\(\int (c+d x)^3 \csc (a+b x) \, dx\) [23]

Optimal result

Integrand size = 14, antiderivative size = 185 \[ \int (c+d x)^3 \csc (a+b x) \, dx=-\frac {2 (c+d x)^3 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^3 \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^3 \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )}{b^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.19 \[ \int (c+d x)^3 \csc (a+b x) \, dx=\frac {-2 b^3 (c+d x)^3 \text {arctanh}(\cos (a+b x)+i \sin (a+b x))+3 i d \left (b^2 (c+d x)^2 \operatorname {PolyLog}(2,-\cos (a+b x)-i \sin (a+b x))+2 i b d (c+d x) \operatorname {PolyLog}(3,-\cos (a+b x)-i \sin (a+b x))-2 d^2 \operatorname {PolyLog}(4,-\cos (a+b x)-i \sin (a+b x))\right )-3 i d \left (b^2 (c+d x)^2 \operatorname {PolyLog}(2,\cos (a+b x)+i \sin (a+b x))+2 i b d (c+d x) \operatorname {PolyLog}(3,\cos (a+b x)+i \sin (a+b x))-2 d^2 \operatorname {PolyLog}(4,\cos (a+b x)+i \sin (a+b x))\right )}{b^4} \] Input:

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

Output:

(-2*b^3*(c + d*x)^3*ArcTanh[Cos[a + b*x] + I*Sin[a + b*x]] + (3*I)*d*(b^2* 
(c + d*x)^2*PolyLog[2, -Cos[a + b*x] - I*Sin[a + b*x]] + (2*I)*b*d*(c + d* 
x)*PolyLog[3, -Cos[a + b*x] - I*Sin[a + b*x]] - 2*d^2*PolyLog[4, -Cos[a + 
b*x] - I*Sin[a + b*x]]) - (3*I)*d*(b^2*(c + d*x)^2*PolyLog[2, Cos[a + b*x] 
 + I*Sin[a + b*x]] + (2*I)*b*d*(c + d*x)*PolyLog[3, Cos[a + b*x] + I*Sin[a 
 + b*x]] - 2*d^2*PolyLog[4, Cos[a + b*x] + I*Sin[a + b*x]]))/b^4
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4671, 3011, 7163, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4671

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

\(\displaystyle -\frac {2 (c+d x)^3 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b}-\frac {2 i d \left (\frac {d \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b}\right )}{b}\right )}{b}-\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b}-\frac {2 i d \left (\frac {d \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )}{b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b}\right )}{b}\right )}{b}\)

Input:

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

Output:

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

rule 7163

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 632 vs. \(2 (167 ) = 334\).

Time = 0.94 (sec) , antiderivative size = 633, normalized size of antiderivative = 3.42

method	result	size
risch	\(-\frac {6 c \,d^{2} a^{2} \operatorname {arctanh}\left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {d^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{3}}{b}+\frac {d^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{3}}{b}+\frac {d^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{3}}{b^{4}}-\frac {d^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a^{3}}{b^{4}}-\frac {6 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {6 d^{3} \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {6 c \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 c \,d^{2} \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{3} a^{3} \operatorname {arctanh}\left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 i d^{3} \operatorname {polylog}\left (4, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 i d^{3} \operatorname {polylog}\left (4, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 i c \,d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {6 i c \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {6 c^{2} d a \,\operatorname {arctanh}\left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {3 c^{2} d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}+\frac {3 c^{2} d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {3 c \,d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {3 c \,d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {3 c^{2} d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{2}}+\frac {3 c^{2} d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {3 c \,d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a^{2}}{b^{3}}-\frac {3 c \,d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}+\frac {3 i c^{2} d \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 i d^{3} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 i c^{2} d \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 c^{3} \operatorname {arctanh}\left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}\)	\(633\)

Input:

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

Output:

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

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. 820 vs. \(2 (161) = 322\).

Time = 0.12 (sec) , antiderivative size = 820, normalized size of antiderivative = 4.43 \[ \int (c+d x)^3 \csc (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

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. 716 vs. \(2 (161) = 322\).

Time = 0.14 (sec) , antiderivative size = 716, normalized size of antiderivative = 3.87 \[ \int (c+d x)^3 \csc (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(2*c^3*log(cot(b*x + a) + csc(b*x + a)) - 6*a*c^2*d*log(cot(b*x + a) 
+ csc(b*x + a))/b + 6*a^2*c*d^2*log(cot(b*x + a) + csc(b*x + a))/b^2 - 2*a 
^3*d^3*log(cot(b*x + a) + csc(b*x + a))/b^3 + (12*I*d^3*polylog(4, -e^(I*b 
*x + I*a)) - 12*I*d^3*polylog(4, e^(I*b*x + I*a)) - 2*(-I*(b*x + a)^3*d^3 
+ 3*(-I*b*c*d^2 + I*a*d^3)*(b*x + a)^2 + 3*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 - 
 I*a^2*d^3)*(b*x + a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - 2*(-I*(b* 
x + a)^3*d^3 + 3*(-I*b*c*d^2 + I*a*d^3)*(b*x + a)^2 + 3*(-I*b^2*c^2*d + 2* 
I*a*b*c*d^2 - I*a^2*d^3)*(b*x + a))*arctan2(sin(b*x + a), -cos(b*x + a) + 
1) - 6*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + I*(b*x + a)^2*d^3 + I*a^2*d^3 + 2*(I 
*b*c*d^2 - I*a*d^3)*(b*x + a))*dilog(-e^(I*b*x + I*a)) - 6*(-I*b^2*c^2*d + 
 2*I*a*b*c*d^2 - I*(b*x + a)^2*d^3 - I*a^2*d^3 + 2*(-I*b*c*d^2 + I*a*d^3)* 
(b*x + a))*dilog(e^(I*b*x + I*a)) + ((b*x + a)^3*d^3 + 3*(b*c*d^2 - a*d^3) 
*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + a))*log(cos(b* 
x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - ((b*x + a)^3*d^3 + 3*(b* 
c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + 
a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 12*(b*c*d^ 
2 + (b*x + a)*d^3 - a*d^3)*polylog(3, -e^(I*b*x + I*a)) - 12*(b*c*d^2 + (b 
*x + a)*d^3 - a*d^3)*polylog(3, e^(I*b*x + I*a)))/b^3)/b
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(int(csc(a + b*x)*x**3,x)*b*d**3 + 3*int(csc(a + b*x)*x**2,x)*b*c*d**2 + 3 
*int(csc(a + b*x)*x,x)*b*c**2*d + log(tan((a + b*x)/2))*c**3)/b