\(\int (c+d x)^3 \csc ^3(a+b x) \, dx\) [33]

Optimal result

Integrand size = 16, antiderivative size = 309 \[ \int (c+d x)^3 \csc ^3(a+b x) \, dx=-\frac {6 d^2 (c+d x) \text {arctanh}\left (e^{i (a+b x)}\right )}{b^3}-\frac {(c+d x)^3 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}-\frac {3 d (c+d x)^2 \csc (a+b x)}{2 b^2}-\frac {(c+d x)^3 \cot (a+b x) \csc (a+b x)}{2 b}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^4}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^4}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^4}+\frac {3 i d^3 \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )}{b^4} \] Output:

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

Mathematica [A] (verified)

Time = 5.95 (sec) , antiderivative size = 528, normalized size of antiderivative = 1.71 \[ \int (c+d x)^3 \csc ^3(a+b x) \, dx=-\frac {b^2 (c+d x)^2 (3 d+b (c+d x) \cot (a+b x)) \csc (a+b x)-b^3 c^3 \log \left (1-e^{i (a+b x)}\right )-6 b c d^2 \log \left (1-e^{i (a+b x)}\right )-3 b^3 c^2 d x \log \left (1-e^{i (a+b x)}\right )-6 b d^3 x \log \left (1-e^{i (a+b x)}\right )-3 b^3 c d^2 x^2 \log \left (1-e^{i (a+b x)}\right )-b^3 d^3 x^3 \log \left (1-e^{i (a+b x)}\right )+b^3 c^3 \log \left (1+e^{i (a+b x)}\right )+6 b c d^2 \log \left (1+e^{i (a+b x)}\right )+3 b^3 c^2 d x \log \left (1+e^{i (a+b x)}\right )+6 b d^3 x \log \left (1+e^{i (a+b x)}\right )+3 b^3 c d^2 x^2 \log \left (1+e^{i (a+b x)}\right )+b^3 d^3 x^3 \log \left (1+e^{i (a+b x)}\right )-3 i d \left (2 d^2+b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )+3 i d \left (2 d^2+b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )+6 b c d^2 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )+6 b d^3 x \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )-6 b c d^2 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )-6 b d^3 x \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )+6 i d^3 \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )-6 i d^3 \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )}{2 b^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3042, 4674, 3042, 4671, 2715, 2838, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4674

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4671

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

\(\displaystyle \frac {1}{2} \left (\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b}-\frac {2 i d \left (\frac {d \int e^{-i (a+b x)} \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b}\right )}{b}\right )}{b}-\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b}-\frac {2 i d \left (\frac {d \int e^{-i (a+b x)} \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b}\right )}{b}\right )}{b}-\frac {2 (c+d x)^3 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}\right )+\frac {3 d^2 \left (-\frac {2 (c+d x) \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {i d \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {i d \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}\right )}{b^2}-\frac {3 d (c+d x)^2 \csc (a+b x)}{2 b^2}-\frac {(c+d x)^3 \cot (a+b x) \csc (a+b x)}{2 b}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {3 d^2 \left (-\frac {2 (c+d x) \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {i d \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {i d \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}\right )}{b^2}+\frac {1}{2} \left (-\frac {2 (c+d x)^3 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b}-\frac {2 i d \left (\frac {d \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b}\right )}{b}\right )}{b}-\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b}-\frac {2 i d \left (\frac {d \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )}{b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b}\right )}{b}\right )}{b}\right )-\frac {3 d (c+d x)^2 \csc (a+b x)}{2 b^2}-\frac {(c+d x)^3 \cot (a+b x) \csc (a+b x)}{2 b}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2715

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

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 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

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 4674

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbo 
l] :> Simp[(-b^2)*(c + d*x)^m*Cot[e + f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n 
- 1))), x] + (-Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^ 
2*(n - 1)*(n - 2))), x] + Simp[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))) 
Int[(c + d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Simp[b^2*((n - 2)/ 
(n - 1))   Int[(c + d*x)^m*(b*Csc[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c 
, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]
 

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. 1055 vs. \(2 (275 ) = 550\).

Time = 1.26 (sec) , antiderivative size = 1056, normalized size of antiderivative = 3.42

method	result	size
risch	\(\text {Expression too large to display}\)	\(1056\)

Input:

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

Output:

3/b^4*d^3*ln(1-exp(I*(b*x+a)))*a+1/2/b*d^3*ln(1-exp(I*(b*x+a)))*x^3+1/2/b^ 
4*d^3*ln(1-exp(I*(b*x+a)))*a^3-3/b^3*d^3*polylog(3,-exp(I*(b*x+a)))*x+3/b^ 
3*d^3*polylog(3,exp(I*(b*x+a)))*x-3/b^3*d^3*ln(exp(I*(b*x+a))+1)*x+3/b^3*d 
^3*ln(1-exp(I*(b*x+a)))*x-3/b^3*c*d^2*polylog(3,-exp(I*(b*x+a)))+3/b^3*c*d 
^2*polylog(3,exp(I*(b*x+a)))-6/b^3*c*d^2*arctanh(exp(I*(b*x+a)))+3/2/b^2*c 
^2*d*ln(1-exp(I*(b*x+a)))*a-3/2/b*c*d^2*ln(exp(I*(b*x+a))+1)*x^2+3/2/b*c*d 
^2*ln(1-exp(I*(b*x+a)))*x^2-3/2/b*c^2*d*ln(exp(I*(b*x+a))+1)*x-3/2/b^2*c^2 
*d*ln(exp(I*(b*x+a))+1)*a+3/2/b*c^2*d*ln(1-exp(I*(b*x+a)))*x-3/b^3*c*d^2*a 
^2*arctanh(exp(I*(b*x+a)))+3/2/b^3*c*d^2*ln(exp(I*(b*x+a))+1)*a^2-3/2/b^3* 
c*d^2*ln(1-exp(I*(b*x+a)))*a^2+3/b^2*c^2*d*a*arctanh(exp(I*(b*x+a)))+3/2*I 
/b^2*d^3*polylog(2,-exp(I*(b*x+a)))*x^2-3/2*I/b^2*d^3*polylog(2,exp(I*(b*x 
+a)))*x^2+3/2*I/b^2*c^2*d*polylog(2,-exp(I*(b*x+a)))-3/2*I/b^2*c^2*d*polyl 
og(2,exp(I*(b*x+a)))+3*I*d^3*polylog(2,-exp(I*(b*x+a)))/b^4+3*I*d^3*polylo 
g(4,exp(I*(b*x+a)))/b^4+1/b^4*d^3*a^3*arctanh(exp(I*(b*x+a)))+6/b^4*d^3*a* 
arctanh(exp(I*(b*x+a)))-1/2/b*d^3*ln(exp(I*(b*x+a))+1)*x^3-1/2/b^4*d^3*ln( 
exp(I*(b*x+a))+1)*a^3-3/b^4*d^3*ln(exp(I*(b*x+a))+1)*a+3*I/b^2*c*d^2*polyl 
og(2,-exp(I*(b*x+a)))*x-3*I/b^2*c*d^2*polylog(2,exp(I*(b*x+a)))*x-1/b*c^3* 
arctanh(exp(I*(b*x+a)))-3*I*d^3*polylog(2,exp(I*(b*x+a)))/b^4-3*I*d^3*poly 
log(4,-exp(I*(b*x+a)))/b^4+1/b^2/(exp(2*I*(b*x+a))-1)^2*(d^3*x^3*b*exp(3*I 
*(b*x+a))+3*c*d^2*x^2*b*exp(3*I*(b*x+a))+3*c^2*d*x*b*exp(3*I*(b*x+a))+d...
 

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. 1744 vs. \(2 (265) = 530\).

Time = 0.15 (sec) , antiderivative size = 1744, normalized size of antiderivative = 5.64 \[ \int (c+d x)^3 \csc ^3(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((c + d*x)**3*csc(a + b*x)**3, 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. 3886 vs. \(2 (265) = 530\).

Time = 1.02 (sec) , antiderivative size = 3886, normalized size of antiderivative = 12.58 \[ \int (c+d x)^3 \csc ^3(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(c^3*(2*cos(b*x + a)/(cos(b*x + a)^2 - 1) - log(cos(b*x + a) + 1) + lo 
g(cos(b*x + a) - 1)) - 3*a*c^2*d*(2*cos(b*x + a)/(cos(b*x + a)^2 - 1) - lo 
g(cos(b*x + a) + 1) + log(cos(b*x + a) - 1))/b + 3*a^2*c*d^2*(2*cos(b*x + 
a)/(cos(b*x + a)^2 - 1) - log(cos(b*x + a) + 1) + log(cos(b*x + a) - 1))/b 
^2 - a^3*d^3*(2*cos(b*x + a)/(cos(b*x + a)^2 - 1) - log(cos(b*x + a) + 1) 
+ log(cos(b*x + a) - 1))/b^3 - 4*(2*((b*x + a)^3*d^3 + 6*b*c*d^2 - 6*a*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 + 2 
)*d^3)*(b*x + a) + ((b*x + a)^3*d^3 + 6*b*c*d^2 - 6*a*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 + 2)*d^3)*(b*x + a)) 
*cos(4*b*x + 4*a) - 2*((b*x + a)^3*d^3 + 6*b*c*d^2 - 6*a*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 + 2)*d^3)*(b*x + 
a))*cos(2*b*x + 2*a) - (-I*(b*x + a)^3*d^3 - 6*I*b*c*d^2 + 6*I*a*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 - 2*I)*d^3)*(b*x + a))*sin(4*b*x + 4*a) - 2*(I*(b*x + a)^3*d^3 + 6*I*b 
*c*d^2 - 6*I*a*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 + 2*I)*d^3)*(b*x + a))*sin(2*b*x + 2*a))*arctan2( 
sin(b*x + a), cos(b*x + a) + 1) - 12*(b*c*d^2 - a*d^3 + (b*c*d^2 - a*d^3)* 
cos(4*b*x + 4*a) - 2*(b*c*d^2 - a*d^3)*cos(2*b*x + 2*a) + (I*b*c*d^2 - I*a 
*d^3)*sin(4*b*x + 4*a) + 2*(-I*b*c*d^2 + I*a*d^3)*sin(2*b*x + 2*a))*arctan 
2(sin(b*x + a), cos(b*x + a) - 1) + 2*((b*x + a)^3*d^3 + 3*(b*c*d^2 - a...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

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