\(\int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1116]

Optimal result

Integrand size = 27, antiderivative size = 231 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3}{8} b \left (12 a^2-b^2\right ) x+\frac {3 a \left (a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 d}-\frac {a \left (a^2-17 b^2\right ) \cos (c+d x)}{2 d}-\frac {b \left (2 a^2-21 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{4 a d}-\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 a^2 d}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d} \] Output:

-3/8*b*(12*a^2-b^2)*x+3/2*a*(a^2-2*b^2)*arctanh(cos(d*x+c))/d-1/2*a*(a^2-1 
7*b^2)*cos(d*x+c)/d-1/8*b*(2*a^2-21*b^2)*cos(d*x+c)*sin(d*x+c)/d-1/4*(a^2- 
6*b^2)*cos(d*x+c)*(a+b*sin(d*x+c))^2/a/d-1/4*(a^2-4*b^2)*cos(d*x+c)*(a+b*s 
in(d*x+c))^3/a^2/d-b*cot(d*x+c)*(a+b*sin(d*x+c))^4/a^2/d-1/2*cot(d*x+c)*cs 
c(d*x+c)*(a+b*sin(d*x+c))^4/a/d
 

Mathematica [A] (verified)

Time = 6.86 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.09 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 b \left (-12 a^2+b^2\right ) (c+d x)}{8 d}-\frac {a \left (4 a^2-15 b^2\right ) \cos (c+d x)}{4 d}+\frac {a b^2 \cos (3 (c+d x))}{4 d}-\frac {3 a^2 b \cot \left (\frac {1}{2} (c+d x)\right )}{2 d}-\frac {a^3 \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {3 \left (a^3-2 a b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {3 \left (a^3-2 a b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {a^3 \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {b \left (-3 a^2+b^2\right ) \sin (2 (c+d x))}{4 d}+\frac {b^3 \sin (4 (c+d x))}{32 d}+\frac {3 a^2 b \tan \left (\frac {1}{2} (c+d x)\right )}{2 d} \] Input:

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

Output:

(3*b*(-12*a^2 + b^2)*(c + d*x))/(8*d) - (a*(4*a^2 - 15*b^2)*Cos[c + d*x])/ 
(4*d) + (a*b^2*Cos[3*(c + d*x)])/(4*d) - (3*a^2*b*Cot[(c + d*x)/2])/(2*d) 
- (a^3*Csc[(c + d*x)/2]^2)/(8*d) + (3*(a^3 - 2*a*b^2)*Log[Cos[(c + d*x)/2] 
])/(2*d) - (3*(a^3 - 2*a*b^2)*Log[Sin[(c + d*x)/2]])/(2*d) + (a^3*Sec[(c + 
 d*x)/2]^2)/(8*d) + (b*(-3*a^2 + b^2)*Sin[2*(c + d*x)])/(4*d) + (b^3*Sin[4 
*(c + d*x)])/(32*d) + (3*a^2*b*Tan[(c + d*x)/2])/(2*d)
 

Rubi [A] (verified)

Time = 1.76 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.10, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.593, Rules used = {3042, 3372, 3042, 3528, 27, 3042, 3528, 3042, 3512, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3372

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

\(\displaystyle -\frac {\frac {3}{2} \left (\frac {1}{3} \int \csc (c+d x) (a+b \sin (c+d x)) \left (13 b \sin (c+d x) a^3-\left (2 a^2-21 b^2\right ) \sin ^2(c+d x) a^2+6 \left (a^2-2 b^2\right ) a^2\right )dx+\frac {a \left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{2 d}}{2 a^2}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3}{2} \left (\frac {1}{3} \int \frac {(a+b \sin (c+d x)) \left (13 b \sin (c+d x) a^3-\left (2 a^2-21 b^2\right ) \sin (c+d x)^2 a^2+6 \left (a^2-2 b^2\right ) a^2\right )}{\sin (c+d x)}dx+\frac {a \left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{2 d}}{2 a^2}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\)

\(\Big \downarrow \) 3512

\(\displaystyle -\frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \int \csc (c+d x) \left (-4 \left (a^2-17 b^2\right ) \sin ^2(c+d x) a^3+12 \left (a^2-2 b^2\right ) a^3+3 b \left (12 a^2-b^2\right ) \sin (c+d x) a^2\right )dx+\frac {a^2 b \left (2 a^2-21 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{2 d}}{2 a^2}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {-4 \left (a^2-17 b^2\right ) \sin (c+d x)^2 a^3+12 \left (a^2-2 b^2\right ) a^3+3 b \left (12 a^2-b^2\right ) \sin (c+d x) a^2}{\sin (c+d x)}dx+\frac {a^2 b \left (2 a^2-21 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{2 d}}{2 a^2}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\)

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (\int 3 \csc (c+d x) \left (4 \left (a^2-2 b^2\right ) a^3+b \left (12 a^2-b^2\right ) \sin (c+d x) a^2\right )dx+\frac {4 a^3 \left (a^2-17 b^2\right ) \cos (c+d x)}{d}\right )+\frac {a^2 b \left (2 a^2-21 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{2 d}}{2 a^2}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \int \csc (c+d x) \left (4 \left (a^2-2 b^2\right ) a^3+b \left (12 a^2-b^2\right ) \sin (c+d x) a^2\right )dx+\frac {4 a^3 \left (a^2-17 b^2\right ) \cos (c+d x)}{d}\right )+\frac {a^2 b \left (2 a^2-21 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{2 d}}{2 a^2}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \int \frac {4 \left (a^2-2 b^2\right ) a^3+b \left (12 a^2-b^2\right ) \sin (c+d x) a^2}{\sin (c+d x)}dx+\frac {4 a^3 \left (a^2-17 b^2\right ) \cos (c+d x)}{d}\right )+\frac {a^2 b \left (2 a^2-21 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{2 d}}{2 a^2}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\)

\(\Big \downarrow \) 3214

\(\displaystyle -\frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (4 a^3 \left (a^2-2 b^2\right ) \int \csc (c+d x)dx+a^2 b x \left (12 a^2-b^2\right )\right )+\frac {4 a^3 \left (a^2-17 b^2\right ) \cos (c+d x)}{d}\right )+\frac {a^2 b \left (2 a^2-21 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{2 d}}{2 a^2}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (4 a^3 \left (a^2-2 b^2\right ) \int \csc (c+d x)dx+a^2 b x \left (12 a^2-b^2\right )\right )+\frac {4 a^3 \left (a^2-17 b^2\right ) \cos (c+d x)}{d}\right )+\frac {a^2 b \left (2 a^2-21 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{2 d}}{2 a^2}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\)

\(\Big \downarrow \) 4257

\(\displaystyle -\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac {3}{2} \left (\frac {a \left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}+\frac {1}{3} \left (\frac {a^2 b \left (2 a^2-21 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} \left (3 \left (a^2 b x \left (12 a^2-b^2\right )-\frac {4 a^3 \left (a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{d}\right )+\frac {4 a^3 \left (a^2-17 b^2\right ) \cos (c+d x)}{d}\right )\right )\right )}{2 a^2}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\)

Input:

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

Output:

-((b*Cot[c + d*x]*(a + b*Sin[c + d*x])^4)/(a^2*d)) - (Cot[c + d*x]*Csc[c + 
 d*x]*(a + b*Sin[c + d*x])^4)/(2*a*d) - (((a^2 - 4*b^2)*Cos[c + d*x]*(a + 
b*Sin[c + d*x])^3)/(2*d) + (3*((a*(a^2 - 6*b^2)*Cos[c + d*x]*(a + b*Sin[c 
+ d*x])^2)/(3*d) + ((3*(a^2*b*(12*a^2 - b^2)*x - (4*a^3*(a^2 - 2*b^2)*ArcT 
anh[Cos[c + d*x]])/d) + (4*a^3*(a^2 - 17*b^2)*Cos[c + d*x])/d)/2 + (a^2*b* 
(2*a^2 - 21*b^2)*Cos[c + d*x]*Sin[c + d*x])/(2*d))/3))/2)/(2*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 3512

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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 3528

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)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 = 3.83 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{5}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{3}}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+3 a^{2} b \left (-\frac {\cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a \,b^{2} \left (\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(198\)
default	\(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{5}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{3}}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+3 a^{2} b \left (-\frac {\cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a \,b^{2} \left (\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(198\)
risch	\(-\frac {9 a^{2} b x}{2}+\frac {3 b^{3} x}{8}+\frac {a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{8 d}+\frac {i b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a^{2} \left (i a \,{\mathrm e}^{3 i \left (d x +c \right )}+i a \,{\mathrm e}^{i \left (d x +c \right )}+6 b \,{\mathrm e}^{2 i \left (d x +c \right )}-6 b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {15 \,{\mathrm e}^{i \left (d x +c \right )} a \,b^{2}}{8 d}-\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {15 \,{\mathrm e}^{-i \left (d x +c \right )} a \,b^{2}}{8 d}+\frac {3 i b \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 d}-\frac {i b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a \,b^{2} {\mathrm e}^{-3 i \left (d x +c \right )}}{8 d}-\frac {3 i b \,{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{d}+\frac {b^{3} \sin \left (4 d x +4 c \right )}{32 d}\)	\(358\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.13 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {8 \, a b^{2} \cos \left (d x + c\right )^{5} - 3 \, {\left (12 \, a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right )^{2} - 8 \, {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (12 \, a^{2} b - b^{3}\right )} d x + 12 \, {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right ) - 6 \, {\left (a^{3} - 2 \, a b^{2} - {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 6 \, {\left (a^{3} - 2 \, a b^{2} - {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (2 \, b^{3} \cos \left (d x + c\right )^{5} - {\left (12 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (12 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)^3*(a+b*sin(d*x+c))^3,x, algorithm="fricas" 
)
 

Output:

1/8*(8*a*b^2*cos(d*x + c)^5 - 3*(12*a^2*b - b^3)*d*x*cos(d*x + c)^2 - 8*(a 
^3 - 2*a*b^2)*cos(d*x + c)^3 + 3*(12*a^2*b - b^3)*d*x + 12*(a^3 - 2*a*b^2) 
*cos(d*x + c) - 6*(a^3 - 2*a*b^2 - (a^3 - 2*a*b^2)*cos(d*x + c)^2)*log(1/2 
*cos(d*x + c) + 1/2) + 6*(a^3 - 2*a*b^2 - (a^3 - 2*a*b^2)*cos(d*x + c)^2)* 
log(-1/2*cos(d*x + c) + 1/2) + (2*b^3*cos(d*x + c)^5 - (12*a^2*b - b^3)*co 
s(d*x + c)^3 + 3*(12*a^2*b - b^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + 
 c)^2 - d)
 

Sympy [F]

\[ \int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \cos {\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx \] Input:

integrate(cos(d*x+c)*cot(d*x+c)**3*(a+b*sin(d*x+c))**3,x)
 

Output:

Integral((a + b*sin(c + d*x))**3*cos(c + d*x)*cot(c + d*x)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {48 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2} b - 16 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a b^{2} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b^{3} - 8 \, a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{32 \, d} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)^3*(a+b*sin(d*x+c))^3,x, algorithm="maxima" 
)
 

Output:

-1/32*(48*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c)^3 + tan(d*x 
+ c)))*a^2*b - 16*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) 
+ 1) + 3*log(cos(d*x + c) - 1))*a*b^2 - (12*d*x + 12*c + sin(4*d*x + 4*c) 
+ 8*sin(2*d*x + 2*c))*b^3 - 8*a^3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - 4 
*cos(d*x + c) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)))/d
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.73 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, {\left (12 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )} - 12 \, {\left (a^{3} - 2 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {18 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {2 \, {\left (12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 48 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 96 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 80 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{3} + 32 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \] Input:

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

Output:

1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^2*b*tan(1/2*d*x + 1/2*c) - 3*(12*a^ 
2*b - b^3)*(d*x + c) - 12*(a^3 - 2*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + 
 (18*a^3*tan(1/2*d*x + 1/2*c)^2 - 36*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 12*a^2 
*b*tan(1/2*d*x + 1/2*c) - a^3)/tan(1/2*d*x + 1/2*c)^2 + 2*(12*a^2*b*tan(1/ 
2*d*x + 1/2*c)^7 - 5*b^3*tan(1/2*d*x + 1/2*c)^7 - 8*a^3*tan(1/2*d*x + 1/2* 
c)^6 + 48*a*b^2*tan(1/2*d*x + 1/2*c)^6 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 
 3*b^3*tan(1/2*d*x + 1/2*c)^5 - 24*a^3*tan(1/2*d*x + 1/2*c)^4 + 96*a*b^2*t 
an(1/2*d*x + 1/2*c)^4 - 12*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 3*b^3*tan(1/2*d* 
x + 1/2*c)^3 - 24*a^3*tan(1/2*d*x + 1/2*c)^2 + 80*a*b^2*tan(1/2*d*x + 1/2* 
c)^2 - 12*a^2*b*tan(1/2*d*x + 1/2*c) + 5*b^3*tan(1/2*d*x + 1/2*c) - 8*a^3 
+ 32*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
 

Mupad [B] (verification not implemented)

Time = 34.05 (sec) , antiderivative size = 718, normalized size of antiderivative = 3.11 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx =\text {Too large to display} \] Input:

int(cos(c + d*x)*cot(c + d*x)^3*(a + b*sin(c + d*x))^3,x)
 

Output:

(a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (tan(c/2 + (d*x)/2)^2*(32*a*b^2 - 10*a^ 
3) + tan(c/2 + (d*x)/2)^8*(48*a*b^2 - (17*a^3)/2) + tan(c/2 + (d*x)/2)^4*( 
80*a*b^2 - 27*a^3) + tan(c/2 + (d*x)/2)^6*(96*a*b^2 - 26*a^3) + tan(c/2 + 
(d*x)/2)^9*(6*a^2*b - 5*b^3) - tan(c/2 + (d*x)/2)^7*(12*a^2*b - 3*b^3) - t 
an(c/2 + (d*x)/2)^3*(36*a^2*b - 5*b^3) - tan(c/2 + (d*x)/2)^5*(48*a^2*b + 
3*b^3) - a^3/2 - 6*a^2*b*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 
16*tan(c/2 + (d*x)/2)^4 + 24*tan(c/2 + (d*x)/2)^6 + 16*tan(c/2 + (d*x)/2)^ 
8 + 4*tan(c/2 + (d*x)/2)^10)) + (log(tan(c/2 + (d*x)/2))*(3*a*b^2 - (3*a^3 
)/2))/d + (3*a^2*b*tan(c/2 + (d*x)/2))/(2*d) + (3*b*atan(((3*b*(12*a^2 - b 
^2)*(tan(c/2 + (d*x)/2)*(6*a*b^2 - 3*a^3) - 9*a^2*b + (3*b^3)/4 - (b*tan(c 
/2 + (d*x)/2)*(12*a^2 - b^2)*9i)/4))/8 + (3*b*(12*a^2 - b^2)*(tan(c/2 + (d 
*x)/2)*(6*a*b^2 - 3*a^3) - 9*a^2*b + (3*b^3)/4 + (b*tan(c/2 + (d*x)/2)*(12 
*a^2 - b^2)*9i)/4))/8)/(2*tan(c/2 + (d*x)/2)*((9*b^6)/16 - (27*a^2*b^4)/2 
+ 81*a^4*b^2) + (9*a*b^5)/2 + 27*a^5*b - (225*a^3*b^3)/4 - (b*(12*a^2 - b^ 
2)*(tan(c/2 + (d*x)/2)*(6*a*b^2 - 3*a^3) - 9*a^2*b + (3*b^3)/4 - (b*tan(c/ 
2 + (d*x)/2)*(12*a^2 - b^2)*9i)/4)*3i)/8 + (b*(12*a^2 - b^2)*(tan(c/2 + (d 
*x)/2)*(6*a*b^2 - 3*a^3) - 9*a^2*b + (3*b^3)/4 + (b*tan(c/2 + (d*x)/2)*(12 
*a^2 - b^2)*9i)/4)*3i)/8))*(12*a^2 - b^2))/(4*d)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.15 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b^{3}-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b +5 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{3}-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}+32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -4 \cos \left (d x +c \right ) a^{3}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{3}+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a \,b^{2}+12 \sin \left (d x +c \right )^{2} a^{3}-36 \sin \left (d x +c \right )^{2} a^{2} b d x -32 \sin \left (d x +c \right )^{2} a \,b^{2}+3 \sin \left (d x +c \right )^{2} b^{3} d x}{8 \sin \left (d x +c \right )^{2} d} \] Input:

int(cos(d*x+c)*cot(d*x+c)^3*(a+b*sin(d*x+c))^3,x)
 

Output:

( - 2*cos(c + d*x)*sin(c + d*x)**5*b**3 - 8*cos(c + d*x)*sin(c + d*x)**4*a 
*b**2 - 12*cos(c + d*x)*sin(c + d*x)**3*a**2*b + 5*cos(c + d*x)*sin(c + d* 
x)**3*b**3 - 8*cos(c + d*x)*sin(c + d*x)**2*a**3 + 32*cos(c + d*x)*sin(c + 
 d*x)**2*a*b**2 - 24*cos(c + d*x)*sin(c + d*x)*a**2*b - 4*cos(c + d*x)*a** 
3 - 12*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**3 + 24*log(tan((c + d*x)/2 
))*sin(c + d*x)**2*a*b**2 + 12*sin(c + d*x)**2*a**3 - 36*sin(c + d*x)**2*a 
**2*b*d*x - 32*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)**2*b**3*d*x)/(8*sin 
(c + d*x)**2*d)