\(\int (c+d x)^4 \csc (a+b x) \sec ^3(a+b x) \, dx\) [310]

Optimal result

Integrand size = 22, antiderivative size = 399 \[ \int (c+d x)^4 \csc (a+b x) \sec ^3(a+b x) \, dx=\frac {2 i d (c+d x)^3}{b^2}+\frac {(c+d x)^4}{2 b}-\frac {2 (c+d x)^4 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}-\frac {6 d^2 (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {6 i d^3 (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^4}+\frac {2 i d (c+d x)^3 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x)^3 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac {3 d^4 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{b^5}-\frac {3 d^2 (c+d x)^2 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{b^3}+\frac {3 d^2 (c+d x)^2 \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{b^3}-\frac {3 i d^3 (c+d x) \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{b^4}+\frac {3 i d^3 (c+d x) \operatorname {PolyLog}\left (4,e^{2 i (a+b x)}\right )}{b^4}+\frac {3 d^4 \operatorname {PolyLog}\left (5,-e^{2 i (a+b x)}\right )}{2 b^5}-\frac {3 d^4 \operatorname {PolyLog}\left (5,e^{2 i (a+b x)}\right )}{2 b^5}-\frac {2 d (c+d x)^3 \tan (a+b x)}{b^2}+\frac {(c+d x)^4 \tan ^2(a+b x)}{2 b} \] Output:

-2*I*d*(d*x+c)^3*polylog(2,exp(2*I*(b*x+a)))/b^2+1/2*(d*x+c)^4/b-2*(d*x+c) 
^4*arctanh(exp(2*I*(b*x+a)))/b-6*d^2*(d*x+c)^2*ln(1+exp(2*I*(b*x+a)))/b^3+ 
3*I*d^3*(d*x+c)*polylog(4,exp(2*I*(b*x+a)))/b^4+2*I*d*(d*x+c)^3/b^2+6*I*d^ 
3*(d*x+c)*polylog(2,-exp(2*I*(b*x+a)))/b^4-3*d^4*polylog(3,-exp(2*I*(b*x+a 
)))/b^5-3*d^2*(d*x+c)^2*polylog(3,-exp(2*I*(b*x+a)))/b^3+3*d^2*(d*x+c)^2*p 
olylog(3,exp(2*I*(b*x+a)))/b^3-3*I*d^3*(d*x+c)*polylog(4,-exp(2*I*(b*x+a)) 
)/b^4+2*I*d*(d*x+c)^3*polylog(2,-exp(2*I*(b*x+a)))/b^2+3/2*d^4*polylog(5,- 
exp(2*I*(b*x+a)))/b^5-3/2*d^4*polylog(5,exp(2*I*(b*x+a)))/b^5-2*d*(d*x+c)^ 
3*tan(b*x+a)/b^2+1/2*(d*x+c)^4*tan(b*x+a)^2/b
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2227\) vs. \(2(399)=798\).

Time = 7.01 (sec) , antiderivative size = 2227, normalized size of antiderivative = 5.58 \[ \int (c+d x)^4 \csc (a+b x) \sec ^3(a+b x) \, dx=\text {Result too large to show} \] Input:

Integrate[(c + d*x)^4*Csc[a + b*x]*Sec[a + b*x]^3,x]
 

Output:

-((c^2*d^2*E^(I*a)*Csc[a]*((2*b^3*x^3)/E^((2*I)*a) + (3*I)*b^2*(1 - E^((-2 
*I)*a))*x^2*Log[1 - E^((-I)*(a + b*x))] + (3*I)*b^2*(1 - E^((-2*I)*a))*x^2 
*Log[1 + E^((-I)*(a + b*x))] - 6*b*(1 - E^((-2*I)*a))*x*PolyLog[2, -E^((-I 
)*(a + b*x))] - 6*b*(1 - E^((-2*I)*a))*x*PolyLog[2, E^((-I)*(a + b*x))] + 
(6*I)*(1 - E^((-2*I)*a))*PolyLog[3, -E^((-I)*(a + b*x))] + (6*I)*(1 - E^(( 
-2*I)*a))*PolyLog[3, E^((-I)*(a + b*x))]))/b^3) - (c*d^3*E^(I*a)*Csc[a]*(( 
b^4*x^4)/E^((2*I)*a) + (2*I)*b^3*(1 - E^((-2*I)*a))*x^3*Log[1 - E^((-I)*(a 
 + b*x))] + (2*I)*b^3*(1 - E^((-2*I)*a))*x^3*Log[1 + E^((-I)*(a + b*x))] - 
 6*b^2*(1 - E^((-2*I)*a))*x^2*PolyLog[2, -E^((-I)*(a + b*x))] - 6*b^2*(1 - 
 E^((-2*I)*a))*x^2*PolyLog[2, E^((-I)*(a + b*x))] + (12*I)*b*(1 - E^((-2*I 
)*a))*x*PolyLog[3, -E^((-I)*(a + b*x))] + (12*I)*b*(1 - E^((-2*I)*a))*x*Po 
lyLog[3, E^((-I)*(a + b*x))] + 12*(1 - E^((-2*I)*a))*PolyLog[4, -E^((-I)*( 
a + b*x))] + 12*(1 - E^((-2*I)*a))*PolyLog[4, E^((-I)*(a + b*x))]))/b^4 - 
(d^4*E^(I*a)*Csc[a]*((2*b^5*x^5)/E^((2*I)*a) + (5*I)*b^4*(1 - E^((-2*I)*a) 
)*x^4*Log[1 - E^((-I)*(a + b*x))] + (5*I)*b^4*(1 - E^((-2*I)*a))*x^4*Log[1 
 + E^((-I)*(a + b*x))] - 20*b^3*(1 - E^((-2*I)*a))*x^3*PolyLog[2, -E^((-I) 
*(a + b*x))] - 20*b^3*(1 - E^((-2*I)*a))*x^3*PolyLog[2, E^((-I)*(a + b*x)) 
] + (60*I)*b^2*(1 - E^((-2*I)*a))*x^2*PolyLog[3, -E^((-I)*(a + b*x))] + (6 
0*I)*b^2*(1 - E^((-2*I)*a))*x^2*PolyLog[3, E^((-I)*(a + b*x))] + 120*b*(1 
- E^((-2*I)*a))*x*PolyLog[4, -E^((-I)*(a + b*x))] + 120*b*(1 - E^((-2*I...
 

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, 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.

\(\displaystyle \int (c+d x)^4 \csc (a+b x) \sec ^3(a+b x) \, dx\)

\(\Big \downarrow \) 4920

\(\displaystyle -4 d \int \frac {1}{2} (c+d x)^3 \left (\frac {\tan ^2(a+b x)}{b}+\frac {2 \log (\tan (a+b x))}{b}\right )dx+\frac {(c+d x)^4 \tan ^2(a+b x)}{2 b}+\frac {(c+d x)^4 \log (\tan (a+b x))}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -2 d \int (c+d x)^3 \left (\frac {\tan ^2(a+b x)}{b}+\frac {2 \log (\tan (a+b x))}{b}\right )dx+\frac {(c+d x)^4 \tan ^2(a+b x)}{2 b}+\frac {(c+d x)^4 \log (\tan (a+b x))}{b}\)

\(\Big \downarrow \) 7292

\(\displaystyle -2 d \int \frac {(c+d x)^3 \left (\tan ^2(a+b x)+2 \log (\tan (a+b x))\right )}{b}dx+\frac {(c+d x)^4 \tan ^2(a+b x)}{2 b}+\frac {(c+d x)^4 \log (\tan (a+b x))}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 d \int (c+d x)^3 \left (\tan ^2(a+b x)+2 \log (\tan (a+b x))\right )dx}{b}+\frac {(c+d x)^4 \tan ^2(a+b x)}{2 b}+\frac {(c+d x)^4 \log (\tan (a+b x))}{b}\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {2 d \int \left (\tan ^2(a+b x) (c+d x)^3+2 \log (\tan (a+b x)) (c+d x)^3\right )dx}{b}+\frac {(c+d x)^4 \tan ^2(a+b x)}{2 b}+\frac {(c+d x)^4 \log (\tan (a+b x))}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 d \left (\frac {(c+d x)^4 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{d}+\frac {3 d^3 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 d^3 \operatorname {PolyLog}\left (5,-e^{2 i (a+b x)}\right )}{4 b^4}+\frac {3 d^3 \operatorname {PolyLog}\left (5,e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^3}+\frac {3 i d^2 (c+d x) \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i d^2 (c+d x) \operatorname {PolyLog}\left (4,e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}-\frac {i (c+d x)^3 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b}+\frac {i (c+d x)^3 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b}+\frac {(c+d x)^3 \tan (a+b x)}{b}+\frac {(c+d x)^4 \log (\tan (a+b x))}{2 d}-\frac {i (c+d x)^3}{b}-\frac {(c+d x)^4}{4 d}\right )}{b}+\frac {(c+d x)^4 \tan ^2(a+b x)}{2 b}+\frac {(c+d x)^4 \log (\tan (a+b x))}{b}\)

Input:

Int[(c + d*x)^4*Csc[a + b*x]*Sec[a + b*x]^3,x]
 

Output:

((c + d*x)^4*Log[Tan[a + b*x]])/b + ((c + d*x)^4*Tan[a + b*x]^2)/(2*b) - ( 
2*d*(((-I)*(c + d*x)^3)/b - (c + d*x)^4/(4*d) + ((c + d*x)^4*ArcTanh[E^((2 
*I)*(a + b*x))])/d + (3*d*(c + d*x)^2*Log[1 + E^((2*I)*(a + b*x))])/b^2 + 
((c + d*x)^4*Log[Tan[a + b*x]])/(2*d) - ((3*I)*d^2*(c + d*x)*PolyLog[2, -E 
^((2*I)*(a + b*x))])/b^3 - (I*(c + d*x)^3*PolyLog[2, -E^((2*I)*(a + b*x))] 
)/b + (I*(c + d*x)^3*PolyLog[2, E^((2*I)*(a + b*x))])/b + (3*d^3*PolyLog[3 
, -E^((2*I)*(a + b*x))])/(2*b^4) + (3*d*(c + d*x)^2*PolyLog[3, -E^((2*I)*( 
a + b*x))])/(2*b^2) - (3*d*(c + d*x)^2*PolyLog[3, E^((2*I)*(a + b*x))])/(2 
*b^2) + (((3*I)/2)*d^2*(c + d*x)*PolyLog[4, -E^((2*I)*(a + b*x))])/b^3 - ( 
((3*I)/2)*d^2*(c + d*x)*PolyLog[4, E^((2*I)*(a + b*x))])/b^3 - (3*d^3*Poly 
Log[5, -E^((2*I)*(a + b*x))])/(4*b^4) + (3*d^3*PolyLog[5, E^((2*I)*(a + b* 
x))])/(4*b^4) + ((c + d*x)^3*Tan[a + b*x])/b))/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. 1728 vs. \(2 (361 ) = 722\).

Time = 0.68 (sec) , antiderivative size = 1729, normalized size of antiderivative = 4.33

method	result	size
risch	\(\text {Expression too large to display}\)	\(1729\)

Input:

int((d*x+c)^4*csc(b*x+a)*sec(b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

2*(b*d^4*x^4*exp(2*I*(b*x+a))+4*b*c*d^3*x^3*exp(2*I*(b*x+a))+6*b*c^2*d^2*x 
^2*exp(2*I*(b*x+a))+4*b*c^3*d*x*exp(2*I*(b*x+a))-2*I*d^4*x^3*exp(2*I*(b*x+ 
a))+b*c^4*exp(2*I*(b*x+a))-6*I*c*d^3*x^2*exp(2*I*(b*x+a))-6*I*c^2*d^2*x*ex 
p(2*I*(b*x+a))-2*I*d^4*x^3-2*I*c^3*d*exp(2*I*(b*x+a))-6*I*c*d^3*x^2-6*I*c^ 
2*d^2*x-2*I*c^3*d)/b^2/(exp(2*I*(b*x+a))+1)^2+12/b^3*c^2*d^2*ln(exp(I*(b*x 
+a)))+12/b^5*a^2*d^4*ln(exp(I*(b*x+a)))+1/b*d^4*ln(exp(I*(b*x+a))+1)*x^4+1 
/b*d^4*ln(1-exp(I*(b*x+a)))*x^4-1/b^5*a^4*d^4*ln(1-exp(I*(b*x+a)))-1/b*d^4 
*ln(exp(2*I*(b*x+a))+1)*x^4-6/b^3*c^2*d^2*ln(exp(2*I*(b*x+a))+1)+1/b^5*a^4 
*d^4*ln(exp(I*(b*x+a))-1)-6/b^3*d^4*ln(exp(2*I*(b*x+a))+1)*x^2+12/b^3*d^4* 
polylog(3,exp(I*(b*x+a)))*x^2-3/b^3*d^4*polylog(3,-exp(2*I*(b*x+a)))*x^2+1 
2/b^3*d^4*polylog(3,-exp(I*(b*x+a)))*x^2-3/b^3*c^2*d^2*polylog(3,-exp(2*I* 
(b*x+a)))+12/b^3*c^2*d^2*polylog(3,-exp(I*(b*x+a)))+12/b^3*c^2*d^2*polylog 
(3,exp(I*(b*x+a)))-8*I/b^5*a^3*d^4+4*I/b^2*d^4*x^3+6*I/b^2*c^2*d^2*polylog 
(2,-exp(2*I*(b*x+a)))*x-12*I/b^2*c^2*d^2*polylog(2,-exp(I*(b*x+a)))*x-12*I 
/b^2*c^2*d^2*polylog(2,exp(I*(b*x+a)))*x+24*I/b^3*c*d^3*a*x+6*I/b^2*c*d^3* 
polylog(2,-exp(2*I*(b*x+a)))*x^2-12*I/b^2*c*d^3*polylog(2,-exp(I*(b*x+a))) 
*x^2-12*I/b^2*c*d^3*polylog(2,exp(I*(b*x+a)))*x^2-4*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)))+6*I/b^4*c*d^3*pol 
ylog(2,-exp(2*I*(b*x+a)))+6*I/b^4*d^4*polylog(2,-exp(2*I*(b*x+a)))*x-12*I/ 
b^4*a^2*d^4*x-3*I/b^4*c*d^3*polylog(4,-exp(2*I*(b*x+a)))+24*I/b^4*c*d^3...
 

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. 3324 vs. \(2 (352) = 704\).

Time = 0.34 (sec) , antiderivative size = 3324, normalized size of antiderivative = 8.33 \[ \int (c+d x)^4 \csc (a+b x) \sec ^3(a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^4*csc(b*x+a)*sec(b*x+a)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int (c+d x)^4 \csc (a+b x) \sec ^3(a+b x) \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**4*csc(b*x+a)*sec(b*x+a)**3,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. 8853 vs. \(2 (352) = 704\).

Time = 4.20 (sec) , antiderivative size = 8853, normalized size of antiderivative = 22.19 \[ \int (c+d x)^4 \csc (a+b x) \sec ^3(a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^4*csc(b*x+a)*sec(b*x+a)^3,x, algorithm="maxima")
 

Output:

-1/2*(c^4*(1/(sin(b*x + a)^2 - 1) + log(sin(b*x + a)^2 - 1) - log(sin(b*x 
+ a)^2)) - 4*a*c^3*d*(1/(sin(b*x + a)^2 - 1) + log(sin(b*x + a)^2 - 1) - l 
og(sin(b*x + a)^2))/b + 6*a^2*c^2*d^2*(1/(sin(b*x + a)^2 - 1) + log(sin(b* 
x + a)^2 - 1) - log(sin(b*x + a)^2))/b^2 - 4*a^3*c*d^3*(1/(sin(b*x + a)^2 
- 1) + log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b^3 + a^4*d^4*(1/(si 
n(b*x + a)^2 - 1) + log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b^4 + 2 
*(24*b^3*c^3*d - 72*a*b^2*c^2*d^2 + 72*a^2*b*c*d^3 - 24*a^3*d^4 + 4*(3*(b* 
x + a)^4*d^4 + 9*b^2*c^2*d^2 - 18*a*b*c*d^3 + 9*a^2*d^4 + 8*(b*c*d^3 - a*d 
^4)*(b*x + a)^3 + 9*(b^2*c^2*d^2 - 2*a*b*c*d^3 + (a^2 + 1)*d^4)*(b*x + a)^ 
2 + 6*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*(a^2 + 1)*b*c*d^3 - (a^3 + 3*a)*d^4 
)*(b*x + a) + (3*(b*x + a)^4*d^4 + 9*b^2*c^2*d^2 - 18*a*b*c*d^3 + 9*a^2*d^ 
4 + 8*(b*c*d^3 - a*d^4)*(b*x + a)^3 + 9*(b^2*c^2*d^2 - 2*a*b*c*d^3 + (a^2 
+ 1)*d^4)*(b*x + a)^2 + 6*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*(a^2 + 1)*b*c*d 
^3 - (a^3 + 3*a)*d^4)*(b*x + a))*cos(4*b*x + 4*a) + 2*(3*(b*x + a)^4*d^4 + 
 9*b^2*c^2*d^2 - 18*a*b*c*d^3 + 9*a^2*d^4 + 8*(b*c*d^3 - a*d^4)*(b*x + a)^ 
3 + 9*(b^2*c^2*d^2 - 2*a*b*c*d^3 + (a^2 + 1)*d^4)*(b*x + a)^2 + 6*(b^3*c^3 
*d - 3*a*b^2*c^2*d^2 + 3*(a^2 + 1)*b*c*d^3 - (a^3 + 3*a)*d^4)*(b*x + a))*c 
os(2*b*x + 2*a) - (-3*I*(b*x + a)^4*d^4 - 9*I*b^2*c^2*d^2 + 18*I*a*b*c*d^3 
 - 9*I*a^2*d^4 + 8*(-I*b*c*d^3 + I*a*d^4)*(b*x + a)^3 + 9*(-I*b^2*c^2*d^2 
+ 2*I*a*b*c*d^3 + (-I*a^2 - I)*d^4)*(b*x + a)^2 + 6*(-I*b^3*c^3*d + 3*I...
 

Giac [F]

\[ \int (c+d x)^4 \csc (a+b x) \sec ^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{4} \csc \left (b x + a\right ) \sec \left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*x+c)^4*csc(b*x+a)*sec(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate((d*x + c)^4*csc(b*x + a)*sec(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^4 \csc (a+b x) \sec ^3(a+b x) \, dx=\text {Hanged} \] Input:

int((c + d*x)^4/(cos(a + b*x)^3*sin(a + b*x)),x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int (c+d x)^4 \csc (a+b x) \sec ^3(a+b x) \, dx=\frac {2 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x^{4}d x \right ) \sin \left (b x +a \right )^{2} b \,d^{4}-2 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x^{4}d x \right ) b \,d^{4}+8 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x^{3}d x \right ) \sin \left (b x +a \right )^{2} b c \,d^{3}-8 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x^{3}d x \right ) b c \,d^{3}+12 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x^{2}d x \right ) \sin \left (b x +a \right )^{2} b \,c^{2} d^{2}-12 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x^{2}d x \right ) b \,c^{2} d^{2}+8 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x d x \right ) \sin \left (b x +a \right )^{2} b \,c^{3} d -8 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x d x \right ) b \,c^{3} d -2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{2} c^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) c^{4}-2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{2} c^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) c^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2} c^{4}-2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) c^{4}-\sin \left (b x +a \right )^{2} c^{4}}{2 b \left (\sin \left (b x +a \right )^{2}-1\right )} \] Input:

int((d*x+c)^4*csc(b*x+a)*sec(b*x+a)^3,x)
 

Output:

(2*int(csc(a + b*x)*sec(a + b*x)**3*x**4,x)*sin(a + b*x)**2*b*d**4 - 2*int 
(csc(a + b*x)*sec(a + b*x)**3*x**4,x)*b*d**4 + 8*int(csc(a + b*x)*sec(a + 
b*x)**3*x**3,x)*sin(a + b*x)**2*b*c*d**3 - 8*int(csc(a + b*x)*sec(a + b*x) 
**3*x**3,x)*b*c*d**3 + 12*int(csc(a + b*x)*sec(a + b*x)**3*x**2,x)*sin(a + 
 b*x)**2*b*c**2*d**2 - 12*int(csc(a + b*x)*sec(a + b*x)**3*x**2,x)*b*c**2* 
d**2 + 8*int(csc(a + b*x)*sec(a + b*x)**3*x,x)*sin(a + b*x)**2*b*c**3*d - 
8*int(csc(a + b*x)*sec(a + b*x)**3*x,x)*b*c**3*d - 2*log(tan((a + b*x)/2) 
- 1)*sin(a + b*x)**2*c**4 + 2*log(tan((a + b*x)/2) - 1)*c**4 - 2*log(tan(( 
a + b*x)/2) + 1)*sin(a + b*x)**2*c**4 + 2*log(tan((a + b*x)/2) + 1)*c**4 + 
 2*log(tan((a + b*x)/2))*sin(a + b*x)**2*c**4 - 2*log(tan((a + b*x)/2))*c* 
*4 - sin(a + b*x)**2*c**4)/(2*b*(sin(a + b*x)**2 - 1))