\(\int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1115]

Optimal result

Integrand size = 29, antiderivative size = 229 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3}{8} a \left (4 a^2-3 b^2\right ) x-\frac {3 a^2 b \text {arctanh}(\cos (c+d x))}{d}+\frac {\left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{10 b d}+\frac {a \left (2 a^2+83 b^2\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {\left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{20 b d}+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{20 a b d}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d} \] Output:

-3/8*a*(4*a^2-3*b^2)*x-3*a^2*b*arctanh(cos(d*x+c))/d+1/10*(a^4+56*a^2*b^2- 
2*b^4)*cos(d*x+c)/b/d+1/40*a*(2*a^2+83*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/20*( 
a^2+28*b^2)*cos(d*x+c)*(a+b*sin(d*x+c))^2/b/d+1/20*(a^2+20*b^2)*cos(d*x+c) 
*(a+b*sin(d*x+c))^3/a/b/d-1/5*cos(d*x+c)*(a+b*sin(d*x+c))^4/b/d-cot(d*x+c) 
*(a+b*sin(d*x+c))^4/a/d
 

Mathematica [A] (verified)

Time = 2.82 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.85 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-240 a^3 c+180 a b^2 c-240 a^3 d x+180 a b^2 d x-20 b \left (-30 a^2+b^2\right ) \cos (c+d x)+10 \left (4 a^2 b-b^3\right ) \cos (3 (c+d x))-2 b^3 \cos (5 (c+d x))-80 a^3 \cot \left (\frac {1}{2} (c+d x)\right )-480 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+480 a^2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-40 a^3 \sin (2 (c+d x))+120 a b^2 \sin (2 (c+d x))+15 a b^2 \sin (4 (c+d x))+80 a^3 \tan \left (\frac {1}{2} (c+d x)\right )}{160 d} \] Input:

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

Output:

(-240*a^3*c + 180*a*b^2*c - 240*a^3*d*x + 180*a*b^2*d*x - 20*b*(-30*a^2 + 
b^2)*Cos[c + d*x] + 10*(4*a^2*b - b^3)*Cos[3*(c + d*x)] - 2*b^3*Cos[5*(c + 
 d*x)] - 80*a^3*Cot[(c + d*x)/2] - 480*a^2*b*Log[Cos[(c + d*x)/2]] + 480*a 
^2*b*Log[Sin[(c + d*x)/2]] - 40*a^3*Sin[2*(c + d*x)] + 120*a*b^2*Sin[2*(c 
+ d*x)] + 15*a*b^2*Sin[4*(c + d*x)] + 80*a^3*Tan[(c + d*x)/2])/(160*d)
 

Rubi [A] (verified)

Time = 1.72 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.586, Rules used = {3042, 3373, 25, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^4 (a+b \sin (c+d x))^3}{\sin (c+d x)^2}dx\)

\(\Big \downarrow \) 3373

\(\displaystyle -\frac {\int -\csc (c+d x) (a+b \sin (c+d x))^3 \left (15 b^2-6 a \sin (c+d x) b-\left (a^2+20 b^2\right ) \sin ^2(c+d x)\right )dx}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \csc (c+d x) (a+b \sin (c+d x))^3 \left (15 b^2-6 a \sin (c+d x) b-\left (a^2+20 b^2\right ) \sin ^2(c+d x)\right )dx}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(a+b \sin (c+d x))^3 \left (15 b^2-6 a \sin (c+d x) b-\left (a^2+20 b^2\right ) \sin (c+d x)^2\right )}{\sin (c+d x)}dx}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3528

\(\displaystyle \frac {\frac {1}{4} \int 3 \csc (c+d x) (a+b \sin (c+d x))^2 \left (-9 b \sin (c+d x) a^2+20 b^2 a-\left (a^2+28 b^2\right ) \sin ^2(c+d x) a\right )dx+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3}{4} \int \csc (c+d x) (a+b \sin (c+d x))^2 \left (-9 b \sin (c+d x) a^2+20 b^2 a-\left (a^2+28 b^2\right ) \sin ^2(c+d x) a\right )dx+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \int \frac {(a+b \sin (c+d x))^2 \left (-9 b \sin (c+d x) a^2+20 b^2 a-\left (a^2+28 b^2\right ) \sin (c+d x)^2 a\right )}{\sin (c+d x)}dx+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3528

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{3} \int \csc (c+d x) (a+b \sin (c+d x)) \left (60 a^2 b^2-a \left (29 a^2-4 b^2\right ) \sin (c+d x) b-a^2 \left (2 a^2+83 b^2\right ) \sin ^2(c+d x)\right )dx+\frac {a \left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{3} \int \frac {(a+b \sin (c+d x)) \left (60 a^2 b^2-a \left (29 a^2-4 b^2\right ) \sin (c+d x) b-a^2 \left (2 a^2+83 b^2\right ) \sin (c+d x)^2\right )}{\sin (c+d x)}dx+\frac {a \left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3512

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \csc (c+d x) \left (120 b^2 a^3-15 b \left (4 a^2-3 b^2\right ) \sin (c+d x) a^2-4 \left (a^4+56 b^2 a^2-2 b^4\right ) \sin ^2(c+d x) a\right )dx+\frac {a^2 b \left (2 a^2+83 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {120 b^2 a^3-15 b \left (4 a^2-3 b^2\right ) \sin (c+d x) a^2-4 \left (a^4+56 b^2 a^2-2 b^4\right ) \sin (c+d x)^2 a}{\sin (c+d x)}dx+\frac {a^2 b \left (2 a^2+83 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\int 15 \csc (c+d x) \left (8 a^3 b^2-a^2 b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )dx+\frac {4 a \left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{d}\right )+\frac {a^2 b \left (2 a^2+83 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \int \csc (c+d x) \left (8 a^3 b^2-a^2 b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )dx+\frac {4 a \left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{d}\right )+\frac {a^2 b \left (2 a^2+83 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \int \frac {8 a^3 b^2-a^2 b \left (4 a^2-3 b^2\right ) \sin (c+d x)}{\sin (c+d x)}dx+\frac {4 a \left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{d}\right )+\frac {a^2 b \left (2 a^2+83 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \left (8 a^3 b^2 \int \csc (c+d x)dx-a^2 b x \left (4 a^2-3 b^2\right )\right )+\frac {4 a \left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{d}\right )+\frac {a^2 b \left (2 a^2+83 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \left (8 a^3 b^2 \int \csc (c+d x)dx-a^2 b x \left (4 a^2-3 b^2\right )\right )+\frac {4 a \left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{d}\right )+\frac {a^2 b \left (2 a^2+83 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a \left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}\right )+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 d}+\frac {3}{4} \left (\frac {a \left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{3 d}+\frac {1}{3} \left (\frac {a^2 b \left (2 a^2+83 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} \left (\frac {4 a \left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{d}+15 \left (-\frac {8 a^3 b^2 \text {arctanh}(\cos (c+d x))}{d}-a^2 b x \left (4 a^2-3 b^2\right )\right )\right )\right )\right )}{5 a b}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\)

Input:

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

Output:

-1/5*(Cos[c + d*x]*(a + b*Sin[c + d*x])^4)/(b*d) - (Cot[c + d*x]*(a + b*Si 
n[c + d*x])^4)/(a*d) + (((a^2 + 20*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^ 
3)/(4*d) + (3*((a*(a^2 + 28*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(3*d 
) + ((15*(-(a^2*b*(4*a^2 - 3*b^2)*x) - (8*a^3*b^2*ArcTanh[Cos[c + d*x]])/d 
) + (4*a*(a^4 + 56*a^2*b^2 - 2*b^4)*Cos[c + d*x])/d)/2 + (a^2*b*(2*a^2 + 8 
3*b^2)*Cos[c + d*x]*Sin[c + d*x])/(2*d))/3))/4)/(5*a*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 3373

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[Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 2)/(b*d 
^2*f*(m + n + 4))), x] + Simp[1/(a*b*d*(n + 1)*(m + n + 4))   Int[(a + b*Si 
n[e + f*x])^m*(d*Sin[e + f*x])^(n + 1)*Simp[a^2*(n + 1)*(n + 2) - b^2*(m + 
n + 2)*(m + n + 4) + a*b*(m + 3)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2* 
(m + n + 3)*(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] && NeQ[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 = 6.93 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{2} b \left (\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {b^{3} \cos \left (d x +c \right )^{5}}{5}}{d}\)	\(152\)
default	\(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{2} b \left (\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {b^{3} \cos \left (d x +c \right )^{5}}{5}}{d}\)	\(152\)
risch	\(-\frac {3 a^{3} x}{2}+\frac {9 a \,b^{2} x}{8}-\frac {2 i a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} a \,b^{2}}{8 d}+\frac {15 b \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{8 d}-\frac {b^{3} {\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {15 b \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 d}-\frac {b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{16 d}-\frac {i a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )} a \,b^{2}}{8 d}+\frac {i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {b^{3} \cos \left (5 d x +5 c \right )}{80 d}+\frac {3 a \,b^{2} \sin \left (4 d x +4 c \right )}{32 d}+\frac {b \cos \left (3 d x +3 c \right ) a^{2}}{4 d}-\frac {b^{3} \cos \left (3 d x +3 c \right )}{16 d}\)	\(293\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {30 \, a b^{2} \cos \left (d x + c\right )^{5} + 60 \, a^{2} b \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 60 \, a^{2} b \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 5 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right ) + {\left (8 \, b^{3} \cos \left (d x + c\right )^{5} - 40 \, a^{2} b \cos \left (d x + c\right )^{3} - 120 \, a^{2} b \cos \left (d x + c\right ) + 15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} d x\right )} \sin \left (d x + c\right )}{40 \, d \sin \left (d x + c\right )} \] Input:

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

Output:

-1/40*(30*a*b^2*cos(d*x + c)^5 + 60*a^2*b*log(1/2*cos(d*x + c) + 1/2)*sin( 
d*x + c) - 60*a^2*b*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 5*(4*a^3 - 
 3*a*b^2)*cos(d*x + c)^3 + 15*(4*a^3 - 3*a*b^2)*cos(d*x + c) + (8*b^3*cos( 
d*x + c)^5 - 40*a^2*b*cos(d*x + c)^3 - 120*a^2*b*cos(d*x + c) + 15*(4*a^3 
- 3*a*b^2)*d*x)*sin(d*x + c))/(d*sin(d*x + c))
 

Sympy [F]

\[ \int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \cos ^{2}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx \] Input:

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

Output:

Integral((a + b*sin(c + d*x))**3*cos(c + d*x)**2*cot(c + d*x)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.62 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {32 \, b^{3} \cos \left (d x + c\right )^{5} + 80 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{3} - 80 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} b - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2}}{160 \, d} \] Input:

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

Output:

-1/160*(32*b^3*cos(d*x + c)^5 + 80*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)/( 
tan(d*x + c)^3 + tan(d*x + c)))*a^3 - 80*(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^2*b - 15*(12*d*x 
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a*b^2)/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.51 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {120 \, a^{2} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )} - \frac {20 \, {\left (6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 240 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 40 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 40 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 880 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 80 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 40 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 560 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 160 \, a^{2} b - 8 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{40 \, d} \] Input:

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

Output:

1/40*(120*a^2*b*log(abs(tan(1/2*d*x + 1/2*c))) + 20*a^3*tan(1/2*d*x + 1/2* 
c) - 15*(4*a^3 - 3*a*b^2)*(d*x + c) - 20*(6*a^2*b*tan(1/2*d*x + 1/2*c) + a 
^3)/tan(1/2*d*x + 1/2*c) + 2*(20*a^3*tan(1/2*d*x + 1/2*c)^9 - 75*a*b^2*tan 
(1/2*d*x + 1/2*c)^9 + 240*a^2*b*tan(1/2*d*x + 1/2*c)^8 - 40*b^3*tan(1/2*d* 
x + 1/2*c)^8 + 40*a^3*tan(1/2*d*x + 1/2*c)^7 - 30*a*b^2*tan(1/2*d*x + 1/2* 
c)^7 + 720*a^2*b*tan(1/2*d*x + 1/2*c)^6 + 880*a^2*b*tan(1/2*d*x + 1/2*c)^4 
 - 80*b^3*tan(1/2*d*x + 1/2*c)^4 - 40*a^3*tan(1/2*d*x + 1/2*c)^3 + 30*a*b^ 
2*tan(1/2*d*x + 1/2*c)^3 + 560*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 20*a^3*tan(1 
/2*d*x + 1/2*c) + 75*a*b^2*tan(1/2*d*x + 1/2*c) + 160*a^2*b - 8*b^3)/(tan( 
1/2*d*x + 1/2*c)^2 + 1)^5)/d
 

Mupad [B] (verification not implemented)

Time = 34.10 (sec) , antiderivative size = 674, normalized size of antiderivative = 2.94 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx =\text {Too large to display} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)*(16*a^2*b - (4*b^3)/5) - tan(c/2 + (d*x)/2)^8*(3*a*b^2 
 + a^3) - 10*a^3*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^4*(3*a*b^2 - 14 
*a^3) + tan(c/2 + (d*x)/2)^2*((15*a*b^2)/2 - 7*a^3) - tan(c/2 + (d*x)/2)^1 
0*((15*a*b^2)/2 - a^3) + tan(c/2 + (d*x)/2)^9*(24*a^2*b - 4*b^3) + tan(c/2 
 + (d*x)/2)^5*(88*a^2*b - 8*b^3) - a^3 + 56*a^2*b*tan(c/2 + (d*x)/2)^3 + 7 
2*a^2*b*tan(c/2 + (d*x)/2)^7)/(d*(2*tan(c/2 + (d*x)/2) + 10*tan(c/2 + (d*x 
)/2)^3 + 20*tan(c/2 + (d*x)/2)^5 + 20*tan(c/2 + (d*x)/2)^7 + 10*tan(c/2 + 
(d*x)/2)^9 + 2*tan(c/2 + (d*x)/2)^11)) + (a^3*tan(c/2 + (d*x)/2))/(2*d) + 
(3*a*atan(((3*a*(4*a^2 - 3*b^2)*((9*a*b^2)/4 - 3*a^3 - (a*tan(c/2 + (d*x)/ 
2)*(4*a^2 - 3*b^2)*9i)/4 + 6*a^2*b*tan(c/2 + (d*x)/2)))/8 + (3*a*(4*a^2 - 
3*b^2)*((9*a*b^2)/4 - 3*a^3 + (a*tan(c/2 + (d*x)/2)*(4*a^2 - 3*b^2)*9i)/4 
+ 6*a^2*b*tan(c/2 + (d*x)/2)))/8)/(2*tan(c/2 + (d*x)/2)*(9*a^6 + (81*a^2*b 
^4)/16 - (27*a^4*b^2)/2) - 18*a^5*b + (27*a^3*b^3)/2 - (a*(4*a^2 - 3*b^2)* 
((9*a*b^2)/4 - 3*a^3 - (a*tan(c/2 + (d*x)/2)*(4*a^2 - 3*b^2)*9i)/4 + 6*a^2 
*b*tan(c/2 + (d*x)/2))*3i)/8 + (a*(4*a^2 - 3*b^2)*((9*a*b^2)/4 - 3*a^3 + ( 
a*tan(c/2 + (d*x)/2)*(4*a^2 - 3*b^2)*9i)/4 + 6*a^2*b*tan(c/2 + (d*x)/2))*3 
i)/8))*(4*a^2 - 3*b^2))/(4*d) + (3*a^2*b*log(tan(c/2 + (d*x)/2)))/d
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.09 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b^{3}-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}-40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b +16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{3}-20 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}+75 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}+160 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{3}-40 \cos \left (d x +c \right ) a^{3}+120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a^{2} b -60 \sin \left (d x +c \right ) a^{3} d x -160 \sin \left (d x +c \right ) a^{2} b +45 \sin \left (d x +c \right ) a \,b^{2} d x +8 \sin \left (d x +c \right ) b^{3}}{40 \sin \left (d x +c \right ) d} \] Input:

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

Output:

( - 8*cos(c + d*x)*sin(c + d*x)**5*b**3 - 30*cos(c + d*x)*sin(c + d*x)**4* 
a*b**2 - 40*cos(c + d*x)*sin(c + d*x)**3*a**2*b + 16*cos(c + d*x)*sin(c + 
d*x)**3*b**3 - 20*cos(c + d*x)*sin(c + d*x)**2*a**3 + 75*cos(c + d*x)*sin( 
c + d*x)**2*a*b**2 + 160*cos(c + d*x)*sin(c + d*x)*a**2*b - 8*cos(c + d*x) 
*sin(c + d*x)*b**3 - 40*cos(c + d*x)*a**3 + 120*log(tan((c + d*x)/2))*sin( 
c + d*x)*a**2*b - 60*sin(c + d*x)*a**3*d*x - 160*sin(c + d*x)*a**2*b + 45* 
sin(c + d*x)*a*b**2*d*x + 8*sin(c + d*x)*b**3)/(40*sin(c + d*x)*d)