\(\int (c+d x)^3 \csc (a+b x) \sec ^3(a+b x) \, dx\) [311]

Optimal result

Integrand size = 22, antiderivative size = 325 \[ \int (c+d x)^3 \csc (a+b x) \sec ^3(a+b x) \, dx=\frac {3 i d (c+d x)^2}{2 b^2}+\frac {(c+d x)^3}{2 b}-\frac {2 (c+d x)^3 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}-\frac {3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}+\frac {3 i d^3 \operatorname {PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac {(c+d x)^3 \tan ^2(a+b x)}{2 b} \] Output:

3/2*I*d*(d*x+c)^2/b^2+1/2*(d*x+c)^3/b-2*(d*x+c)^3*arctanh(exp(2*I*(b*x+a)) 
)/b-3*d^2*(d*x+c)*ln(1+exp(2*I*(b*x+a)))/b^3+3/2*I*d^3*polylog(2,-exp(2*I* 
(b*x+a)))/b^4+3/2*I*d*(d*x+c)^2*polylog(2,-exp(2*I*(b*x+a)))/b^2-3/2*I*d*( 
d*x+c)^2*polylog(2,exp(2*I*(b*x+a)))/b^2-3/2*d^2*(d*x+c)*polylog(3,-exp(2* 
I*(b*x+a)))/b^3+3/2*d^2*(d*x+c)*polylog(3,exp(2*I*(b*x+a)))/b^3-3/4*I*d^3* 
polylog(4,-exp(2*I*(b*x+a)))/b^4+3/4*I*d^3*polylog(4,exp(2*I*(b*x+a)))/b^4 
-3/2*d*(d*x+c)^2*tan(b*x+a)/b^2+1/2*(d*x+c)^3*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. \(1535\) vs. \(2(325)=650\).

Time = 6.59 (sec) , antiderivative size = 1535, normalized size of antiderivative = 4.72 \[ \int (c+d x)^3 \csc (a+b x) \sec ^3(a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(c*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 - (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*Poly 
Log[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))]))/(4*b^4) 
+ (x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*Csc[a]*Sec[a])/4 - ((I/4) 
*c*d^2*(2*b^2*x^2*(2*b*x - (3*I)*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)*(a + 
b*x))]) + 6*b*(1 + E^((2*I)*a))*x*PolyLog[2, -E^((-2*I)*(a + b*x))] - (3*I 
)*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/(b^3*E^(I*a 
)) - ((I/8)*d^3*E^(I*a)*((2*b^4*x^4)/E^((2*I)*a) - (4*I)*b^3*(1 + E^((-2*I 
)*a))*x^3*Log[1 + E^((-2*I)*(a + b*x))] + 6*b^2*(1 + E^((-2*I)*a))*x^2*Pol 
yLog[2, -E^((-2*I)*(a + b*x))] - (6*I)*b*(1 + E^((-2*I)*a))*x*PolyLog[3...
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.07, 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.

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

\(\Big \downarrow \) 4920

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((c + d*x)^3*Log[Tan[a + b*x]])/b + ((c + d*x)^3*Tan[a + b*x]^2)/(2*b) - ( 
3*d*(((-I)*(c + d*x)^2)/b - (c + d*x)^3/(3*d) + (4*(c + d*x)^3*ArcTanh[E^( 
(2*I)*(a + b*x))])/(3*d) + (2*d*(c + d*x)*Log[1 + E^((2*I)*(a + b*x))])/b^ 
2 + (2*(c + d*x)^3*Log[Tan[a + b*x]])/(3*d) - (I*d^2*PolyLog[2, -E^((2*I)* 
(a + b*x))])/b^3 - (I*(c + d*x)^2*PolyLog[2, -E^((2*I)*(a + b*x))])/b + (I 
*(c + d*x)^2*PolyLog[2, E^((2*I)*(a + b*x))])/b + (d*(c + d*x)*PolyLog[3, 
-E^((2*I)*(a + b*x))])/b^2 - (d*(c + d*x)*PolyLog[3, E^((2*I)*(a + b*x))]) 
/b^2 + ((I/2)*d^2*PolyLog[4, -E^((2*I)*(a + b*x))])/b^3 - ((I/2)*d^2*PolyL 
og[4, E^((2*I)*(a + b*x))])/b^3 + ((c + d*x)^2*Tan[a + b*x])/b))/(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. 1114 vs. \(2 (279 ) = 558\).

Time = 0.56 (sec) , antiderivative size = 1115, normalized size of antiderivative = 3.43

method	result	size
risch	\(\text {Expression too large to display}\)	\(1115\)

Input:

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

Output:

-3/b^3*c*d^2*ln(1-exp(I*(b*x+a)))*a^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+(2*b*d^3*x^3*exp(2*I*(b*x 
+a))-3*I*d^3*x^2*exp(2*I*(b*x+a))+6*b*c*d^2*x^2*exp(2*I*(b*x+a))-6*I*c*d^2 
*x*exp(2*I*(b*x+a))+6*b*c^2*d*x*exp(2*I*(b*x+a))-3*I*c^2*d*exp(2*I*(b*x+a) 
)-3*I*d^3*x^2+2*b*c^3*exp(2*I*(b*x+a))-6*I*c*d^2*x-3*I*c^2*d)/b^2/(exp(2*I 
*(b*x+a))+1)^2-3/b^3*d^2*c*ln(exp(2*I*(b*x+a))+1)-3/b^3*d^3*ln(exp(2*I*(b* 
x+a))+1)*x+3*I/b^2*d^3*x^2+3*I/b^4*d^3*a^2+3/2*I/b^2*d^3*polylog(2,-exp(2* 
I*(b*x+a)))*x^2+3/2*I/b^2*c^2*d*polylog(2,-exp(2*I*(b*x+a)))-1/b*d^3*ln(ex 
p(2*I*(b*x+a))+1)*x^3-3/2/b^3*c*d^2*polylog(3,-exp(2*I*(b*x+a)))-3/2/b^3*d 
^3*polylog(3,-exp(2*I*(b*x+a)))*x-3/b*c^2*d*ln(exp(2*I*(b*x+a))+1)*x-3/b*c 
*d^2*ln(exp(2*I*(b*x+a))+1)*x^2+6*I/b^3*d^3*a*x+6*d^2/b^3*c*ln(exp(I*(b*x+ 
a)))-6*d^3/b^4*a*ln(exp(I*(b*x+a)))+6*I/b^4*d^3*polylog(4,-exp(I*(b*x+a))) 
+3*I/b^2*c*d^2*polylog(2,-exp(2*I*(b*x+a)))*x-1/b^4*d^3*a^3*ln(exp(I*(b*x+ 
a))-1)+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)))+1/b*d^3*ln(1-exp(I*(b*x+a)))*x^3+1/b*d^3*ln(exp(I*(b*x+a))+1)*x^ 
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+1/b^4*d^3*ln(1-exp(I*(b*x+a)))*a^3+3/b^3*c*d^2*a^2*ln(exp(I*(b*x+a))- 
1)-3/b^2*c^2*d*a*ln(exp(I*(b*x+a))-1)+3/b^2*d*c^2*ln(1-exp(I*(b*x+a)))*a+3 
/b*c*d^2*ln(exp(I*(b*x+a))+1)*x^2+3/b*c*d^2*ln(1-exp(I*(b*x+a)))*x^2+3/b*d 
*c^2*ln(exp(I*(b*x+a))+1)*x+3/b*d*c^2*ln(1-exp(I*(b*x+a)))*x-3*I/b^2*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. 2268 vs. \(2 (270) = 540\).

Time = 0.24 (sec) , antiderivative size = 2268, normalized size of antiderivative = 6.98 \[ \int (c+d x)^3 \csc (a+b x) \sec ^3(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(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 + 6*I*d^3*cos 
(b*x + a)^2*polylog(4, cos(b*x + a) + I*sin(b*x + a)) - 6*I*d^3*cos(b*x + 
a)^2*polylog(4, cos(b*x + a) - I*sin(b*x + a)) + 6*I*d^3*cos(b*x + a)^2*po 
lylog(4, I*cos(b*x + a) + sin(b*x + a)) - 6*I*d^3*cos(b*x + a)^2*polylog(4 
, I*cos(b*x + a) - sin(b*x + a)) - 6*I*d^3*cos(b*x + a)^2*polylog(4, -I*co 
s(b*x + a) + sin(b*x + a)) + 6*I*d^3*cos(b*x + a)^2*polylog(4, -I*cos(b*x 
+ a) - sin(b*x + a)) - 6*I*d^3*cos(b*x + a)^2*polylog(4, -cos(b*x + a) + I 
*sin(b*x + a)) + 6*I*d^3*cos(b*x + a)^2*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)*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)*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 + I*d^3)*cos(b*x + a)^2*dil 
og(I*cos(b*x + a) + 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 - I*d^3)*cos(b*x + a)^2*dilog(I*cos(b*x + a) - 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 - I*d^3)*cos(b*x + a)^2 
*dilog(-I*cos(b*x + a) + 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 + I*d^3)*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - 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)*cos(b*x + a)^2*di 
log(-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)*cos(b*x + a)^2*dilog(-cos(b*x + a) - I*sin(b*x + a)) + (b...
 

Sympy [F(-1)]

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

integrate((d*x+c)**3*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. 4954 vs. \(2 (270) = 540\).

Time = 1.24 (sec) , antiderivative size = 4954, normalized size of antiderivative = 15.24 \[ \int (c+d x)^3 \csc (a+b x) \sec ^3(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(c^3*(1/(sin(b*x + a)^2 - 1) + log(sin(b*x + a)^2 - 1) - log(sin(b*x 
+ a)^2)) - 3*a*c^2*d*(1/(sin(b*x + a)^2 - 1) + log(sin(b*x + a)^2 - 1) - l 
og(sin(b*x + a)^2))/b + 3*a^2*c*d^2*(1/(sin(b*x + a)^2 - 1) + log(sin(b*x 
+ a)^2 - 1) - log(sin(b*x + a)^2))/b^2 - a^3*d^3*(1/(sin(b*x + a)^2 - 1) + 
 log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b^3 + 2*(18*b^2*c^2*d - 36 
*a*b*c*d^2 + 18*a^2*d^3 + 2*(4*(b*x + a)^3*d^3 + 9*b*c*d^2 - 9*a*d^3 + 9*( 
b*c*d^2 - a*d^3)*(b*x + a)^2 + 9*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 + 1)*d^3) 
*(b*x + a) + (4*(b*x + a)^3*d^3 + 9*b*c*d^2 - 9*a*d^3 + 9*(b*c*d^2 - a*d^3 
)*(b*x + a)^2 + 9*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 + 1)*d^3)*(b*x + a))*cos 
(4*b*x + 4*a) + 2*(4*(b*x + a)^3*d^3 + 9*b*c*d^2 - 9*a*d^3 + 9*(b*c*d^2 - 
a*d^3)*(b*x + a)^2 + 9*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 + 1)*d^3)*(b*x + a) 
)*cos(2*b*x + 2*a) - (-4*I*(b*x + a)^3*d^3 - 9*I*b*c*d^2 + 9*I*a*d^3 + 9*( 
-I*b*c*d^2 + I*a*d^3)*(b*x + a)^2 + 9*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 + (-I* 
a^2 - I)*d^3)*(b*x + a))*sin(4*b*x + 4*a) - 2*(-4*I*(b*x + a)^3*d^3 - 9*I* 
b*c*d^2 + 9*I*a*d^3 + 9*(-I*b*c*d^2 + I*a*d^3)*(b*x + a)^2 + 9*(-I*b^2*c^2 
*d + 2*I*a*b*c*d^2 + (-I*a^2 - I)*d^3)*(b*x + a))*sin(2*b*x + 2*a))*arctan 
2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 6*((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) + 
 ((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))*cos(4*b*x + 4*a) + 2*((b*x + a)^3*d^3 + 3*...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int (c+d x)^3 \csc (a+b x) \sec ^3(a+b x) \, dx=\frac {2 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x^{3}d x \right ) \sin \left (b x +a \right )^{2} b \,d^{3}-2 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x^{3}d x \right ) b \,d^{3}+6 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x^{2}d x \right ) \sin \left (b x +a \right )^{2} b c \,d^{2}-6 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x^{2}d x \right ) b c \,d^{2}+6 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x d x \right ) \sin \left (b x +a \right )^{2} b \,c^{2} d -6 \left (\int \csc \left (b x +a \right ) \sec \left (b x +a \right )^{3} x d x \right ) b \,c^{2} d -2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{2} c^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) c^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{2} c^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) c^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2} c^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) c^{3}-\sin \left (b x +a \right )^{2} c^{3}}{2 b \left (\sin \left (b x +a \right )^{2}-1\right )} \] Input:

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

Output:

(2*int(csc(a + b*x)*sec(a + b*x)**3*x**3,x)*sin(a + b*x)**2*b*d**3 - 2*int 
(csc(a + b*x)*sec(a + b*x)**3*x**3,x)*b*d**3 + 6*int(csc(a + b*x)*sec(a + 
b*x)**3*x**2,x)*sin(a + b*x)**2*b*c*d**2 - 6*int(csc(a + b*x)*sec(a + b*x) 
**3*x**2,x)*b*c*d**2 + 6*int(csc(a + b*x)*sec(a + b*x)**3*x,x)*sin(a + b*x 
)**2*b*c**2*d - 6*int(csc(a + b*x)*sec(a + b*x)**3*x,x)*b*c**2*d - 2*log(t 
an((a + b*x)/2) - 1)*sin(a + b*x)**2*c**3 + 2*log(tan((a + b*x)/2) - 1)*c* 
*3 - 2*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**2*c**3 + 2*log(tan((a + b*x 
)/2) + 1)*c**3 + 2*log(tan((a + b*x)/2))*sin(a + b*x)**2*c**3 - 2*log(tan( 
(a + b*x)/2))*c**3 - sin(a + b*x)**2*c**3)/(2*b*(sin(a + b*x)**2 - 1))