\(\int (c+d x)^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx\) [279]

Optimal result

Integrand size = 24, antiderivative size = 601 \[ \int (c+d x)^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\frac {12 i c d^2 x \arctan \left (e^{i (a+b x)}\right )}{b^2}+\frac {6 i d^3 x^2 \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {6 d^3 x \text {arctanh}\left (e^{i (a+b x)}\right )}{b^3}-\frac {3 (c+d x)^3 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}-\frac {3 c d^2 \text {arctanh}(\cos (a+b x))}{b^3}-\frac {3 c^2 d \text {arctanh}(\sin (a+b x))}{b^2}-\frac {3 c^2 d \csc (a+b x)}{2 b^2}-\frac {3 c d^2 x \csc (a+b x)}{b^2}-\frac {3 d^3 x^2 \csc (a+b x)}{2 b^2}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^4}+\frac {9 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}-\frac {6 i c d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^3 x \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {6 i c d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^3 x \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^4}-\frac {9 i d (c+d x)^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}-\frac {9 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {6 d^3 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^4}-\frac {6 d^3 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^4}+\frac {9 d^2 (c+d x) \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac {9 i d^3 \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^4}+\frac {9 i d^3 \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )}{b^4}+\frac {3 (c+d x)^3 \sec (a+b x)}{2 b}-\frac {(c+d x)^3 \csc ^2(a+b x) \sec (a+b x)}{2 b} \] Output:

6*I*c*d^2*polylog(2,I*exp(I*(b*x+a)))/b^3+6*I*d^3*x*polylog(2,I*exp(I*(b*x 
+a)))/b^3+6*I*d^3*x^2*arctan(exp(I*(b*x+a)))/b^2-9*d^2*(d*x+c)*polylog(3,- 
exp(I*(b*x+a)))/b^3+9*d^2*(d*x+c)*polylog(3,exp(I*(b*x+a)))/b^3-3*c^2*d*ar 
ctanh(sin(b*x+a))/b^2-3*c*d^2*arctanh(cos(b*x+a))/b^3-1/2*(d*x+c)^3*csc(b* 
x+a)^2*sec(b*x+a)/b-3/2*c^2*d*csc(b*x+a)/b^2-3/2*d^3*x^2*csc(b*x+a)/b^2-6* 
d^3*x*arctanh(exp(I*(b*x+a)))/b^3-9*I*d^3*polylog(4,-exp(I*(b*x+a)))/b^4+3 
*I*d^3*polylog(2,-exp(I*(b*x+a)))/b^4+9*I*d^3*polylog(4,exp(I*(b*x+a)))/b^ 
4+12*I*c*d^2*x*arctan(exp(I*(b*x+a)))/b^2+9/2*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*(d*x+c)^3*sec(b 
*x+a)/b-3*c*d^2*x*csc(b*x+a)/b^2-6*I*c*d^2*polylog(2,-I*exp(I*(b*x+a)))/b^ 
3-6*I*d^3*x*polylog(2,-I*exp(I*(b*x+a)))/b^3-9/2*I*d*(d*x+c)^2*polylog(2,e 
xp(I*(b*x+a)))/b^2+6*d^3*polylog(3,-I*exp(I*(b*x+a)))/b^4-6*d^3*polylog(3, 
I*exp(I*(b*x+a)))/b^4-3*(d*x+c)^3*arctanh(exp(I*(b*x+a)))/b
 

Mathematica [A] (verified)

Time = 7.37 (sec) , antiderivative size = 907, normalized size of antiderivative = 1.51 \[ \int (c+d x)^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=-\frac {3 d \left (-2 i b^2 c^2 \arctan \left (e^{i (a+b x)}\right )+2 b^2 c d x \log \left (1-i e^{i (a+b x)}\right )+b^2 d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )-2 b^2 c d x \log \left (1+i e^{i (a+b x)}\right )-b^2 d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+2 i b d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )-2 i b d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )-2 d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )+2 d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )\right )}{b^4}+\frac {3 \left (b^3 c^3 \log \left (1-e^{i (a+b x)}\right )+2 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 )+2 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 )-2 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 )-2 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 )+i d \left (2 d^2+3 b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )-i d \left (2 d^2+3 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 )\right )}{2 b^4}-\frac {\csc ^2(a+b x) \sec (a+b x) \left (-b c^3-3 b c^2 d x-3 b c d^2 x^2-b d^3 x^3+3 b c^3 \cos (2 a+2 b x)+9 b c^2 d x \cos (2 a+2 b x)+9 b c d^2 x^2 \cos (2 a+2 b x)+3 b d^3 x^3 \cos (2 a+2 b x)+3 c^2 d \sin (2 a+2 b x)+6 c d^2 x \sin (2 a+2 b x)+3 d^3 x^2 \sin (2 a+2 b x)\right )}{4 b^2} \] Input:

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

Output:

(-3*d*((-2*I)*b^2*c^2*ArcTan[E^(I*(a + b*x))] + 2*b^2*c*d*x*Log[1 - I*E^(I 
*(a + b*x))] + b^2*d^2*x^2*Log[1 - I*E^(I*(a + b*x))] - 2*b^2*c*d*x*Log[1 
+ I*E^(I*(a + b*x))] - b^2*d^2*x^2*Log[1 + I*E^(I*(a + b*x))] + (2*I)*b*d* 
(c + d*x)*PolyLog[2, (-I)*E^(I*(a + b*x))] - (2*I)*b*d*(c + d*x)*PolyLog[2 
, I*E^(I*(a + b*x))] - 2*d^2*PolyLog[3, (-I)*E^(I*(a + b*x))] + 2*d^2*Poly 
Log[3, I*E^(I*(a + b*x))]))/b^4 + (3*(b^3*c^3*Log[1 - E^(I*(a + b*x))] + 2 
*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))] 
 + 2*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))] - 2*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))] - 2*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))] + I*d*(2*d^2 + 3*b 
^2*(c + d*x)^2)*PolyLog[2, -E^(I*(a + b*x))] - I*d*(2*d^2 + 3*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))]))/(2*b^4) - (Csc[a + 
 b*x]^2*Sec[a + b*x]*(-(b*c^3) - 3*b*c^2*d*x - 3*b*c*d^2*x^2 - b*d^3*x^3 + 
 3*b*c^3*Cos[2*a + 2*b*x] + 9*b*c^2*d*x*Cos[2*a + 2*b*x] + 9*b*c*d^2*x^2*C 
os[2*a + 2*b*x] + 3*b*d^3*x^3*Cos[2*a + 2*b*x] + 3*c^2*d*Sin[2*a + 2*b*...
 

Rubi [A] (verified)

Time = 2.37 (sec) , antiderivative size = 623, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4920, 27, 7292, 27, 7293, 2009}

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.

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

\(\Big \downarrow \) 4920

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 d \left (\frac {8 i c d x \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {4 i d^2 x^2 \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {2 c d \text {arctanh}(\cos (a+b x))}{b^2}-\frac {4 d^2 x \text {arctanh}\left (e^{i (a+b x)}\right )}{b^2}-\frac {2 c^2 \text {arctanh}(\sin (a+b x))}{b}-\frac {2 (c+d x)^3 \text {arctanh}\left (e^{i (a+b x)}\right )}{d}+\frac {(c+d x)^3 \text {arctanh}(\cos (a+b x))}{d}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^3}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^3}+\frac {4 d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}-\frac {4 d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^2 \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )}{b^3}-\frac {6 d (c+d x) \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^2}+\frac {6 d (c+d x) \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^2}-\frac {4 i c d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}+\frac {4 i c d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {4 i d^2 x \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}+\frac {4 i d^2 x \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {c^2 \csc (a+b x)}{b}+\frac {3 i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b}-\frac {3 i (c+d x)^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b}-\frac {2 c d x \csc (a+b x)}{b}-\frac {d^2 x^2 \csc (a+b x)}{b}\right )}{2 b}-\frac {3 (c+d x)^3 \text {arctanh}(\cos (a+b x))}{2 b}+\frac {3 (c+d x)^3 \sec (a+b x)}{2 b}-\frac {(c+d x)^3 \csc ^2(a+b x) \sec (a+b x)}{2 b}\)

Input:

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

Output:

(-3*(c + d*x)^3*ArcTanh[Cos[a + b*x]])/(2*b) + (3*d*(((8*I)*c*d*x*ArcTan[E 
^(I*(a + b*x))])/b + ((4*I)*d^2*x^2*ArcTan[E^(I*(a + b*x))])/b - (4*d^2*x* 
ArcTanh[E^(I*(a + b*x))])/b^2 - (2*(c + d*x)^3*ArcTanh[E^(I*(a + b*x))])/d 
 - (2*c*d*ArcTanh[Cos[a + b*x]])/b^2 + ((c + d*x)^3*ArcTanh[Cos[a + b*x]]) 
/d - (2*c^2*ArcTanh[Sin[a + b*x]])/b - (c^2*Csc[a + b*x])/b - (2*c*d*x*Csc 
[a + b*x])/b - (d^2*x^2*Csc[a + b*x])/b + ((2*I)*d^2*PolyLog[2, -E^(I*(a + 
 b*x))])/b^3 + ((3*I)*(c + d*x)^2*PolyLog[2, -E^(I*(a + b*x))])/b - ((4*I) 
*c*d*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^2 - ((4*I)*d^2*x*PolyLog[2, (-I)* 
E^(I*(a + b*x))])/b^2 + ((4*I)*c*d*PolyLog[2, I*E^(I*(a + b*x))])/b^2 + (( 
4*I)*d^2*x*PolyLog[2, I*E^(I*(a + b*x))])/b^2 - ((2*I)*d^2*PolyLog[2, E^(I 
*(a + b*x))])/b^3 - ((3*I)*(c + d*x)^2*PolyLog[2, E^(I*(a + b*x))])/b - (6 
*d*(c + d*x)*PolyLog[3, -E^(I*(a + b*x))])/b^2 + (4*d^2*PolyLog[3, (-I)*E^ 
(I*(a + b*x))])/b^3 - (4*d^2*PolyLog[3, I*E^(I*(a + b*x))])/b^3 + (6*d*(c 
+ d*x)*PolyLog[3, E^(I*(a + b*x))])/b^2 - ((6*I)*d^2*PolyLog[4, -E^(I*(a + 
 b*x))])/b^3 + ((6*I)*d^2*PolyLog[4, E^(I*(a + b*x))])/b^3))/(2*b) + (3*(c 
 + d*x)^3*Sec[a + b*x])/(2*b) - ((c + d*x)^3*Csc[a + b*x]^2*Sec[a + b*x])/ 
(2*b)
 

Defintions of rubi rules used

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4920

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> Module[{u = IntHide[Csc[a + b*x]^n*Sec[a + b* 
x]^p, x]}, Simp[(c + d*x)^m   u, x] - Simp[d*m   Int[(c + d*x)^(m - 1)*u, x 
], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, 
p]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1612 vs. \(2 (535 ) = 1070\).

Time = 1.06 (sec) , antiderivative size = 1613, normalized size of antiderivative = 2.68

method	result	size
risch	\(\text {Expression too large to display}\)	\(1613\)

Input:

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

Output:

-9/2/b^3*c*d^2*ln(1-exp(I*(b*x+a)))*a^2+3/b^4*a^2*d^3*ln(1-I*exp(I*(b*x+a) 
))+1/b^2/(exp(2*I*(b*x+a))-1)^2/(exp(2*I*(b*x+a))+1)*(3*d^3*x^3*b*exp(5*I* 
(b*x+a))+9*c*d^2*x^2*b*exp(5*I*(b*x+a))+9*c^2*d*x*b*exp(5*I*(b*x+a))-2*d^3 
*x^3*b*exp(3*I*(b*x+a))+3*b*c^3*exp(5*I*(b*x+a))-6*c*d^2*x^2*b*exp(3*I*(b* 
x+a))+3*I*d^3*x^2*exp(I*(b*x+a))-6*c^2*d*x*b*exp(3*I*(b*x+a))+3*d^3*x^3*b* 
exp(I*(b*x+a))+3*I*c^2*d*exp(I*(b*x+a))-2*b*c^3*exp(3*I*(b*x+a))+9*c*d^2*x 
^2*b*exp(I*(b*x+a))-3*I*c^2*d*exp(5*I*(b*x+a))+9*c^2*d*x*b*exp(I*(b*x+a))+ 
3*b*c^3*exp(I*(b*x+a))-3*I*d^3*x^2*exp(5*I*(b*x+a))+6*I*c*d^2*x*exp(I*(b*x 
+a))-6*I*c*d^2*x*exp(5*I*(b*x+a)))+3*d^2/b^3*c*ln(exp(I*(b*x+a))-1)-3*d^2/ 
b^3*c*ln(exp(I*(b*x+a))+1)-3*d^3/b^3*ln(exp(I*(b*x+a))+1)*x+3*d^3/b^3*ln(1 
-exp(I*(b*x+a)))*x+3*d^3/b^4*ln(1-exp(I*(b*x+a)))*a-3*d^3/b^4*a*ln(exp(I*( 
b*x+a))-1)+6/b^2*c*d^2*ln(I*exp(I*(b*x+a))+1)*x-6/b^2*c*d^2*ln(1-I*exp(I*( 
b*x+a)))*x+6/b^3*c*d^2*ln(I*exp(I*(b*x+a))+1)*a-6/b^3*c*d^2*ln(1-I*exp(I*( 
b*x+a)))*a+6*I/b^2*c^2*d*arctan(exp(I*(b*x+a)))+6*I/b^4*a*d^3*dilog(I*exp( 
I*(b*x+a))+1)-6*I/b^4*a*d^3*dilog(1-I*exp(I*(b*x+a)))+9/2*I/b^2*d^3*polylo 
g(2,-exp(I*(b*x+a)))*x^2+9/2*I/b^2*d*c^2*polylog(2,-exp(I*(b*x+a)))-9/2*I/ 
b^2*d*c^2*polylog(2,exp(I*(b*x+a)))-9/2*I/b^2*d^3*polylog(2,exp(I*(b*x+a)) 
)*x^2-12*I/b^3*a*d^2*c*arctan(exp(I*(b*x+a)))-3/b^4*a^2*d^3*ln(I*exp(I*(b* 
x+a))+1)+3/b^2*d^3*ln(I*exp(I*(b*x+a))+1)*x^2-3/b^2*d^3*ln(1-I*exp(I*(b*x+ 
a)))*x^2-3/2/b^4*d^3*a^3*ln(exp(I*(b*x+a))-1)-9/b^3*c*d^2*polylog(3,-ex...
 

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. 3173 vs. \(2 (509) = 1018\).

Time = 0.23 (sec) , antiderivative size = 3173, normalized size of antiderivative = 5.28 \[ \int (c+d x)^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^3*csc(b*x+a)^3*sec(b*x+a)^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

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

integrate((d*x+c)**3*csc(b*x+a)**3*sec(b*x+a)**2,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. 8032 vs. \(2 (509) = 1018\).

Time = 4.88 (sec) , antiderivative size = 8032, normalized size of antiderivative = 13.36 \[ \int (c+d x)^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^3*csc(b*x+a)^3*sec(b*x+a)^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

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

Output:

integrate((d*x + c)^3*csc(b*x + a)^3*sec(b*x + a)^2, x)
 

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int (c+d x)^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\frac {8 \cos \left (b x +a \right ) \left (\int \csc \left (b x +a \right )^{3} \sec \left (b x +a \right )^{2} x^{3}d x \right ) \sin \left (b x +a \right )^{2} b \,d^{3}+24 \cos \left (b x +a \right ) \left (\int \csc \left (b x +a \right )^{3} \sec \left (b x +a \right )^{2} x^{2}d x \right ) \sin \left (b x +a \right )^{2} b c \,d^{2}+24 \cos \left (b x +a \right ) \left (\int \csc \left (b x +a \right )^{3} \sec \left (b x +a \right )^{2} x d x \right ) \sin \left (b x +a \right )^{2} b \,c^{2} d +12 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2} c^{3}-9 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} c^{3}+12 \sin \left (b x +a \right )^{2} c^{3}-4 c^{3}}{8 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} b} \] Input:

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

Output:

(8*cos(a + b*x)*int(csc(a + b*x)**3*sec(a + b*x)**2*x**3,x)*sin(a + b*x)** 
2*b*d**3 + 24*cos(a + b*x)*int(csc(a + b*x)**3*sec(a + b*x)**2*x**2,x)*sin 
(a + b*x)**2*b*c*d**2 + 24*cos(a + b*x)*int(csc(a + b*x)**3*sec(a + b*x)** 
2*x,x)*sin(a + b*x)**2*b*c**2*d + 12*cos(a + b*x)*log(tan((a + b*x)/2))*si 
n(a + b*x)**2*c**3 - 9*cos(a + b*x)*sin(a + b*x)**2*c**3 + 12*sin(a + b*x) 
**2*c**3 - 4*c**3)/(8*cos(a + b*x)*sin(a + b*x)**2*b)