\(\int x^2 \csc ^3(a+b x) \sec ^2(a+b x) \, dx\) [286]

Optimal result

Integrand size = 20, antiderivative size = 235 \[ \int x^2 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\frac {4 i x \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {3 x^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}-\frac {\text {arctanh}(\cos (a+b x))}{b^3}-\frac {x \csc (a+b x)}{b^2}+\frac {3 i x \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {2 i \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {3 i x \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac {3 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {3 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac {3 x^2 \sec (a+b x)}{2 b}-\frac {x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b} \] Output:

4*I*x*arctan(exp(I*(b*x+a)))/b^2-3*x^2*arctanh(exp(I*(b*x+a)))/b-arctanh(c 
os(b*x+a))/b^3-x*csc(b*x+a)/b^2+3*I*x*polylog(2,-exp(I*(b*x+a)))/b^2-2*I*p 
olylog(2,-I*exp(I*(b*x+a)))/b^3+2*I*polylog(2,I*exp(I*(b*x+a)))/b^3-3*I*x* 
polylog(2,exp(I*(b*x+a)))/b^2-3*polylog(3,-exp(I*(b*x+a)))/b^3+3*polylog(3 
,exp(I*(b*x+a)))/b^3+3/2*x^2*sec(b*x+a)/b-1/2*x^2*csc(b*x+a)^2*sec(b*x+a)/ 
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. \(613\) vs. \(2(235)=470\).

Time = 6.52 (sec) , antiderivative size = 613, normalized size of antiderivative = 2.61 \[ \int x^2 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=-\frac {x^2 \csc ^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b}-\frac {2 \left (\left (-a+\frac {\pi }{2}-b x\right ) \left (\log \left (1-e^{i \left (-a+\frac {\pi }{2}-b x\right )}\right )-\log \left (1+e^{i \left (-a+\frac {\pi }{2}-b x\right )}\right )\right )-\left (-a+\frac {\pi }{2}\right ) \log \left (\tan \left (\frac {1}{2} \left (-a+\frac {\pi }{2}-b x\right )\right )\right )+i \left (\operatorname {PolyLog}\left (2,-e^{i \left (-a+\frac {\pi }{2}-b x\right )}\right )-\operatorname {PolyLog}\left (2,e^{i \left (-a+\frac {\pi }{2}-b x\right )}\right )\right )\right )}{b^3}+\frac {2 \log (1-\cos (a+b x)-i \sin (a+b x))+3 b^2 x^2 \log (1-\cos (a+b x)-i \sin (a+b x))-2 \log (1+\cos (a+b x)+i \sin (a+b x))-3 b^2 x^2 \log (1+\cos (a+b x)+i \sin (a+b x))+6 i b x \operatorname {PolyLog}(2,-\cos (a+b x)-i \sin (a+b x))-6 i b x \operatorname {PolyLog}(2,\cos (a+b x)+i \sin (a+b x))-6 \operatorname {PolyLog}(3,-\cos (a+b x)-i \sin (a+b x))+6 \operatorname {PolyLog}(3,\cos (a+b x)+i \sin (a+b x))}{2 b^3}+\frac {x^2 \sec ^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b}+\frac {x \csc (a) \sec (a) (-\cos (a)+b x \sin (a))}{b^2}+\frac {x \csc \left (\frac {a}{2}\right ) \csc \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {b x}{2}\right )}{2 b^2}-\frac {x \sec \left (\frac {a}{2}\right ) \sec \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {b x}{2}\right )}{2 b^2}+\frac {x^2 \sin \left (\frac {b x}{2}\right )}{b \left (\cos \left (\frac {a}{2}\right )-\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right )-\sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}-\frac {x^2 \sin \left (\frac {b x}{2}\right )}{b \left (\cos \left (\frac {a}{2}\right )+\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right )+\sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right )} \] Input:

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

Output:

-1/8*(x^2*Csc[a/2 + (b*x)/2]^2)/b - (2*((-a + Pi/2 - b*x)*(Log[1 - E^(I*(- 
a + Pi/2 - b*x))] - Log[1 + E^(I*(-a + Pi/2 - b*x))]) - (-a + Pi/2)*Log[Ta 
n[(-a + Pi/2 - b*x)/2]] + I*(PolyLog[2, -E^(I*(-a + Pi/2 - b*x))] - PolyLo 
g[2, E^(I*(-a + Pi/2 - b*x))])))/b^3 + (2*Log[1 - Cos[a + b*x] - I*Sin[a + 
 b*x]] + 3*b^2*x^2*Log[1 - Cos[a + b*x] - I*Sin[a + b*x]] - 2*Log[1 + Cos[ 
a + b*x] + I*Sin[a + b*x]] - 3*b^2*x^2*Log[1 + Cos[a + b*x] + I*Sin[a + b* 
x]] + (6*I)*b*x*PolyLog[2, -Cos[a + b*x] - I*Sin[a + b*x]] - (6*I)*b*x*Pol 
yLog[2, Cos[a + b*x] + I*Sin[a + b*x]] - 6*PolyLog[3, -Cos[a + b*x] - I*Si 
n[a + b*x]] + 6*PolyLog[3, Cos[a + b*x] + I*Sin[a + b*x]])/(2*b^3) + (x^2* 
Sec[a/2 + (b*x)/2]^2)/(8*b) + (x*Csc[a]*Sec[a]*(-Cos[a] + b*x*Sin[a]))/b^2 
 + (x*Csc[a/2]*Csc[a/2 + (b*x)/2]*Sin[(b*x)/2])/(2*b^2) - (x*Sec[a/2]*Sec[ 
a/2 + (b*x)/2]*Sin[(b*x)/2])/(2*b^2) + (x^2*Sin[(b*x)/2])/(b*(Cos[a/2] - S 
in[a/2])*(Cos[a/2 + (b*x)/2] - Sin[a/2 + (b*x)/2])) - (x^2*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 = 0.71 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 4920

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2010

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (208 ) = 416\).

Time = 0.58 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.83

method	result	size
risch	\(\frac {x \left (3 b x \,{\mathrm e}^{5 i \left (b x +a \right )}-2 b x \,{\mathrm e}^{3 i \left (b x +a \right )}-2 i {\mathrm e}^{5 i \left (b x +a \right )}+3 b x \,{\mathrm e}^{i \left (b x +a \right )}+2 i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}-\frac {3 a^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {2 \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}+\frac {2 \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{3}}-\frac {3 \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {3 \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}+\frac {2 i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i \operatorname {dilog}\left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{2 b^{3}}-\frac {2 \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{2 b}+\frac {3 i x \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{2 b}-\frac {3 i x \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {4 i a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}\)	\(429\)

Input:

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

Output:

x/b^2/(exp(2*I*(b*x+a))-1)^2/(exp(2*I*(b*x+a))+1)*(3*b*x*exp(5*I*(b*x+a))- 
2*b*x*exp(3*I*(b*x+a))-2*I*exp(5*I*(b*x+a))+3*b*x*exp(I*(b*x+a))+2*I*exp(I 
*(b*x+a)))+1/b^3*ln(exp(I*(b*x+a))-1)-3/2/b^3*a^2*ln(1-exp(I*(b*x+a)))-2/b 
^3*ln(1-I*exp(I*(b*x+a)))*a+2/b^3*ln(I*exp(I*(b*x+a))+1)*a-3*polylog(3,-ex 
p(I*(b*x+a)))/b^3+3*polylog(3,exp(I*(b*x+a)))/b^3-1/b^3*ln(exp(I*(b*x+a))+ 
1)+2*I/b^3*dilog(1-I*exp(I*(b*x+a)))-2*I/b^3*dilog(I*exp(I*(b*x+a))+1)+3/2 
/b^3*a^2*ln(exp(I*(b*x+a))-1)-2/b^2*ln(1-I*exp(I*(b*x+a)))*x-3/2/b*ln(exp( 
I*(b*x+a))+1)*x^2+3*I*x*polylog(2,-exp(I*(b*x+a)))/b^2+3/2/b*ln(1-exp(I*(b 
*x+a)))*x^2-3*I*x*polylog(2,exp(I*(b*x+a)))/b^2+2/b^2*ln(I*exp(I*(b*x+a))+ 
1)*x-4*I/b^3*a*arctan(exp(I*(b*x+a)))
 

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. 1237 vs. \(2 (197) = 394\).

Time = 0.15 (sec) , antiderivative size = 1237, normalized size of antiderivative = 5.26 \[ \int x^2 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(6*b^2*x^2*cos(b*x + a)^2 - 4*b^2*x^2 + 4*b*x*cos(b*x + a)*sin(b*x + a 
) - 6*(I*b*x*cos(b*x + a)^3 - I*b*x*cos(b*x + a))*dilog(cos(b*x + a) + I*s 
in(b*x + a)) - 6*(-I*b*x*cos(b*x + a)^3 + I*b*x*cos(b*x + a))*dilog(cos(b* 
x + a) - I*sin(b*x + a)) - 4*(-I*cos(b*x + a)^3 + I*cos(b*x + a))*dilog(I* 
cos(b*x + a) + sin(b*x + a)) - 4*(-I*cos(b*x + a)^3 + I*cos(b*x + a))*dilo 
g(I*cos(b*x + a) - sin(b*x + a)) - 4*(I*cos(b*x + a)^3 - I*cos(b*x + a))*d 
ilog(-I*cos(b*x + a) + sin(b*x + a)) - 4*(I*cos(b*x + a)^3 - I*cos(b*x + a 
))*dilog(-I*cos(b*x + a) - sin(b*x + a)) - 6*(I*b*x*cos(b*x + a)^3 - I*b*x 
*cos(b*x + a))*dilog(-cos(b*x + a) + I*sin(b*x + a)) - 6*(-I*b*x*cos(b*x + 
 a)^3 + I*b*x*cos(b*x + a))*dilog(-cos(b*x + a) - I*sin(b*x + a)) - ((3*b^ 
2*x^2 + 2)*cos(b*x + a)^3 - (3*b^2*x^2 + 2)*cos(b*x + a))*log(cos(b*x + a) 
 + I*sin(b*x + a) + 1) + 4*(a*cos(b*x + a)^3 - a*cos(b*x + a))*log(cos(b*x 
 + a) + I*sin(b*x + a) + I) - ((3*b^2*x^2 + 2)*cos(b*x + a)^3 - (3*b^2*x^2 
 + 2)*cos(b*x + a))*log(cos(b*x + a) - I*sin(b*x + a) + 1) - 4*(a*cos(b*x 
+ a)^3 - a*cos(b*x + a))*log(cos(b*x + a) - I*sin(b*x + a) + I) - 4*((b*x 
+ a)*cos(b*x + a)^3 - (b*x + a)*cos(b*x + a))*log(I*cos(b*x + a) + sin(b*x 
 + a) + 1) + 4*((b*x + a)*cos(b*x + a)^3 - (b*x + a)*cos(b*x + a))*log(I*c 
os(b*x + a) - sin(b*x + a) + 1) - 4*((b*x + a)*cos(b*x + a)^3 - (b*x + a)* 
cos(b*x + a))*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + 4*((b*x + a)*cos(b 
*x + a)^3 - (b*x + a)*cos(b*x + a))*log(-I*cos(b*x + a) - sin(b*x + a) ...
 

Sympy [F]

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

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

Output:

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

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. 2205 vs. \(2 (197) = 394\).

Time = 0.55 (sec) , antiderivative size = 2205, normalized size of antiderivative = 9.38 \[ \int x^2 \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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