\(\int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1070]

Optimal result

Integrand size = 29, antiderivative size = 138 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-3 a b^2 x+\frac {b \left (3 a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 d}+\frac {11 b^3 \cos (c+d x)}{6 d}+\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{3 d}-\frac {b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d} \] Output:

-3*a*b^2*x+1/2*b*(3*a^2-2*b^2)*arctanh(cos(d*x+c))/d+11/6*b^3*cos(d*x+c)/d 
+1/3*a*(a^2-3*b^2)*cot(d*x+c)/d-1/2*b*cot(d*x+c)*csc(d*x+c)*(a+b*sin(d*x+c 
))^2/d-1/3*cot(d*x+c)*csc(d*x+c)^2*(a+b*sin(d*x+c))^3/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(615\) vs. \(2(138)=276\).

Time = 7.77 (sec) , antiderivative size = 615, normalized size of antiderivative = 4.46 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 a b^2 (c+d x) (b+a \csc (c+d x))^3 \sin ^3(c+d x)}{d (a+b \sin (c+d x))^3}+\frac {b^3 \cos (c+d x) (b+a \csc (c+d x))^3 \sin ^3(c+d x)}{d (a+b \sin (c+d x))^3}+\frac {\left (a^3 \cos \left (\frac {1}{2} (c+d x)\right )-9 a b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right ) \csc \left (\frac {1}{2} (c+d x)\right ) (b+a \csc (c+d x))^3 \sin ^3(c+d x)}{6 d (a+b \sin (c+d x))^3}-\frac {3 a^2 b \csc ^2\left (\frac {1}{2} (c+d x)\right ) (b+a \csc (c+d x))^3 \sin ^3(c+d x)}{8 d (a+b \sin (c+d x))^3}-\frac {a^3 \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (b+a \csc (c+d x))^3 \sin ^3(c+d x)}{24 d (a+b \sin (c+d x))^3}+\frac {\left (3 a^2 b-2 b^3\right ) (b+a \csc (c+d x))^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin ^3(c+d x)}{2 d (a+b \sin (c+d x))^3}+\frac {\left (-3 a^2 b+2 b^3\right ) (b+a \csc (c+d x))^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin ^3(c+d x)}{2 d (a+b \sin (c+d x))^3}+\frac {3 a^2 b (b+a \csc (c+d x))^3 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sin ^3(c+d x)}{8 d (a+b \sin (c+d x))^3}+\frac {(b+a \csc (c+d x))^3 \sec \left (\frac {1}{2} (c+d x)\right ) \left (-a^3 \sin \left (\frac {1}{2} (c+d x)\right )+9 a b^2 \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin ^3(c+d x)}{6 d (a+b \sin (c+d x))^3}+\frac {a^3 (b+a \csc (c+d x))^3 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sin ^3(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{24 d (a+b \sin (c+d x))^3} \] Input:

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

Output:

(-3*a*b^2*(c + d*x)*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(d*(a + b*Sin[c 
 + d*x])^3) + (b^3*Cos[c + d*x]*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(d* 
(a + b*Sin[c + d*x])^3) + ((a^3*Cos[(c + d*x)/2] - 9*a*b^2*Cos[(c + d*x)/2 
])*Csc[(c + d*x)/2]*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(6*d*(a + b*Sin 
[c + d*x])^3) - (3*a^2*b*Csc[(c + d*x)/2]^2*(b + a*Csc[c + d*x])^3*Sin[c + 
 d*x]^3)/(8*d*(a + b*Sin[c + d*x])^3) - (a^3*Cot[(c + d*x)/2]*Csc[(c + d*x 
)/2]^2*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(24*d*(a + b*Sin[c + d*x])^3 
) + ((3*a^2*b - 2*b^3)*(b + a*Csc[c + d*x])^3*Log[Cos[(c + d*x)/2]]*Sin[c 
+ d*x]^3)/(2*d*(a + b*Sin[c + d*x])^3) + ((-3*a^2*b + 2*b^3)*(b + a*Csc[c 
+ d*x])^3*Log[Sin[(c + d*x)/2]]*Sin[c + d*x]^3)/(2*d*(a + b*Sin[c + d*x])^ 
3) + (3*a^2*b*(b + a*Csc[c + d*x])^3*Sec[(c + d*x)/2]^2*Sin[c + d*x]^3)/(8 
*d*(a + b*Sin[c + d*x])^3) + ((b + a*Csc[c + d*x])^3*Sec[(c + d*x)/2]*(-(a 
^3*Sin[(c + d*x)/2]) + 9*a*b^2*Sin[(c + d*x)/2])*Sin[c + d*x]^3)/(6*d*(a + 
 b*Sin[c + d*x])^3) + (a^3*(b + a*Csc[c + d*x])^3*Sec[(c + d*x)/2]^2*Sin[c 
 + d*x]^3*Tan[(c + d*x)/2])/(24*d*(a + b*Sin[c + d*x])^3)
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.586, Rules used = {3042, 3368, 3042, 3527, 3042, 3526, 25, 3042, 3510, 25, 3042, 3502, 27, 3042, 3214, 3042, 4257}

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

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3368

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int -\csc ^2(c+d x) (a+b \sin (c+d x)) \left (11 b^2 \sin ^2(c+d x)+7 a b \sin (c+d x)+2 \left (a^2-3 b^2\right )\right )dx-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (-\frac {1}{2} \int \csc ^2(c+d x) (a+b \sin (c+d x)) \left (11 b^2 \sin ^2(c+d x)+7 a b \sin (c+d x)+2 \left (a^2-3 b^2\right )\right )dx-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-\frac {1}{2} \int \frac {(a+b \sin (c+d x)) \left (11 b^2 \sin (c+d x)^2+7 a b \sin (c+d x)+2 \left (a^2-3 b^2\right )\right )}{\sin (c+d x)^2}dx-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3510

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int -\csc (c+d x) \left (11 \sin ^2(c+d x) b^3+18 a \sin (c+d x) b^2+3 \left (3 a^2-2 b^2\right ) b\right )dx+\frac {2 a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}\right )-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {2 a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}-\int \csc (c+d x) \left (11 \sin ^2(c+d x) b^3+18 a \sin (c+d x) b^2+3 \left (3 a^2-2 b^2\right ) b\right )dx\right )-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {2 a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}-\int \frac {11 \sin (c+d x)^2 b^3+18 a \sin (c+d x) b^2+3 \left (3 a^2-2 b^2\right ) b}{\sin (c+d x)}dx\right )-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (-\int 3 \csc (c+d x) \left (6 a \sin (c+d x) b^2+\left (3 a^2-2 b^2\right ) b\right )dx+\frac {2 a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}+\frac {11 b^3 \cos (c+d x)}{d}\right )-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (-3 \int \csc (c+d x) \left (6 a \sin (c+d x) b^2+\left (3 a^2-2 b^2\right ) b\right )dx+\frac {2 a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}+\frac {11 b^3 \cos (c+d x)}{d}\right )-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (-3 \int \frac {6 a \sin (c+d x) b^2+\left (3 a^2-2 b^2\right ) b}{\sin (c+d x)}dx+\frac {2 a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}+\frac {11 b^3 \cos (c+d x)}{d}\right )-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (-3 \left (b \left (3 a^2-2 b^2\right ) \int \csc (c+d x)dx+6 a b^2 x\right )+\frac {2 a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}+\frac {11 b^3 \cos (c+d x)}{d}\right )-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (-3 \left (b \left (3 a^2-2 b^2\right ) \int \csc (c+d x)dx+6 a b^2 x\right )+\frac {2 a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}+\frac {11 b^3 \cos (c+d x)}{d}\right )-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (-3 \left (6 a b^2 x-\frac {b \left (3 a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{d}\right )+\frac {2 a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}+\frac {11 b^3 \cos (c+d x)}{d}\right )-\frac {3 b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\right )-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\)

Input:

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

Output:

-1/3*(Cot[c + d*x]*Csc[c + d*x]^2*(a + b*Sin[c + d*x])^3)/d + ((-3*(6*a*b^ 
2*x - (b*(3*a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/d) + (11*b^3*Cos[c + d*x]) 
/d + (2*a*(a^2 - 3*b^2)*Cot[c + d*x])/d)/2 - (3*b*Cot[c + d*x]*Csc[c + d*x 
]*(a + b*Sin[c + d*x])^2)/(2*d))/3
 

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 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 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3510

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 

rule 3526

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 
 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* 
d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 
1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x 
] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f 
*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d 
, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 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 4257

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

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {-\frac {a^{3} \cos \left (d x +c \right )^{3}}{3 \sin \left (d x +c \right )^{3}}+3 a^{2} b \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+3 a \,b^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+b^{3} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(127\)
default	\(\frac {-\frac {a^{3} \cos \left (d x +c \right )^{3}}{3 \sin \left (d x +c \right )^{3}}+3 a^{2} b \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+3 a \,b^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+b^{3} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(127\)
risch	\(-3 a \,b^{2} x +\frac {b^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a \left (6 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-18 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+36 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 i a^{2}-18 i b^{2}-9 a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\)	\(227\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.67 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {36 \, a b^{2} \cos \left (d x + c\right ) + 4 \, {\left (a^{3} - 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 6 \, {\left (6 \, a b^{2} d x \cos \left (d x + c\right )^{2} - 2 \, b^{3} \cos \left (d x + c\right )^{3} - 6 \, a b^{2} d x - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \] Input:

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

Output:

1/12*(36*a*b^2*cos(d*x + c) + 4*(a^3 - 9*a*b^2)*cos(d*x + c)^3 - 3*(3*a^2* 
b - 2*b^3 - (3*a^2*b - 2*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)* 
sin(d*x + c) + 3*(3*a^2*b - 2*b^3 - (3*a^2*b - 2*b^3)*cos(d*x + c)^2)*log( 
-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 6*(6*a*b^2*d*x*cos(d*x + c)^2 - 2* 
b^3*cos(d*x + c)^3 - 6*a*b^2*d*x - (3*a^2*b - 2*b^3)*cos(d*x + c))*sin(d*x 
 + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.86 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {4 \, a^{3} \cot \left (d x + c\right )^{3} + 36 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a b^{2} - 9 \, a^{2} b {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, b^{3} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \] Input:

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

Output:

-1/12*(4*a^3*cot(d*x + c)^3 + 36*(d*x + c + 1/tan(d*x + c))*a*b^2 - 9*a^2* 
b*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) + 1) - log(cos(d 
*x + c) - 1)) - 6*b^3*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d* 
x + c) - 1)))/d
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.61 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 72 \, {\left (d x + c\right )} a b^{2} - 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {48 \, b^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - 12 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {66 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \] Input:

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

Output:

1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 + 9*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 72*(d* 
x + c)*a*b^2 - 3*a^3*tan(1/2*d*x + 1/2*c) + 36*a*b^2*tan(1/2*d*x + 1/2*c) 
+ 48*b^3/(tan(1/2*d*x + 1/2*c)^2 + 1) - 12*(3*a^2*b - 2*b^3)*log(abs(tan(1 
/2*d*x + 1/2*c))) + (66*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 44*b^3*tan(1/2*d*x 
+ 1/2*c)^3 + 3*a^3*tan(1/2*d*x + 1/2*c)^2 - 36*a*b^2*tan(1/2*d*x + 1/2*c)^ 
2 - 9*a^2*b*tan(1/2*d*x + 1/2*c) - a^3)/tan(1/2*d*x + 1/2*c)^3)/d
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 35.52 (sec) , antiderivative size = 477, normalized size of antiderivative = 3.46 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {\frac {a^3\,\cos \left (c+d\,x\right )}{4}-\frac {3\,b^3\,\sin \left (c+d\,x\right )}{4}+\frac {a^3\,\cos \left (3\,c+3\,d\,x\right )}{12}-\frac {b^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {b^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {b^3\,\sin \left (4\,c+4\,d\,x\right )}{8}-\frac {3\,a\,b^2\,\cos \left (3\,c+3\,d\,x\right )}{4}-\frac {3\,b^3\,\sin \left (c+d\,x\right )\,\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}+\frac {3\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {b^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,a\,b^2\,\cos \left (c+d\,x\right )}{4}-\frac {3\,a^2\,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 )\,\sin \left (3\,c+3\,d\,x\right )}{8}-\frac {9\,a\,b^2\,\mathrm {atan}\left (\frac {3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+6\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{-3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}\right )\,\sin \left (c+d\,x\right )}{2}+\frac {3\,a\,b^2\,\mathrm {atan}\left (\frac {3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+6\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{-3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}\right )\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {9\,a^2\,b\,\sin \left (c+d\,x\right )\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{8}}{d\,{\sin \left (c+d\,x\right )}^3} \] Input:

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

Output:

-((a^3*cos(c + d*x))/4 - (3*b^3*sin(c + d*x))/4 + (a^3*cos(3*c + 3*d*x))/1 
2 - (b^3*sin(2*c + 2*d*x))/4 + (b^3*sin(3*c + 3*d*x))/4 + (b^3*sin(4*c + 4 
*d*x))/8 - (3*a*b^2*cos(3*c + 3*d*x))/4 - (3*b^3*sin(c + d*x)*log(sin(c/2 
+ (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (3*a^2*b*sin(2*c + 2*d*x))/4 + (b^3*lo 
g(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x))/4 + (3*a*b^2*co 
s(c + d*x))/4 - (3*a^2*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3* 
c + 3*d*x))/8 - (9*a*b^2*atan((3*a^2*sin(c/2 + (d*x)/2) - 2*b^2*sin(c/2 + 
(d*x)/2) + 6*a*b*cos(c/2 + (d*x)/2))/(2*b^2*cos(c/2 + (d*x)/2) - 3*a^2*cos 
(c/2 + (d*x)/2) + 6*a*b*sin(c/2 + (d*x)/2)))*sin(c + d*x))/2 + (3*a*b^2*at 
an((3*a^2*sin(c/2 + (d*x)/2) - 2*b^2*sin(c/2 + (d*x)/2) + 6*a*b*cos(c/2 + 
(d*x)/2))/(2*b^2*cos(c/2 + (d*x)/2) - 3*a^2*cos(c/2 + (d*x)/2) + 6*a*b*sin 
(c/2 + (d*x)/2)))*sin(3*c + 3*d*x))/2 + (9*a^2*b*sin(c + d*x)*log(sin(c/2 
+ (d*x)/2)/cos(c/2 + (d*x)/2)))/8)/(d*sin(c + d*x)^3)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.38 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{3}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-72 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-36 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -8 \cos \left (d x +c \right ) a^{3}-36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a^{2} b +24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} b^{3}+9 \sin \left (d x +c \right )^{3} a^{2} b -72 \sin \left (d x +c \right )^{3} a \,b^{2} d x -24 \sin \left (d x +c \right )^{3} b^{3}}{24 \sin \left (d x +c \right )^{3} d} \] Input:

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

Output:

(24*cos(c + d*x)*sin(c + d*x)**3*b**3 + 8*cos(c + d*x)*sin(c + d*x)**2*a** 
3 - 72*cos(c + d*x)*sin(c + d*x)**2*a*b**2 - 36*cos(c + d*x)*sin(c + d*x)* 
a**2*b - 8*cos(c + d*x)*a**3 - 36*log(tan((c + d*x)/2))*sin(c + d*x)**3*a* 
*2*b + 24*log(tan((c + d*x)/2))*sin(c + d*x)**3*b**3 + 9*sin(c + d*x)**3*a 
**2*b - 72*sin(c + d*x)**3*a*b**2*d*x - 24*sin(c + d*x)**3*b**3)/(24*sin(c 
 + d*x)**3*d)