\(\int (c+d x)^3 \cot (a+b x) \csc (a+b x) \, dx\) [40]

Optimal result

Integrand size = 20, antiderivative size = 146 \[ \int (c+d x)^3 \cot (a+b x) \csc (a+b x) \, dx=-\frac {6 d (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^3 \csc (a+b x)}{b}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^3}-\frac {6 d^3 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^4}+\frac {6 d^3 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^4} \] Output:

-6*d*(d*x+c)^2*arctanh(exp(I*(b*x+a)))/b^2-(d*x+c)^3*csc(b*x+a)/b+6*I*d^2* 
(d*x+c)*polylog(2,-exp(I*(b*x+a)))/b^3-6*I*d^2*(d*x+c)*polylog(2,exp(I*(b* 
x+a)))/b^3-6*d^3*polylog(3,-exp(I*(b*x+a)))/b^4+6*d^3*polylog(3,exp(I*(b*x 
+a)))/b^4
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(311\) vs. \(2(146)=292\).

Time = 0.82 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.13 \[ \int (c+d x)^3 \cot (a+b x) \csc (a+b x) \, dx=-\frac {b^3 c^3 \csc (a+b x)+3 b^3 c^2 d x \csc (a+b x)+3 b^3 c d^2 x^2 \csc (a+b x)+b^3 d^3 x^3 \csc (a+b x)-3 b^2 c^2 d \log \left (1-e^{i (a+b x)}\right )-6 b^2 c d^2 x \log \left (1-e^{i (a+b x)}\right )-3 b^2 d^3 x^2 \log \left (1-e^{i (a+b x)}\right )+3 b^2 c^2 d \log \left (1+e^{i (a+b x)}\right )+6 b^2 c d^2 x \log \left (1+e^{i (a+b x)}\right )+3 b^2 d^3 x^2 \log \left (1+e^{i (a+b x)}\right )-6 i b d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )+6 i b d^2 (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )+6 d^3 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )-6 d^3 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^4} \] Input:

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

Output:

-((b^3*c^3*Csc[a + b*x] + 3*b^3*c^2*d*x*Csc[a + b*x] + 3*b^3*c*d^2*x^2*Csc 
[a + b*x] + b^3*d^3*x^3*Csc[a + b*x] - 3*b^2*c^2*d*Log[1 - E^(I*(a + b*x)) 
] - 6*b^2*c*d^2*x*Log[1 - E^(I*(a + b*x))] - 3*b^2*d^3*x^2*Log[1 - E^(I*(a 
 + b*x))] + 3*b^2*c^2*d*Log[1 + E^(I*(a + b*x))] + 6*b^2*c*d^2*x*Log[1 + E 
^(I*(a + b*x))] + 3*b^2*d^3*x^2*Log[1 + E^(I*(a + b*x))] - (6*I)*b*d^2*(c 
+ d*x)*PolyLog[2, -E^(I*(a + b*x))] + (6*I)*b*d^2*(c + d*x)*PolyLog[2, E^( 
I*(a + b*x))] + 6*d^3*PolyLog[3, -E^(I*(a + b*x))] - 6*d^3*PolyLog[3, E^(I 
*(a + b*x))])/b^4)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4910, 3042, 4671, 3011, 2720, 7143}

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.

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

\(\Big \downarrow \) 4910

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4671

\(\displaystyle -\frac {(c+d x)^3 \csc (a+b x)}{b}+\frac {3 d \left (-\frac {2 d \int (c+d x) \log \left (1-e^{i (a+b x)}\right )dx}{b}+\frac {2 d \int (c+d x) \log \left (1+e^{i (a+b x)}\right )dx}{b}-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {(c+d x)^3 \csc (a+b x)}{b}+\frac {3 d \left (\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b}-\frac {i d \int \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )dx}{b}\right )}{b}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b}-\frac {i d \int \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )dx}{b}\right )}{b}-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {(c+d x)^3 \csc (a+b x)}{b}+\frac {3 d \left (\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b}-\frac {d \int e^{-i (a+b x)} \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}\right )}{b}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b}-\frac {d \int e^{-i (a+b x)} \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}\right )}{b}-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 7143

\(\displaystyle -\frac {(c+d x)^3 \csc (a+b x)}{b}+\frac {3 d \left (-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b}-\frac {d \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^2}\right )}{b}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b}-\frac {d \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^2}\right )}{b}\right )}{b}\)

Input:

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

Output:

-(((c + d*x)^3*Csc[a + b*x])/b) + (3*d*((-2*(c + d*x)^2*ArcTanh[E^(I*(a + 
b*x))])/b + (2*d*((I*(c + d*x)*PolyLog[2, -E^(I*(a + b*x))])/b - (d*PolyLo 
g[3, -E^(I*(a + b*x))])/b^2))/b - (2*d*((I*(c + d*x)*PolyLog[2, E^(I*(a + 
b*x))])/b - (d*PolyLog[3, E^(I*(a + b*x))])/b^2))/b))/b
 

Defintions of rubi rules used

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4671

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[- 
2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*x))]/f), x] + (-Simp[d*(m/f)   Int[(c + 
d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f)   Int[(c + d*x 
)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG 
tQ[m, 0]
 

rule 4910

Int[Cot[(a_.) + (b_.)*(x_)]^(p_.)*Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d 
_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Csc[a + b*x]^n/(b*n)), x 
] + Simp[d*(m/(b*n))   Int[(c + d*x)^(m - 1)*Csc[a + b*x]^n, x], x] /; Free 
Q[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 432 vs. \(2 (134 ) = 268\).

Time = 0.35 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.97

method	result	size
risch	\(\frac {6 i d^{2} c \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 d^{2} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{3}}-\frac {2 i \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right ) {\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {6 d^{2} c \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {6 i d^{2} c \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 d \,c^{2} \operatorname {arctanh}\left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {6 d^{3} a^{2} \operatorname {arctanh}\left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 d^{2} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}+\frac {6 d^{2} c \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a^{2}}{b^{4}}-\frac {3 d^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{4}}+\frac {12 d^{2} c a \,\operatorname {arctanh}\left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 i d^{3} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {6 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b^{2}}-\frac {6 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 d^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}+\frac {6 d^{3} \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}\)	\(433\)

Input:

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

Output:

6*I*d^2/b^3*c*polylog(2,-exp(I*(b*x+a)))-6*d^2/b^3*c*ln(exp(I*(b*x+a))+1)* 
a-2*I*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)*exp(I*(b*x+a))/b/(exp(2*I*(b*x+a 
))-1)+6*d^2/b^3*c*ln(1-exp(I*(b*x+a)))*a-6*I*d^2/b^3*c*polylog(2,exp(I*(b* 
x+a)))-6*d/b^2*c^2*arctanh(exp(I*(b*x+a)))-6*d^3/b^4*a^2*arctanh(exp(I*(b* 
x+a)))-6*d^2/b^2*c*ln(exp(I*(b*x+a))+1)*x+6*d^2/b^2*c*ln(1-exp(I*(b*x+a))) 
*x+3*d^3/b^4*ln(exp(I*(b*x+a))+1)*a^2-3*d^3/b^4*ln(1-exp(I*(b*x+a)))*a^2+1 
2*d^2/b^3*c*a*arctanh(exp(I*(b*x+a)))-6*I*d^3/b^3*polylog(2,exp(I*(b*x+a)) 
)*x+6*I*d^3/b^3*polylog(2,-exp(I*(b*x+a)))*x-3*d^3/b^2*ln(exp(I*(b*x+a))+1 
)*x^2-6*d^3*polylog(3,-exp(I*(b*x+a)))/b^4+3*d^3/b^2*ln(1-exp(I*(b*x+a)))* 
x^2+6*d^3*polylog(3,exp(I*(b*x+a)))/b^4
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 669 vs. \(2 (130) = 260\).

Time = 0.11 (sec) , antiderivative size = 669, normalized size of antiderivative = 4.58 \[ \int (c+d x)^3 \cot (a+b x) \csc (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 + 6*b^3*c^2*d*x + 2*b^3*c^3 - 6*d^3* 
polylog(3, cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) - 6*d^3*polylog(3, 
cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + 6*d^3*polylog(3, -cos(b*x + 
a) + I*sin(b*x + a))*sin(b*x + a) + 6*d^3*polylog(3, -cos(b*x + a) - I*sin 
(b*x + a))*sin(b*x + a) + 6*(I*b*d^3*x + I*b*c*d^2)*dilog(cos(b*x + a) + I 
*sin(b*x + a))*sin(b*x + a) + 6*(-I*b*d^3*x - I*b*c*d^2)*dilog(cos(b*x + a 
) - I*sin(b*x + a))*sin(b*x + a) + 6*(I*b*d^3*x + I*b*c*d^2)*dilog(-cos(b* 
x + a) + I*sin(b*x + a))*sin(b*x + a) + 6*(-I*b*d^3*x - I*b*c*d^2)*dilog(- 
cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + 3*(b^2*d^3*x^2 + 2*b^2*c*d^2 
*x + b^2*c^2*d)*log(cos(b*x + a) + I*sin(b*x + a) + 1)*sin(b*x + a) + 3*(b 
^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*log(cos(b*x + a) - I*sin(b*x + a) 
+ 1)*sin(b*x + a) - 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*log(-1/2*cos(b*x 
 + a) + 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a) - 3*(b^2*c^2*d - 2*a*b*c*d^ 
2 + a^2*d^3)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a 
) - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*log(-cos(b*x + 
 a) + I*sin(b*x + a) + 1)*sin(b*x + a) - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 
2*a*b*c*d^2 - a^2*d^3)*log(-cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a 
))/(b^4*sin(b*x + a))
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1770 vs. \(2 (130) = 260\).

Time = 0.18 (sec) , antiderivative size = 1770, normalized size of antiderivative = 12.12 \[ \int (c+d x)^3 \cot (a+b x) \csc (a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(3*(4*(b*x + a)*cos(b*x + a)*sin(2*b*x + 2*a) - 4*(b*x + a)*cos(2*b*x 
 + 2*a)*sin(b*x + a) + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2* 
b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) 
- (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*log(c 
os(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 4*(b*x + a)*sin(b*x 
 + a))*c^2*d/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a 
) + 1)*b) - 6*(4*(b*x + a)*cos(b*x + a)*sin(2*b*x + 2*a) - 4*(b*x + a)*cos 
(2*b*x + 2*a)*sin(b*x + a) + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2* 
cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) 
 + 1) - (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1) 
*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 4*(b*x + a)*s 
in(b*x + a))*a*c*d^2/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b 
*x + 2*a) + 1)*b^2) + 3*(4*(b*x + a)*cos(b*x + a)*sin(2*b*x + 2*a) - 4*(b* 
x + a)*cos(2*b*x + 2*a)*sin(b*x + a) + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2 
*a)^2 - 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*co 
s(b*x + a) + 1) - (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 
 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 4*( 
b*x + a)*sin(b*x + a))*a^2*d^3/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 
 2*cos(2*b*x + 2*a) + 1)*b^3) + 2*c^3/sin(b*x + a) - 6*a*c^2*d/(b*sin(b*x 
+ a)) + 6*a^2*c*d^2/(b^2*sin(b*x + a)) - 2*a^3*d^3/(b^3*sin(b*x + a)) -...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^3 \cot (a+b x) \csc (a+b x) \, dx=\frac {3 \cos \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b c \,d^{2} x^{2}+\cos \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,d^{3} x^{3}+3 \left (\int \frac {x^{2}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}d x \right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,d^{3}+6 \left (\int \frac {x}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}d x \right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b c \,d^{2}+3 \left (\int \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) x^{2}d x \right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,d^{3}+6 \left (\int \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) x d x \right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b c \,d^{2}+6 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) c^{2} d -3 \sin \left (b x +a \right ) b c \,d^{2} x^{2}-\sin \left (b x +a \right ) b \,d^{3} x^{3}-2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,c^{3}-6 \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,c^{2} d x -3 \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b c \,d^{2} x^{2}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,d^{3} x^{3}}{2 \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b^{2}} \] Input:

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

Output:

(3*cos(a + b*x)*tan((a + b*x)/2)*b*c*d**2*x**2 + cos(a + b*x)*tan((a + b*x 
)/2)*b*d**3*x**3 + 3*int(x**2/tan((a + b*x)/2),x)*sin(a + b*x)*tan((a + b* 
x)/2)*b*d**3 + 6*int(x/tan((a + b*x)/2),x)*sin(a + b*x)*tan((a + b*x)/2)*b 
*c*d**2 + 3*int(tan((a + b*x)/2)*x**2,x)*sin(a + b*x)*tan((a + b*x)/2)*b*d 
**3 + 6*int(tan((a + b*x)/2)*x,x)*sin(a + b*x)*tan((a + b*x)/2)*b*c*d**2 + 
 6*log(tan((a + b*x)/2))*sin(a + b*x)*tan((a + b*x)/2)*c**2*d - 3*sin(a + 
b*x)*b*c*d**2*x**2 - sin(a + b*x)*b*d**3*x**3 - 2*tan((a + b*x)/2)*b*c**3 
- 6*tan((a + b*x)/2)*b*c**2*d*x - 3*tan((a + b*x)/2)*b*c*d**2*x**2 - tan(( 
a + b*x)/2)*b*d**3*x**3)/(2*sin(a + b*x)*tan((a + b*x)/2)*b**2)