\(\int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1071]

Optimal result

Integrand size = 29, antiderivative size = 152 \[ \int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=-b^3 x+\frac {a \left (a^2+12 b^2\right ) \text {arctanh}(\cos (c+d x))}{8 d}+\frac {b \left (2 a^2-b^2\right ) \cot (c+d x)}{2 d}+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d} \] Output:

-b^3*x+1/8*a*(a^2+12*b^2)*arctanh(cos(d*x+c))/d+1/2*b*(2*a^2-b^2)*cot(d*x+ 
c)/d+1/8*a*(a^2-2*b^2)*cot(d*x+c)*csc(d*x+c)/d-1/4*b*cot(d*x+c)*csc(d*x+c) 
^2*(a+b*sin(d*x+c))^2/d-1/4*cot(d*x+c)*csc(d*x+c)^3*(a+b*sin(d*x+c))^3/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(690\) vs. \(2(152)=304\).

Time = 7.36 (sec) , antiderivative size = 690, normalized size of antiderivative = 4.54 \[ \int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {b^3 (c+d x) (b+a \csc (c+d x))^3 \sin ^3(c+d x)}{d (a+b \sin (c+d x))^3}+\frac {\left (a^2 b \cos \left (\frac {1}{2} (c+d x)\right )-b^3 \cos \left (\frac {1}{2} (c+d x)\right )\right ) \csc \left (\frac {1}{2} (c+d x)\right ) (b+a \csc (c+d x))^3 \sin ^3(c+d x)}{2 d (a+b \sin (c+d x))^3}+\frac {\left (a^3-12 a b^2\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (b+a \csc (c+d x))^3 \sin ^3(c+d x)}{32 d (a+b \sin (c+d x))^3}-\frac {a^2 b \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (b+a \csc (c+d x))^3 \sin ^3(c+d x)}{8 d (a+b \sin (c+d x))^3}-\frac {a^3 \csc ^4\left (\frac {1}{2} (c+d x)\right ) (b+a \csc (c+d x))^3 \sin ^3(c+d x)}{64 d (a+b \sin (c+d x))^3}+\frac {\left (a^3+12 a b^2\right ) (b+a \csc (c+d x))^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin ^3(c+d x)}{8 d (a+b \sin (c+d x))^3}+\frac {\left (-a^3-12 a b^2\right ) (b+a \csc (c+d x))^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin ^3(c+d x)}{8 d (a+b \sin (c+d x))^3}+\frac {\left (-a^3+12 a b^2\right ) (b+a \csc (c+d x))^3 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sin ^3(c+d x)}{32 d (a+b \sin (c+d x))^3}+\frac {a^3 (b+a \csc (c+d x))^3 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sin ^3(c+d x)}{64 d (a+b \sin (c+d x))^3}+\frac {(b+a \csc (c+d x))^3 \sec \left (\frac {1}{2} (c+d x)\right ) \left (-a^2 b \sin \left (\frac {1}{2} (c+d x)\right )+b^3 \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin ^3(c+d x)}{2 d (a+b \sin (c+d x))^3}+\frac {a^2 b (b+a \csc (c+d x))^3 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sin ^3(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{8 d (a+b \sin (c+d x))^3} \] Input:

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

Output:

-((b^3*(c + d*x)*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(d*(a + b*Sin[c + 
d*x])^3)) + ((a^2*b*Cos[(c + d*x)/2] - b^3*Cos[(c + d*x)/2])*Csc[(c + d*x) 
/2]*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(2*d*(a + b*Sin[c + d*x])^3) + 
((a^3 - 12*a*b^2)*Csc[(c + d*x)/2]^2*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3 
)/(32*d*(a + b*Sin[c + d*x])^3) - (a^2*b*Cot[(c + d*x)/2]*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*Csc[(c + d*x)/2]^4*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(64*d*(a + b 
*Sin[c + d*x])^3) + ((a^3 + 12*a*b^2)*(b + a*Csc[c + d*x])^3*Log[Cos[(c + 
d*x)/2]]*Sin[c + d*x]^3)/(8*d*(a + b*Sin[c + d*x])^3) + ((-a^3 - 12*a*b^2) 
*(b + a*Csc[c + d*x])^3*Log[Sin[(c + d*x)/2]]*Sin[c + d*x]^3)/(8*d*(a + b* 
Sin[c + d*x])^3) + ((-a^3 + 12*a*b^2)*(b + a*Csc[c + d*x])^3*Sec[(c + d*x) 
/2]^2*Sin[c + d*x]^3)/(32*d*(a + b*Sin[c + d*x])^3) + (a^3*(b + a*Csc[c + 
d*x])^3*Sec[(c + d*x)/2]^4*Sin[c + d*x]^3)/(64*d*(a + b*Sin[c + d*x])^3) + 
 ((b + a*Csc[c + d*x])^3*Sec[(c + d*x)/2]*(-(a^2*b*Sin[(c + d*x)/2]) + b^3 
*Sin[(c + d*x)/2])*Sin[c + d*x]^3)/(2*d*(a + b*Sin[c + d*x])^3) + (a^2*b*( 
b + a*Csc[c + d*x])^3*Sec[(c + d*x)/2]^2*Sin[c + d*x]^3*Tan[(c + d*x)/2])/ 
(8*d*(a + b*Sin[c + d*x])^3)
 

Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.02, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.552, Rules used = {3042, 3368, 3042, 3527, 3042, 3526, 27, 3042, 3510, 25, 3042, 3500, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^2 (a+b \sin (c+d x))^3}{\sin (c+d x)^5}dx\)

\(\Big \downarrow \) 3368

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3510

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int -\csc ^2(c+d x) \left (8 \sin ^2(c+d x) b^3+4 \left (2 a^2-b^2\right ) b+a \left (a^2+12 b^2\right ) \sin (c+d x)\right )dx+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{d}\right )-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \csc ^2(c+d x) \left (8 \sin ^2(c+d x) b^3+4 \left (2 a^2-b^2\right ) b+a \left (a^2+12 b^2\right ) \sin (c+d x)\right )dx+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{d}\right )-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \frac {8 \sin (c+d x)^2 b^3+4 \left (2 a^2-b^2\right ) b+a \left (a^2+12 b^2\right ) \sin (c+d x)}{\sin (c+d x)^2}dx+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{d}\right )-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {4 b \left (2 a^2-b^2\right ) \cot (c+d x)}{d}-\int \csc (c+d x) \left (8 \sin (c+d x) b^3+a \left (a^2+12 b^2\right )\right )dx\right )+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{d}\right )-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {4 b \left (2 a^2-b^2\right ) \cot (c+d x)}{d}-\int \frac {8 \sin (c+d x) b^3+a \left (a^2+12 b^2\right )}{\sin (c+d x)}dx\right )+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{d}\right )-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-a \left (a^2+12 b^2\right ) \int \csc (c+d x)dx+\frac {4 b \left (2 a^2-b^2\right ) \cot (c+d x)}{d}-8 b^3 x\right )+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{d}\right )-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-a \left (a^2+12 b^2\right ) \int \csc (c+d x)dx+\frac {4 b \left (2 a^2-b^2\right ) \cot (c+d x)}{d}-8 b^3 x\right )+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{d}\right )-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {a \left (a^2+12 b^2\right ) \text {arctanh}(\cos (c+d x))}{d}+\frac {4 b \left (2 a^2-b^2\right ) \cot (c+d x)}{d}-8 b^3 x\right )+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{2 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{d}\right )-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}\)

Input:

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

Output:

-1/4*(Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^3)/d + ((-8*b^3*x + 
 (a*(a^2 + 12*b^2)*ArcTanh[Cos[c + d*x]])/d + (4*b*(2*a^2 - b^2)*Cot[c + d 
*x])/d)/2 + (a*(a^2 - 2*b^2)*Cot[c + d*x]*Csc[c + d*x])/(2*d) - (b*Cot[c + 
 d*x]*Csc[c + d*x]^2*(a + b*Sin[c + d*x])^2)/d)/4
 

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 3500

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

rule 3510

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

rule 3526

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

rule 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 = 0.39 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )-\frac {a^{2} b \cos \left (d x +c \right )^{3}}{\sin \left (d x +c \right )^{3}}+3 a \,b^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+b^{3} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(166\)
default	\(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )-\frac {a^{2} b \cos \left (d x +c \right )^{3}}{\sin \left (d x +c \right )^{3}}+3 a \,b^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+b^{3} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(166\)
risch	\(-b^{3} x -\frac {i \left (-7 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+12 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-12 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-12 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-24 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+8 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+12 i a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-i a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+24 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-24 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-i a^{3} {\mathrm e}^{i \left (d x +c \right )}-7 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-8 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+24 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+8 a^{2} b -8 b^{3}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{2 d}\)	\(332\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.74 \[ \int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {16 \, b^{3} d x \cos \left (d x + c\right )^{4} - 32 \, b^{3} d x \cos \left (d x + c\right )^{2} + 16 \, b^{3} d x + 2 \, {\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{3} + 12 \, a b^{2} - 2 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{3} + 12 \, a b^{2} - 2 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 16 \, {\left (b^{3} \cos \left (d x + c\right ) + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{16 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.98 \[ \int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {16 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} b^{3} + a^{3} {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 12 \, a b^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {16 \, a^{2} b}{\tan \left (d x + c\right )^{3}}}{16 \, d} \] Input:

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

Output:

-1/16*(16*(d*x + c + 1/tan(d*x + c))*b^3 + a^3*(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)) - 12*a*b^2*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + l 
og(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) + 16*a^2*b/tan(d*x + c)^3)/d
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.54 \[ \int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 192 \, {\left (d x + c\right )} b^{3} - 72 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, {\left (a^{3} + 12 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {50 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 600 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 72 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 96 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \] Input:

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

Output:

1/192*(3*a^3*tan(1/2*d*x + 1/2*c)^4 + 24*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 72 
*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 192*(d*x + c)*b^3 - 72*a^2*b*tan(1/2*d*x + 
 1/2*c) + 96*b^3*tan(1/2*d*x + 1/2*c) - 24*(a^3 + 12*a*b^2)*log(abs(tan(1/ 
2*d*x + 1/2*c))) + (50*a^3*tan(1/2*d*x + 1/2*c)^4 + 600*a*b^2*tan(1/2*d*x 
+ 1/2*c)^4 + 72*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 96*b^3*tan(1/2*d*x + 1/2*c) 
^3 - 72*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 24*a^2*b*tan(1/2*d*x + 1/2*c) - 3*a 
^3)/tan(1/2*d*x + 1/2*c)^4)/d
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 34.57 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.29 \[ \int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^3\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{8\,d}-\frac {b^3\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {2\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+12\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2+8\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+12\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )}{d}-\frac {3\,a\,b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2\,d}+\frac {3\,a^2\,b\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {3\,a\,b^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {a^2\,b\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,d}+\frac {3\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,d} \] Input:

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

Output:

(a^3*tan(c/2 + (d*x)/2)^4)/(64*d) - (a^3*cot(c/2 + (d*x)/2)^4)/(64*d) - (a 
^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(8*d) - (b^3*cot(c/2 + (d*x 
)/2))/(2*d) + (b^3*tan(c/2 + (d*x)/2))/(2*d) - (2*b^3*atan((8*b^3*cos(c/2 
+ (d*x)/2) + a^3*sin(c/2 + (d*x)/2) + 12*a*b^2*sin(c/2 + (d*x)/2))/(a^3*co 
s(c/2 + (d*x)/2) - 8*b^3*sin(c/2 + (d*x)/2) + 12*a*b^2*cos(c/2 + (d*x)/2)) 
))/d - (3*a*b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(2*d) + (3*a^2 
*b*cot(c/2 + (d*x)/2))/(8*d) - (3*a^2*b*tan(c/2 + (d*x)/2))/(8*d) - (3*a*b 
^2*cot(c/2 + (d*x)/2)^2)/(8*d) - (a^2*b*cot(c/2 + (d*x)/2)^3)/(8*d) + (3*a 
*b^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^2*b*tan(c/2 + (d*x)/2)^3)/(8*d)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.20 \[ \int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b -8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{3}+\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -2 \cos \left (d x +c \right ) a^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a^{3}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a \,b^{2}-8 \sin \left (d x +c \right )^{4} b^{3} d x}{8 \sin \left (d x +c \right )^{4} d} \] Input:

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

Output:

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