\(\int (c+d x)^4 \csc (a+b x) \sec ^2(a+b x) \, dx\) [266]

Optimal result

Integrand size = 22, antiderivative size = 469 \[ \int (c+d x)^4 \csc (a+b x) \sec ^2(a+b x) \, dx=\frac {8 i d (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^4 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {4 i d (c+d x)^3 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {12 i d^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {12 i d^2 (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {4 i d (c+d x)^3 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac {12 d^2 (c+d x)^2 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {24 d^3 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^4}-\frac {24 d^3 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^4}+\frac {12 d^2 (c+d x)^2 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac {24 i d^3 (c+d x) \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^4}+\frac {24 i d^4 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^5}-\frac {24 i d^4 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^5}+\frac {24 i d^3 (c+d x) \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )}{b^4}+\frac {24 d^4 \operatorname {PolyLog}\left (5,-e^{i (a+b x)}\right )}{b^5}-\frac {24 d^4 \operatorname {PolyLog}\left (5,e^{i (a+b x)}\right )}{b^5}+\frac {(c+d x)^4 \sec (a+b x)}{b} \] Output:

-24*I*d^4*polylog(4,I*exp(I*(b*x+a)))/b^5-2*(d*x+c)^4*arctanh(exp(I*(b*x+a 
)))/b+12*I*d^2*(d*x+c)^2*polylog(2,I*exp(I*(b*x+a)))/b^3+8*I*d*(d*x+c)^3*a 
rctan(exp(I*(b*x+a)))/b^2+24*I*d^3*(d*x+c)*polylog(4,exp(I*(b*x+a)))/b^4-2 
4*I*d^3*(d*x+c)*polylog(4,-exp(I*(b*x+a)))/b^4-12*d^2*(d*x+c)^2*polylog(3, 
-exp(I*(b*x+a)))/b^3+24*d^3*(d*x+c)*polylog(3,-I*exp(I*(b*x+a)))/b^4-24*d^ 
3*(d*x+c)*polylog(3,I*exp(I*(b*x+a)))/b^4+12*d^2*(d*x+c)^2*polylog(3,exp(I 
*(b*x+a)))/b^3+24*I*d^4*polylog(4,-I*exp(I*(b*x+a)))/b^5-4*I*d*(d*x+c)^3*p 
olylog(2,exp(I*(b*x+a)))/b^2+4*I*d*(d*x+c)^3*polylog(2,-exp(I*(b*x+a)))/b^ 
2-12*I*d^2*(d*x+c)^2*polylog(2,-I*exp(I*(b*x+a)))/b^3+24*d^4*polylog(5,-ex 
p(I*(b*x+a)))/b^5-24*d^4*polylog(5,exp(I*(b*x+a)))/b^5+(d*x+c)^4*sec(b*x+a 
)/b
 

Mathematica [A] (verified)

Time = 2.16 (sec) , antiderivative size = 694, normalized size of antiderivative = 1.48 \[ \int (c+d x)^4 \csc (a+b x) \sec ^2(a+b x) \, dx=\frac {b^4 (c+d x)^4 \log \left (1-e^{i (a+b x)}\right )-b^4 (c+d x)^4 \log \left (1+e^{i (a+b x)}\right )-4 d \left (-2 i b^3 c^3 \arctan \left (e^{i (a+b x)}\right )+3 b^3 c^2 d x \log \left (1-i e^{i (a+b x)}\right )+3 b^3 c d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )+b^3 d^3 x^3 \log \left (1-i e^{i (a+b x)}\right )-3 b^3 c^2 d x \log \left (1+i e^{i (a+b x)}\right )-3 b^3 c d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )-b^3 d^3 x^3 \log \left (1+i e^{i (a+b x)}\right )+3 i b^2 d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )-3 i b^2 d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )-6 b c d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )-6 b d^3 x \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )+6 b c d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )+6 b d^3 x \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )-6 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )+6 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )\right )+4 i d \left (b^3 (c+d x)^3 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )+3 i b^2 d (c+d x)^2 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )-6 d^2 \left (b (c+d x) \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )+i d \operatorname {PolyLog}\left (5,-e^{i (a+b x)}\right )\right )\right )-4 i d \left (b^3 (c+d x)^3 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )+3 i b^2 d (c+d x)^2 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )-6 d^2 \left (b (c+d x) \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )+i d \operatorname {PolyLog}\left (5,e^{i (a+b x)}\right )\right )\right )+b^4 (c+d x)^4 \sec (a+b x)}{b^5} \] Input:

Integrate[(c + d*x)^4*Csc[a + b*x]*Sec[a + b*x]^2,x]
 

Output:

(b^4*(c + d*x)^4*Log[1 - E^(I*(a + b*x))] - b^4*(c + d*x)^4*Log[1 + E^(I*( 
a + b*x))] - 4*d*((-2*I)*b^3*c^3*ArcTan[E^(I*(a + b*x))] + 3*b^3*c^2*d*x*L 
og[1 - I*E^(I*(a + b*x))] + 3*b^3*c*d^2*x^2*Log[1 - I*E^(I*(a + b*x))] + b 
^3*d^3*x^3*Log[1 - I*E^(I*(a + b*x))] - 3*b^3*c^2*d*x*Log[1 + I*E^(I*(a + 
b*x))] - 3*b^3*c*d^2*x^2*Log[1 + I*E^(I*(a + b*x))] - b^3*d^3*x^3*Log[1 + 
I*E^(I*(a + b*x))] + (3*I)*b^2*d*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(a + b*x 
))] - (3*I)*b^2*d*(c + d*x)^2*PolyLog[2, I*E^(I*(a + b*x))] - 6*b*c*d^2*Po 
lyLog[3, (-I)*E^(I*(a + b*x))] - 6*b*d^3*x*PolyLog[3, (-I)*E^(I*(a + b*x)) 
] + 6*b*c*d^2*PolyLog[3, I*E^(I*(a + b*x))] + 6*b*d^3*x*PolyLog[3, I*E^(I* 
(a + b*x))] - (6*I)*d^3*PolyLog[4, (-I)*E^(I*(a + b*x))] + (6*I)*d^3*PolyL 
og[4, I*E^(I*(a + b*x))]) + (4*I)*d*(b^3*(c + d*x)^3*PolyLog[2, -E^(I*(a + 
 b*x))] + (3*I)*b^2*d*(c + d*x)^2*PolyLog[3, -E^(I*(a + b*x))] - 6*d^2*(b* 
(c + d*x)*PolyLog[4, -E^(I*(a + b*x))] + I*d*PolyLog[5, -E^(I*(a + b*x))]) 
) - (4*I)*d*(b^3*(c + d*x)^3*PolyLog[2, E^(I*(a + b*x))] + (3*I)*b^2*d*(c 
+ d*x)^2*PolyLog[3, E^(I*(a + b*x))] - 6*d^2*(b*(c + d*x)*PolyLog[4, E^(I* 
(a + b*x))] + I*d*PolyLog[5, E^(I*(a + b*x))])) + b^4*(c + d*x)^4*Sec[a + 
b*x])/b^5
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 507, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4920, 25, 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.

\(\displaystyle \int (c+d x)^4 \csc (a+b x) \sec ^2(a+b x) \, dx\)

\(\Big \downarrow \) 4920

\(\displaystyle -4 d \int -(c+d x)^3 \left (\frac {\text {arctanh}(\cos (a+b x))}{b}-\frac {\sec (a+b x)}{b}\right )dx-\frac {(c+d x)^4 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^4 \sec (a+b x)}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle 4 d \int (c+d x)^3 \left (\frac {\text {arctanh}(\cos (a+b x))}{b}-\frac {\sec (a+b x)}{b}\right )dx-\frac {(c+d x)^4 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^4 \sec (a+b x)}{b}\)

\(\Big \downarrow \) 7292

\(\displaystyle 4 d \int \frac {(c+d x)^3 (\text {arctanh}(\cos (a+b x))-\sec (a+b x))}{b}dx-\frac {(c+d x)^4 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^4 \sec (a+b x)}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 d \int (c+d x)^3 (\text {arctanh}(\cos (a+b x))-\sec (a+b x))dx}{b}-\frac {(c+d x)^4 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^4 \sec (a+b x)}{b}\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {4 d \int \left ((c+d x)^3 \text {arctanh}(\cos (a+b x))-(c+d x)^3 \sec (a+b x)\right )dx}{b}-\frac {(c+d x)^4 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^4 \sec (a+b x)}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 d \left (\frac {2 i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {(c+d x)^4 \text {arctanh}\left (e^{i (a+b x)}\right )}{2 d}+\frac {(c+d x)^4 \text {arctanh}(\cos (a+b x))}{4 d}+\frac {6 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}-\frac {6 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4}+\frac {6 d^3 \operatorname {PolyLog}\left (5,-e^{i (a+b x)}\right )}{b^4}-\frac {6 d^3 \operatorname {PolyLog}\left (5,e^{i (a+b x)}\right )}{b^4}+\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}-\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )}{b^3}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^2}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^2}+\frac {i (c+d x)^3 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b}-\frac {i (c+d x)^3 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b}\right )}{b}-\frac {(c+d x)^4 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^4 \sec (a+b x)}{b}\)

Input:

Int[(c + d*x)^4*Csc[a + b*x]*Sec[a + b*x]^2,x]
 

Output:

-(((c + d*x)^4*ArcTanh[Cos[a + b*x]])/b) + (4*d*(((2*I)*(c + d*x)^3*ArcTan 
[E^(I*(a + b*x))])/b - ((c + d*x)^4*ArcTanh[E^(I*(a + b*x))])/(2*d) + ((c 
+ d*x)^4*ArcTanh[Cos[a + b*x]])/(4*d) + (I*(c + d*x)^3*PolyLog[2, -E^(I*(a 
 + b*x))])/b - ((3*I)*d*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^2 
+ ((3*I)*d*(c + d*x)^2*PolyLog[2, I*E^(I*(a + b*x))])/b^2 - (I*(c + d*x)^3 
*PolyLog[2, E^(I*(a + b*x))])/b - (3*d*(c + d*x)^2*PolyLog[3, -E^(I*(a + b 
*x))])/b^2 + (6*d^2*(c + d*x)*PolyLog[3, (-I)*E^(I*(a + b*x))])/b^3 - (6*d 
^2*(c + d*x)*PolyLog[3, I*E^(I*(a + b*x))])/b^3 + (3*d*(c + d*x)^2*PolyLog 
[3, E^(I*(a + b*x))])/b^2 - ((6*I)*d^2*(c + d*x)*PolyLog[4, -E^(I*(a + b*x 
))])/b^3 + ((6*I)*d^3*PolyLog[4, (-I)*E^(I*(a + b*x))])/b^4 - ((6*I)*d^3*P 
olyLog[4, I*E^(I*(a + b*x))])/b^4 + ((6*I)*d^2*(c + d*x)*PolyLog[4, E^(I*( 
a + b*x))])/b^3 + (6*d^3*PolyLog[5, -E^(I*(a + b*x))])/b^4 - (6*d^3*PolyLo 
g[5, E^(I*(a + b*x))])/b^4))/b + ((c + d*x)^4*Sec[a + b*x])/b
 

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 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. 1865 vs. \(2 (422 ) = 844\).

Time = 0.77 (sec) , antiderivative size = 1866, normalized size of antiderivative = 3.98

method	result	size
risch	\(\text {Expression too large to display}\)	\(1866\)

Input:

int((d*x+c)^4*csc(b*x+a)*sec(b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

6/b^3*a^2*d^2*c^2*ln(exp(I*(b*x+a))-1)-12/b^3*d^2*c^2*polylog(3,-exp(I*(b* 
x+a)))+4/b^2*d^4*ln(I*exp(I*(b*x+a))+1)*x^3-4/b^2*d^4*ln(1-I*exp(I*(b*x+a) 
))*x^3-4/b*c^3*d*ln(exp(I*(b*x+a))+1)*x-24/b^3*c*d^3*polylog(3,-exp(I*(b*x 
+a)))*x+4/b*c^3*d*ln(1-exp(I*(b*x+a)))*x-4/b^2*a*c^3*d*ln(exp(I*(b*x+a))-1 
)+12/b^3*d^2*c^2*polylog(3,exp(I*(b*x+a)))+1/b^5*a^4*d^4*ln(exp(I*(b*x+a)) 
-1)+24/b^4*c*d^3*polylog(3,-I*exp(I*(b*x+a)))+6/b*d^2*c^2*ln(1-exp(I*(b*x+ 
a)))*x^2+12*I/b^5*a^2*d^4*dilog(1-I*exp(I*(b*x+a)))-12*I/b^3*d^2*c^2*dilog 
(I*exp(I*(b*x+a))+1)-24*I/b^4*c*d^3*polylog(4,-exp(I*(b*x+a)))-12*I/b^5*a^ 
2*d^4*polylog(2,I*exp(I*(b*x+a)))-12/b^4*c*d^3*ln(I*exp(I*(b*x+a))+1)*a^2- 
1/b^5*a^4*d^4*ln(1-exp(I*(b*x+a)))-12/b^3*d^4*polylog(3,-exp(I*(b*x+a)))*x 
^2+24/b^4*d^4*polylog(3,-I*exp(I*(b*x+a)))*x+12/b^3*d^4*polylog(3,exp(I*(b 
*x+a)))*x^2-1/b*d^4*ln(exp(I*(b*x+a))+1)*x^4-24/b^4*d^4*polylog(3,I*exp(I* 
(b*x+a)))*x+4/b^5*a^3*d^4*ln(I*exp(I*(b*x+a))+1)-4/b^5*a^3*d^4*ln(1-I*exp( 
I*(b*x+a)))+1/b*d^4*ln(1-exp(I*(b*x+a)))*x^4+2*exp(I*(b*x+a))*(d^4*x^4+4*c 
*d^3*x^3+6*c^2*d^2*x^2+4*c^3*d*x+c^4)/b/(exp(2*I*(b*x+a))+1)-6/b^3*d^2*c^2 
*ln(1-exp(I*(b*x+a)))*a^2+4/b^4*c*d^3*a^3*ln(1-exp(I*(b*x+a)))+4/b*c*d^3*l 
n(1-exp(I*(b*x+a)))*x^3-4/b*c*d^3*ln(exp(I*(b*x+a))+1)*x^3+12/b^4*c*d^3*ln 
(1-I*exp(I*(b*x+a)))*a^2+12*I/b^3*d^2*c^2*dilog(1-I*exp(I*(b*x+a)))-24*I/b 
^4*d^4*polylog(4,-exp(I*(b*x+a)))*x-4*I/b^2*c^3*d*polylog(2,exp(I*(b*x+a)) 
)+4*I/b^2*c^3*d*polylog(2,-exp(I*(b*x+a)))+12*I/b^3*d^4*polylog(2,I*exp...
 

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. 2519 vs. \(2 (403) = 806\).

Time = 0.24 (sec) , antiderivative size = 2519, normalized size of antiderivative = 5.37 \[ \int (c+d x)^4 \csc (a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^4*csc(b*x+a)*sec(b*x+a)^2,x, algorithm="fricas")
 

Output:

1/2*(2*b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 + 12*b^4*c^2*d^2*x^2 + 8*b^4*c^3*d*x 
+ 2*b^4*c^4 - 24*d^4*cos(b*x + a)*polylog(5, cos(b*x + a) + I*sin(b*x + a) 
) - 24*d^4*cos(b*x + a)*polylog(5, cos(b*x + a) - I*sin(b*x + a)) + 24*d^4 
*cos(b*x + a)*polylog(5, -cos(b*x + a) + I*sin(b*x + a)) + 24*d^4*cos(b*x 
+ a)*polylog(5, -cos(b*x + a) - I*sin(b*x + a)) - 24*I*d^4*cos(b*x + a)*po 
lylog(4, I*cos(b*x + a) + sin(b*x + a)) - 24*I*d^4*cos(b*x + a)*polylog(4, 
 I*cos(b*x + a) - sin(b*x + a)) + 24*I*d^4*cos(b*x + a)*polylog(4, -I*cos( 
b*x + a) + sin(b*x + a)) + 24*I*d^4*cos(b*x + a)*polylog(4, -I*cos(b*x + a 
) - sin(b*x + a)) - 4*(I*b^3*d^4*x^3 + 3*I*b^3*c*d^3*x^2 + 3*I*b^3*c^2*d^2 
*x + I*b^3*c^3*d)*cos(b*x + a)*dilog(cos(b*x + a) + I*sin(b*x + a)) - 4*(- 
I*b^3*d^4*x^3 - 3*I*b^3*c*d^3*x^2 - 3*I*b^3*c^2*d^2*x - I*b^3*c^3*d)*cos(b 
*x + a)*dilog(cos(b*x + a) - I*sin(b*x + a)) - 12*(-I*b^2*d^4*x^2 - 2*I*b^ 
2*c*d^3*x - I*b^2*c^2*d^2)*cos(b*x + a)*dilog(I*cos(b*x + a) + sin(b*x + a 
)) - 12*(-I*b^2*d^4*x^2 - 2*I*b^2*c*d^3*x - I*b^2*c^2*d^2)*cos(b*x + a)*di 
log(I*cos(b*x + a) - sin(b*x + a)) - 12*(I*b^2*d^4*x^2 + 2*I*b^2*c*d^3*x + 
 I*b^2*c^2*d^2)*cos(b*x + a)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - 12*(I 
*b^2*d^4*x^2 + 2*I*b^2*c*d^3*x + I*b^2*c^2*d^2)*cos(b*x + a)*dilog(-I*cos( 
b*x + a) - sin(b*x + a)) - 4*(I*b^3*d^4*x^3 + 3*I*b^3*c*d^3*x^2 + 3*I*b^3* 
c^2*d^2*x + I*b^3*c^3*d)*cos(b*x + a)*dilog(-cos(b*x + a) + I*sin(b*x + a) 
) - 4*(-I*b^3*d^4*x^3 - 3*I*b^3*c*d^3*x^2 - 3*I*b^3*c^2*d^2*x - I*b^3*c...
 

Sympy [F(-1)]

Timed out. \[ \int (c+d x)^4 \csc (a+b x) \sec ^2(a+b x) \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**4*csc(b*x+a)*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. 5709 vs. \(2 (403) = 806\).

Time = 2.67 (sec) , antiderivative size = 5709, normalized size of antiderivative = 12.17 \[ \int (c+d x)^4 \csc (a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^4*csc(b*x+a)*sec(b*x+a)^2,x, algorithm="maxima")
 

Output:

1/2*(c^4*(2/cos(b*x + a) - log(cos(b*x + a) + 1) + log(cos(b*x + a) - 1)) 
- 4*a*c^3*d*(2/cos(b*x + a) - log(cos(b*x + a) + 1) + log(cos(b*x + a) - 1 
))/b + 6*a^2*c^2*d^2*(2/cos(b*x + a) - log(cos(b*x + a) + 1) + log(cos(b*x 
 + a) - 1))/b^2 - 4*a^3*c*d^3*(2/cos(b*x + a) - log(cos(b*x + a) + 1) + lo 
g(cos(b*x + a) - 1))/b^3 + a^4*d^4*(2/cos(b*x + a) - log(cos(b*x + a) + 1) 
 + log(cos(b*x + a) - 1))/b^4 + 2*(8*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2* 
b*c*d^3 + (b*x + a)^3*d^4 - a^3*d^4 + 3*(b*c*d^3 - a*d^4)*(b*x + a)^2 + 3* 
(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(b*x + a) + (b^3*c^3*d - 3*a*b^2*c^2 
*d^2 + 3*a^2*b*c*d^3 + (b*x + a)^3*d^4 - a^3*d^4 + 3*(b*c*d^3 - a*d^4)*(b* 
x + a)^2 + 3*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(b*x + a))*cos(2*b*x + 
2*a) + (I*b^3*c^3*d - 3*I*a*b^2*c^2*d^2 + 3*I*a^2*b*c*d^3 + I*(b*x + a)^3* 
d^4 - I*a^3*d^4 + 3*(I*b*c*d^3 - I*a*d^4)*(b*x + a)^2 + 3*(I*b^2*c^2*d^2 - 
 2*I*a*b*c*d^3 + I*a^2*d^4)*(b*x + a))*sin(2*b*x + 2*a))*arctan2(cos(b*x + 
 a), sin(b*x + a) + 1) + 8*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 + 
(b*x + a)^3*d^4 - a^3*d^4 + 3*(b*c*d^3 - a*d^4)*(b*x + a)^2 + 3*(b^2*c^2*d 
^2 - 2*a*b*c*d^3 + a^2*d^4)*(b*x + a) + (b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a 
^2*b*c*d^3 + (b*x + a)^3*d^4 - a^3*d^4 + 3*(b*c*d^3 - a*d^4)*(b*x + a)^2 + 
 3*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(b*x + a))*cos(2*b*x + 2*a) + (I* 
b^3*c^3*d - 3*I*a*b^2*c^2*d^2 + 3*I*a^2*b*c*d^3 + I*(b*x + a)^3*d^4 - I*a^ 
3*d^4 + 3*(I*b*c*d^3 - I*a*d^4)*(b*x + a)^2 + 3*(I*b^2*c^2*d^2 - 2*I*a*...
 

Giac [F]

\[ \int (c+d x)^4 \csc (a+b x) \sec ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{4} \csc \left (b x + a\right ) \sec \left (b x + a\right )^{2} \,d x } \] Input:

integrate((d*x+c)^4*csc(b*x+a)*sec(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate((d*x + c)^4*csc(b*x + a)*sec(b*x + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^4 \csc (a+b x) \sec ^2(a+b x) \, dx=\text {Hanged} \] Input:

int((c + d*x)^4/(cos(a + b*x)^2*sin(a + b*x)),x)
                                                                                    
                                                                                    
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int (c+d x)^4 \csc (a+b x) \sec ^2(a+b x) \, dx=\frac {\cos \left (b x +a \right ) \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{2} x^{4}d x \right ) b \,d^{4}+4 \cos \left (b x +a \right ) \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{2} x^{3}d x \right ) b c \,d^{3}+6 \cos \left (b x +a \right ) \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{2} x^{2}d x \right ) b \,c^{2} d^{2}+4 \cos \left (b x +a \right ) \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{2} x d x \right ) b \,c^{3} d +\cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) c^{4}-\cos \left (b x +a \right ) c^{4}+c^{4}}{\cos \left (b x +a \right ) b} \] Input:

int((d*x+c)^4*csc(b*x+a)*sec(b*x+a)^2,x)
 

Output:

(cos(a + b*x)*int(csc(a + b*x)*sec(a + b*x)**2*x**4,x)*b*d**4 + 4*cos(a + 
b*x)*int(csc(a + b*x)*sec(a + b*x)**2*x**3,x)*b*c*d**3 + 6*cos(a + b*x)*in 
t(csc(a + b*x)*sec(a + b*x)**2*x**2,x)*b*c**2*d**2 + 4*cos(a + b*x)*int(cs 
c(a + b*x)*sec(a + b*x)**2*x,x)*b*c**3*d + cos(a + b*x)*log(tan((a + b*x)/ 
2))*c**4 - cos(a + b*x)*c**4 + c**4)/(cos(a + b*x)*b)