\(\int x^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx\) [285]

Optimal result

Integrand size = 20, antiderivative size = 387 \[ \int x^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\frac {6 i x^2 \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {6 x \text {arctanh}\left (e^{i (a+b x)}\right )}{b^3}-\frac {3 x^3 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}-\frac {3 x^2 \csc (a+b x)}{2 b^2}+\frac {3 i \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^4}+\frac {9 i x^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}-\frac {6 i x \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {6 i x \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {3 i \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^4}-\frac {9 i x^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}-\frac {9 x \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {6 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^4}-\frac {6 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^4}+\frac {9 x \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac {9 i \operatorname {PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^4}+\frac {9 i \operatorname {PolyLog}\left (4,e^{i (a+b x)}\right )}{b^4}+\frac {3 x^3 \sec (a+b x)}{2 b}-\frac {x^3 \csc ^2(a+b x) \sec (a+b x)}{2 b} \] Output:

3*I*polylog(2,-exp(I*(b*x+a)))/b^4-6*x*arctanh(exp(I*(b*x+a)))/b^3-3*x^3*a 
rctanh(exp(I*(b*x+a)))/b-3/2*x^2*csc(b*x+a)/b^2-3*I*polylog(2,exp(I*(b*x+a 
)))/b^4+9/2*I*x^2*polylog(2,-exp(I*(b*x+a)))/b^2+6*I*x*polylog(2,I*exp(I*( 
b*x+a)))/b^3-9*I*polylog(4,-exp(I*(b*x+a)))/b^4+9*I*polylog(4,exp(I*(b*x+a 
)))/b^4-9/2*I*x^2*polylog(2,exp(I*(b*x+a)))/b^2-9*x*polylog(3,-exp(I*(b*x+ 
a)))/b^3+6*polylog(3,-I*exp(I*(b*x+a)))/b^4-6*polylog(3,I*exp(I*(b*x+a)))/ 
b^4+9*x*polylog(3,exp(I*(b*x+a)))/b^3-6*I*x*polylog(2,-I*exp(I*(b*x+a)))/b 
^3+6*I*x^2*arctan(exp(I*(b*x+a)))/b^2+3/2*x^3*sec(b*x+a)/b-1/2*x^3*csc(b*x 
+a)^2*sec(b*x+a)/b
 

Mathematica [A] (warning: unable to verify)

Time = 6.74 (sec) , antiderivative size = 672, normalized size of antiderivative = 1.74 \[ \int x^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=-\frac {x^3 \csc ^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b}+\frac {6 \left (i b^2 x^2 \arctan (\cos (a+b x)+i \sin (a+b x))+i b x \operatorname {PolyLog}(2,i \cos (a+b x)-\sin (a+b x))-i b x \operatorname {PolyLog}(2,-i \cos (a+b x)+\sin (a+b x))-\operatorname {PolyLog}(3,i \cos (a+b x)-\sin (a+b x))+\operatorname {PolyLog}(3,-i \cos (a+b x)+\sin (a+b x))\right )}{b^4}+\frac {3 \left (2 b x \log (1-\cos (a+b x)-i \sin (a+b x))+b^3 x^3 \log (1-\cos (a+b x)-i \sin (a+b x))-2 b x \log (1+\cos (a+b x)+i \sin (a+b x))-b^3 x^3 \log (1+\cos (a+b x)+i \sin (a+b x))+i \left (2+3 b^2 x^2\right ) \operatorname {PolyLog}(2,-\cos (a+b x)-i \sin (a+b x))-i \left (2+3 b^2 x^2\right ) \operatorname {PolyLog}(2,\cos (a+b x)+i \sin (a+b x))-6 b x \operatorname {PolyLog}(3,-\cos (a+b x)-i \sin (a+b x))+6 b x \operatorname {PolyLog}(3,\cos (a+b x)+i \sin (a+b x))-6 i \operatorname {PolyLog}(4,-\cos (a+b x)-i \sin (a+b x))+6 i \operatorname {PolyLog}(4,\cos (a+b x)+i \sin (a+b x))\right )}{2 b^4}+\frac {x^3 \sec ^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b}+\frac {x^2 \csc (a) \sec (a) (-3 \cos (a)+2 b x \sin (a))}{2 b^2}+\frac {3 x^2 \csc \left (\frac {a}{2}\right ) \csc \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {b x}{2}\right )}{4 b^2}-\frac {3 x^2 \sec \left (\frac {a}{2}\right ) \sec \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {b x}{2}\right )}{4 b^2}+\frac {x^3 \sin \left (\frac {b x}{2}\right )}{b \left (\cos \left (\frac {a}{2}\right )-\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right )-\sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}-\frac {x^3 \sin \left (\frac {b x}{2}\right )}{b \left (\cos \left (\frac {a}{2}\right )+\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right )+\sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right )} \] Input:

Integrate[x^3*Csc[a + b*x]^3*Sec[a + b*x]^2,x]
 

Output:

-1/8*(x^3*Csc[a/2 + (b*x)/2]^2)/b + (6*(I*b^2*x^2*ArcTan[Cos[a + b*x] + I* 
Sin[a + b*x]] + I*b*x*PolyLog[2, I*Cos[a + b*x] - Sin[a + b*x]] - I*b*x*Po 
lyLog[2, (-I)*Cos[a + b*x] + Sin[a + b*x]] - PolyLog[3, I*Cos[a + b*x] - S 
in[a + b*x]] + PolyLog[3, (-I)*Cos[a + b*x] + Sin[a + b*x]]))/b^4 + (3*(2* 
b*x*Log[1 - Cos[a + b*x] - I*Sin[a + b*x]] + b^3*x^3*Log[1 - Cos[a + b*x] 
- I*Sin[a + b*x]] - 2*b*x*Log[1 + Cos[a + b*x] + I*Sin[a + b*x]] - b^3*x^3 
*Log[1 + Cos[a + b*x] + I*Sin[a + b*x]] + I*(2 + 3*b^2*x^2)*PolyLog[2, -Co 
s[a + b*x] - I*Sin[a + b*x]] - I*(2 + 3*b^2*x^2)*PolyLog[2, Cos[a + b*x] + 
 I*Sin[a + b*x]] - 6*b*x*PolyLog[3, -Cos[a + b*x] - I*Sin[a + b*x]] + 6*b* 
x*PolyLog[3, Cos[a + b*x] + I*Sin[a + b*x]] - (6*I)*PolyLog[4, -Cos[a + b* 
x] - I*Sin[a + b*x]] + (6*I)*PolyLog[4, Cos[a + b*x] + I*Sin[a + b*x]]))/( 
2*b^4) + (x^3*Sec[a/2 + (b*x)/2]^2)/(8*b) + (x^2*Csc[a]*Sec[a]*(-3*Cos[a] 
+ 2*b*x*Sin[a]))/(2*b^2) + (3*x^2*Csc[a/2]*Csc[a/2 + (b*x)/2]*Sin[(b*x)/2] 
)/(4*b^2) - (3*x^2*Sec[a/2]*Sec[a/2 + (b*x)/2]*Sin[(b*x)/2])/(4*b^2) + (x^ 
3*Sin[(b*x)/2])/(b*(Cos[a/2] - Sin[a/2])*(Cos[a/2 + (b*x)/2] - Sin[a/2 + ( 
b*x)/2])) - (x^3*Sin[(b*x)/2])/(b*(Cos[a/2] + Sin[a/2])*(Cos[a/2 + (b*x)/2 
] + Sin[a/2 + (b*x)/2]))
 

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4920, 27, 2010, 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.

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

\(\Big \downarrow \) 4920

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2010

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

\(\Big \downarrow \) 2009

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

Input:

Int[x^3*Csc[a + b*x]^3*Sec[a + b*x]^2,x]
 

Output:

(-3*x^3*ArcTanh[Cos[a + b*x]])/(2*b) + (3*(((4*I)*x^2*ArcTan[E^(I*(a + b*x 
))])/b^2 - (4*x*ArcTanh[E^(I*(a + b*x))])/b^3 - (2*x^3*ArcTanh[E^(I*(a + b 
*x))])/b + (x^3*ArcTanh[Cos[a + b*x]])/b - (x^2*Csc[a + b*x])/b^2 + ((2*I) 
*PolyLog[2, -E^(I*(a + b*x))])/b^4 + ((3*I)*x^2*PolyLog[2, -E^(I*(a + b*x) 
)])/b^2 - ((4*I)*x*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^3 + ((4*I)*x*PolyLo 
g[2, I*E^(I*(a + b*x))])/b^3 - ((2*I)*PolyLog[2, E^(I*(a + b*x))])/b^4 - ( 
(3*I)*x^2*PolyLog[2, E^(I*(a + b*x))])/b^2 - (6*x*PolyLog[3, -E^(I*(a + b* 
x))])/b^3 + (4*PolyLog[3, (-I)*E^(I*(a + b*x))])/b^4 - (4*PolyLog[3, I*E^( 
I*(a + b*x))])/b^4 + (6*x*PolyLog[3, E^(I*(a + b*x))])/b^3 - ((6*I)*PolyLo 
g[4, -E^(I*(a + b*x))])/b^4 + ((6*I)*PolyLog[4, E^(I*(a + b*x))])/b^4))/2 
+ (3*x^3*Sec[a + b*x])/(2*b) - (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 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

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]
 

Maple [F]

\[\int x^{3} \csc \left (b x +a \right )^{3} \sec \left (b x +a \right )^{2}d x\]

Input:

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

Output:

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

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. 1747 vs. \(2 (315) = 630\).

Time = 0.17 (sec) , antiderivative size = 1747, normalized size of antiderivative = 4.51 \[ \int x^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(6*b^3*x^3*cos(b*x + a)^2 - 4*b^3*x^3 + 6*b^2*x^2*cos(b*x + a)*sin(b*x 
 + a) - 3*((3*I*b^2*x^2 + 2*I)*cos(b*x + a)^3 + (-3*I*b^2*x^2 - 2*I)*cos(b 
*x + a))*dilog(cos(b*x + a) + I*sin(b*x + a)) - 3*((-3*I*b^2*x^2 - 2*I)*co 
s(b*x + a)^3 + (3*I*b^2*x^2 + 2*I)*cos(b*x + a))*dilog(cos(b*x + a) - I*si 
n(b*x + a)) - 12*(-I*b*x*cos(b*x + a)^3 + I*b*x*cos(b*x + a))*dilog(I*cos( 
b*x + a) + sin(b*x + a)) - 12*(-I*b*x*cos(b*x + a)^3 + I*b*x*cos(b*x + a)) 
*dilog(I*cos(b*x + a) - sin(b*x + a)) - 12*(I*b*x*cos(b*x + a)^3 - I*b*x*c 
os(b*x + a))*dilog(-I*cos(b*x + a) + sin(b*x + a)) - 12*(I*b*x*cos(b*x + a 
)^3 - I*b*x*cos(b*x + a))*dilog(-I*cos(b*x + a) - sin(b*x + a)) - 3*((3*I* 
b^2*x^2 + 2*I)*cos(b*x + a)^3 + (-3*I*b^2*x^2 - 2*I)*cos(b*x + a))*dilog(- 
cos(b*x + a) + I*sin(b*x + a)) - 3*((-3*I*b^2*x^2 - 2*I)*cos(b*x + a)^3 + 
(3*I*b^2*x^2 + 2*I)*cos(b*x + a))*dilog(-cos(b*x + a) - I*sin(b*x + a)) - 
3*((b^3*x^3 + 2*b*x)*cos(b*x + a)^3 - (b^3*x^3 + 2*b*x)*cos(b*x + a))*log( 
cos(b*x + a) + I*sin(b*x + a) + 1) - 6*(a^2*cos(b*x + a)^3 - a^2*cos(b*x + 
 a))*log(cos(b*x + a) + I*sin(b*x + a) + I) - 3*((b^3*x^3 + 2*b*x)*cos(b*x 
 + a)^3 - (b^3*x^3 + 2*b*x)*cos(b*x + a))*log(cos(b*x + a) - I*sin(b*x + a 
) + 1) + 6*(a^2*cos(b*x + a)^3 - a^2*cos(b*x + a))*log(cos(b*x + a) - I*si 
n(b*x + a) + I) - 6*((b^2*x^2 - a^2)*cos(b*x + a)^3 - (b^2*x^2 - a^2)*cos( 
b*x + a))*log(I*cos(b*x + a) + sin(b*x + a) + 1) + 6*((b^2*x^2 - a^2)*cos( 
b*x + a)^3 - (b^2*x^2 - a^2)*cos(b*x + a))*log(I*cos(b*x + a) - sin(b*x...
 

Sympy [F(-1)]

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

integrate(x**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. 3940 vs. \(2 (315) = 630\).

Time = 0.81 (sec) , antiderivative size = 3940, normalized size of antiderivative = 10.18 \[ \int x^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(a^3*(2*(3*cos(b*x + a)^2 - 2)/(cos(b*x + a)^3 - cos(b*x + a)) - 3*lo 
g(cos(b*x + a) + 1) + 3*log(cos(b*x + a) - 1)) - 4*(12*((b*x + a)^2 - 2*(b 
*x + a)*a + a^2 + ((b*x + a)^2 - 2*(b*x + a)*a + a^2)*cos(6*b*x + 6*a) - ( 
(b*x + a)^2 - 2*(b*x + a)*a + a^2)*cos(4*b*x + 4*a) - ((b*x + a)^2 - 2*(b* 
x + a)*a + a^2)*cos(2*b*x + 2*a) + (I*(b*x + a)^2 - 2*I*(b*x + a)*a + I*a^ 
2)*sin(6*b*x + 6*a) + (-I*(b*x + a)^2 + 2*I*(b*x + a)*a - I*a^2)*sin(4*b*x 
 + 4*a) + (-I*(b*x + a)^2 + 2*I*(b*x + a)*a - I*a^2)*sin(2*b*x + 2*a))*arc 
tan2(cos(b*x + a), sin(b*x + a) + 1) + 12*((b*x + a)^2 - 2*(b*x + a)*a + a 
^2 + ((b*x + a)^2 - 2*(b*x + a)*a + a^2)*cos(6*b*x + 6*a) - ((b*x + a)^2 - 
 2*(b*x + a)*a + a^2)*cos(4*b*x + 4*a) - ((b*x + a)^2 - 2*(b*x + a)*a + a^ 
2)*cos(2*b*x + 2*a) + (I*(b*x + a)^2 - 2*I*(b*x + a)*a + I*a^2)*sin(6*b*x 
+ 6*a) + (-I*(b*x + a)^2 + 2*I*(b*x + a)*a - I*a^2)*sin(4*b*x + 4*a) + (-I 
*(b*x + a)^2 + 2*I*(b*x + a)*a - I*a^2)*sin(2*b*x + 2*a))*arctan2(cos(b*x 
+ a), -sin(b*x + a) + 1) - 6*((b*x + a)^3 - 3*(b*x + a)^2*a + (3*a^2 + 2)* 
(b*x + a) + ((b*x + a)^3 - 3*(b*x + a)^2*a + (3*a^2 + 2)*(b*x + a) - 2*a)* 
cos(6*b*x + 6*a) - ((b*x + a)^3 - 3*(b*x + a)^2*a + (3*a^2 + 2)*(b*x + a) 
- 2*a)*cos(4*b*x + 4*a) - ((b*x + a)^3 - 3*(b*x + a)^2*a + (3*a^2 + 2)*(b* 
x + a) - 2*a)*cos(2*b*x + 2*a) - (-I*(b*x + a)^3 + 3*I*(b*x + a)^2*a + (-3 
*I*a^2 - 2*I)*(b*x + a) + 2*I*a)*sin(6*b*x + 6*a) - (I*(b*x + a)^3 - 3*I*( 
b*x + a)^2*a + (3*I*a^2 + 2*I)*(b*x + a) - 2*I*a)*sin(4*b*x + 4*a) - (I...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^3 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\int \frac {x^3}{{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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