 + b^2/24))/d + (a*b*tan(c/2 + (d*x)/2)^4 
)/(32*d) - (a*b*log(tan(c/2 + (d*x)/2)))/(4*d)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.06 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{2}+20 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{2}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b +4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2}-20 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -12 \cos \left (d x +c \right ) a^{2}-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} a b}{60 \sin \left (d x +c \right )^{5} d} \] Input:

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

Output:

(8*cos(c + d*x)*sin(c + d*x)**4*a**2 + 20*cos(c + d*x)*sin(c + d*x)**4*b** 
2 + 15*cos(c + d*x)*sin(c + d*x)**3*a*b + 4*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 - 15*log(tan((c + d*x)/2))*sin(c + d*x)**5*a*b) 
/(60*sin(c + d*x)**5*d)