\(\int (c+d x)^2 \cos (a+b x) \cot ^2(a+b x) \, dx\) [173]

Optimal result

Integrand size = 22, antiderivative size = 139 \[ \int (c+d x)^2 \cos (a+b x) \cot ^2(a+b x) \, dx=-\frac {4 d (c+d x) \text {arctanh}\left (e^{i (a+b x)}\right )}{b^2}-\frac {2 d (c+d x) \cos (a+b x)}{b^2}-\frac {(c+d x)^2 \csc (a+b x)}{b}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^3}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \sin (a+b x)}{b^3}-\frac {(c+d x)^2 \sin (a+b x)}{b} \] Output:

-4*d*(d*x+c)*arctanh(exp(I*(b*x+a)))/b^2-2*d*(d*x+c)*cos(b*x+a)/b^2-(d*x+c 
)^2*csc(b*x+a)/b+2*I*d^2*polylog(2,-exp(I*(b*x+a)))/b^3-2*I*d^2*polylog(2, 
exp(I*(b*x+a)))/b^3+2*d^2*sin(b*x+a)/b^3-(d*x+c)^2*sin(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. \(310\) vs. \(2(139)=278\).

Time = 2.79 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.23 \[ \int (c+d x)^2 \cos (a+b x) \cot ^2(a+b x) \, dx=-\frac {8 b c d \text {arctanh}\left (\cos (a)-\sin (a) \tan \left (\frac {b x}{2}\right )\right )+2 b^2 (c+d x)^2 \csc (a)-4 d^2 \left (2 \arctan (\tan (a)) \text {arctanh}\left (\cos (a)-\sin (a) \tan \left (\frac {b x}{2}\right )\right )+\frac {\left ((b x+\arctan (\tan (a))) \left (\log \left (1-e^{i (b x+\arctan (\tan (a)))}\right )-\log \left (1+e^{i (b x+\arctan (\tan (a)))}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i (b x+\arctan (\tan (a)))}\right )-i \operatorname {PolyLog}\left (2,e^{i (b x+\arctan (\tan (a)))}\right )\right ) \sec (a)}{\sqrt {\sec ^2(a)}}\right )+2 \cos (b x) \left (2 b d (c+d x) \cos (a)+\left (-2 d^2+b^2 (c+d x)^2\right ) \sin (a)\right )-b^2 (c+d x)^2 \csc \left (\frac {a}{2}\right ) \csc \left (\frac {1}{2} (a+b x)\right ) \sin \left (\frac {b x}{2}\right )+b^2 (c+d x)^2 \sec \left (\frac {a}{2}\right ) \sec \left (\frac {1}{2} (a+b x)\right ) \sin \left (\frac {b x}{2}\right )+2 \left (\left (-2 d^2+b^2 (c+d x)^2\right ) \cos (a)-2 b d (c+d x) \sin (a)\right ) \sin (b x)}{2 b^3} \] Input:

Integrate[(c + d*x)^2*Cos[a + b*x]*Cot[a + b*x]^2,x]
 

Output:

-1/2*(8*b*c*d*ArcTanh[Cos[a] - Sin[a]*Tan[(b*x)/2]] + 2*b^2*(c + d*x)^2*Cs 
c[a] - 4*d^2*(2*ArcTan[Tan[a]]*ArcTanh[Cos[a] - Sin[a]*Tan[(b*x)/2]] + ((( 
b*x + ArcTan[Tan[a]])*(Log[1 - E^(I*(b*x + ArcTan[Tan[a]]))] - Log[1 + E^( 
I*(b*x + ArcTan[Tan[a]]))]) + I*PolyLog[2, -E^(I*(b*x + ArcTan[Tan[a]]))] 
- I*PolyLog[2, E^(I*(b*x + ArcTan[Tan[a]]))])*Sec[a])/Sqrt[Sec[a]^2]) + 2* 
Cos[b*x]*(2*b*d*(c + d*x)*Cos[a] + (-2*d^2 + b^2*(c + d*x)^2)*Sin[a]) - b^ 
2*(c + d*x)^2*Csc[a/2]*Csc[(a + b*x)/2]*Sin[(b*x)/2] + b^2*(c + d*x)^2*Sec 
[a/2]*Sec[(a + b*x)/2]*Sin[(b*x)/2] + 2*((-2*d^2 + b^2*(c + d*x)^2)*Cos[a] 
 - 2*b*d*(c + d*x)*Sin[a])*Sin[b*x])/b^3
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {4908, 3042, 3777, 25, 3042, 3777, 3042, 3117, 4910, 3042, 4671, 2715, 2838}

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.

\(\displaystyle \int (c+d x)^2 \cos (a+b x) \cot ^2(a+b x) \, dx\)

\(\Big \downarrow \) 4908

\(\displaystyle \int (c+d x)^2 \cot (a+b x) \csc (a+b x)dx-\int (c+d x)^2 \cos (a+b x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x)^2 \cot (a+b x) \csc (a+b x)dx-\int (c+d x)^2 \sin \left (a+b x+\frac {\pi }{2}\right )dx\)

\(\Big \downarrow \) 3777

\(\displaystyle -\frac {2 d \int -((c+d x) \sin (a+b x))dx}{b}+\int (c+d x)^2 \cot (a+b x) \csc (a+b x)dx-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 d \int (c+d x) \sin (a+b x)dx}{b}+\int (c+d x)^2 \cot (a+b x) \csc (a+b x)dx-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 d \int (c+d x) \sin (a+b x)dx}{b}+\int (c+d x)^2 \cot (a+b x) \csc (a+b x)dx-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {2 d \left (\frac {d \int \cos (a+b x)dx}{b}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}+\int (c+d x)^2 \cot (a+b x) \csc (a+b x)dx-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 d \left (\frac {d \int \sin \left (a+b x+\frac {\pi }{2}\right )dx}{b}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}+\int (c+d x)^2 \cot (a+b x) \csc (a+b x)dx-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 3117

\(\displaystyle \int (c+d x)^2 \cot (a+b x) \csc (a+b x)dx+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 4910

\(\displaystyle \frac {2 d \int (c+d x) \csc (a+b x)dx}{b}+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}-\frac {(c+d x)^2 \csc (a+b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 d \int (c+d x) \csc (a+b x)dx}{b}+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}-\frac {(c+d x)^2 \csc (a+b x)}{b}\)

\(\Big \downarrow \) 4671

\(\displaystyle \frac {2 d \left (-\frac {d \int \log \left (1-e^{i (a+b x)}\right )dx}{b}+\frac {d \int \log \left (1+e^{i (a+b x)}\right )dx}{b}-\frac {2 (c+d x) \text {arctanh}\left (e^{i (a+b x)}\right )}{b}\right )}{b}+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}-\frac {(c+d x)^2 \csc (a+b x)}{b}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {2 d \left (\frac {i d \int e^{-i (a+b x)} \log \left (1-e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}-\frac {i d \int e^{-i (a+b x)} \log \left (1+e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}-\frac {2 (c+d x) \text {arctanh}\left (e^{i (a+b x)}\right )}{b}\right )}{b}+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}-\frac {(c+d x)^2 \csc (a+b x)}{b}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {2 d \left (-\frac {2 (c+d x) \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {i d \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {i d \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}\right )}{b}+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}-\frac {(c+d x)^2 \csc (a+b x)}{b}\)

Input:

Int[(c + d*x)^2*Cos[a + b*x]*Cot[a + b*x]^2,x]
 

Output:

-(((c + d*x)^2*Csc[a + b*x])/b) + (2*d*((-2*(c + d*x)*ArcTanh[E^(I*(a + b* 
x))])/b + (I*d*PolyLog[2, -E^(I*(a + b*x))])/b^2 - (I*d*PolyLog[2, E^(I*(a 
 + b*x))])/b^2))/b - ((c + d*x)^2*Sin[a + b*x])/b + (2*d*(-(((c + d*x)*Cos 
[a + b*x])/b) + (d*Sin[a + b*x])/b^2))/b
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3117

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; 
 FreeQ[{c, d}, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 4671

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[- 
2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*x))]/f), x] + (-Simp[d*(m/f)   Int[(c + 
d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f)   Int[(c + d*x 
)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG 
tQ[m, 0]
 

rule 4908

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*Cot[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d 
_.)*(x_))^(m_.), x_Symbol] :> -Int[(c + d*x)^m*Cos[a + b*x]^n*Cot[a + b*x]^ 
(p - 2), x] + Int[(c + d*x)^m*Cos[a + b*x]^(n - 2)*Cot[a + b*x]^p, x] /; Fr 
eeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]
 

rule 4910

Int[Cot[(a_.) + (b_.)*(x_)]^(p_.)*Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d 
_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Csc[a + b*x]^n/(b*n)), x 
] + Simp[d*(m/(b*n))   Int[(c + d*x)^(m - 1)*Csc[a + b*x]^n, x], x] /; Free 
Q[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 331 vs. \(2 (131 ) = 262\).

Time = 0.69 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.39

method	result	size
risch	\(\frac {i \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +2 i b \,d^{2} x +b^{2} c^{2}+2 i b c d -2 d^{2}\right ) {\mathrm e}^{i \left (b x +a \right )}}{2 b^{3}}-\frac {i \left (x^{2} d^{2} b^{2}+2 b^{2} c d x -2 i b \,d^{2} x +b^{2} c^{2}-2 i b c d -2 d^{2}\right ) {\mathrm e}^{-i \left (b x +a \right )}}{2 b^{3}}-\frac {2 i \left (x^{2} d^{2}+2 c d x +c^{2}\right ) {\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {4 d c \,\operatorname {arctanh}\left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{3}}+\frac {2 i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {4 d^{2} a \,\operatorname {arctanh}\left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}\)	\(332\)

Input:

int((d*x+c)^2*cos(b*x+a)*cot(b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

1/2*I*(x^2*d^2*b^2+2*b^2*c*d*x+b^2*c^2+2*I*b*d^2*x-2*d^2+2*I*b*c*d)/b^3*ex 
p(I*(b*x+a))-1/2*I*(x^2*d^2*b^2+2*b^2*c*d*x+b^2*c^2-2*I*b*d^2*x-2*d^2-2*I* 
b*c*d)/b^3*exp(-I*(b*x+a))-2*I*(d^2*x^2+2*c*d*x+c^2)*exp(I*(b*x+a))/b/(exp 
(2*I*(b*x+a))-1)-4*d/b^2*c*arctanh(exp(I*(b*x+a)))-2*d^2/b^2*ln(exp(I*(b*x 
+a))+1)*x-2*d^2/b^3*ln(exp(I*(b*x+a))+1)*a+2*I*d^2*polylog(2,-exp(I*(b*x+a 
)))/b^3+2*d^2/b^2*ln(1-exp(I*(b*x+a)))*x+2*d^2/b^3*ln(1-exp(I*(b*x+a)))*a- 
2*I*d^2*polylog(2,exp(I*(b*x+a)))/b^3+4*d^2/b^3*a*arctanh(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. 448 vs. \(2 (127) = 254\).

Time = 0.10 (sec) , antiderivative size = 448, normalized size of antiderivative = 3.22 \[ \int (c+d x)^2 \cos (a+b x) \cot ^2(a+b x) \, dx=-\frac {2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} + i \, d^{2} {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d^{2} {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + i \, d^{2} {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d^{2} {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )^{2} + 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + {\left (b d^{2} x + b c d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + {\left (b d^{2} x + b c d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - {\left (b c d - a d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) \sin \left (b x + a\right ) - {\left (b c d - a d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) \sin \left (b x + a\right ) - {\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - {\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 2 \, d^{2}}{b^{3} \sin \left (b x + a\right )} \] Input:

integrate((d*x+c)^2*cos(b*x+a)*cot(b*x+a)^2,x, algorithm="fricas")
 

Output:

-(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 + I*d^2*dilog(cos(b*x + a) + I*s 
in(b*x + a))*sin(b*x + a) - I*d^2*dilog(cos(b*x + a) - I*sin(b*x + a))*sin 
(b*x + a) + I*d^2*dilog(-cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) - I*d 
^2*dilog(-cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) - (b^2*d^2*x^2 + 2*b 
^2*c*d*x + b^2*c^2 - 2*d^2)*cos(b*x + a)^2 + 2*(b*d^2*x + b*c*d)*cos(b*x + 
 a)*sin(b*x + a) + (b*d^2*x + b*c*d)*log(cos(b*x + a) + I*sin(b*x + a) + 1 
)*sin(b*x + a) + (b*d^2*x + b*c*d)*log(cos(b*x + a) - I*sin(b*x + a) + 1)* 
sin(b*x + a) - (b*c*d - a*d^2)*log(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) 
+ 1/2)*sin(b*x + a) - (b*c*d - a*d^2)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b* 
x + a) + 1/2)*sin(b*x + a) - (b*d^2*x + a*d^2)*log(-cos(b*x + a) + I*sin(b 
*x + a) + 1)*sin(b*x + a) - (b*d^2*x + a*d^2)*log(-cos(b*x + a) - I*sin(b* 
x + a) + 1)*sin(b*x + a) - 2*d^2)/(b^3*sin(b*x + a))
 

Sympy [F]

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

integrate((d*x+c)**2*cos(b*x+a)*cot(b*x+a)**2,x)
 

Output:

Integral((c + d*x)**2*cos(a + b*x)*cot(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. 3199 vs. \(2 (127) = 254\).

Time = 0.96 (sec) , antiderivative size = 3199, normalized size of antiderivative = 23.01 \[ \int (c+d x)^2 \cos (a+b x) \cot ^2(a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^2*cos(b*x+a)*cot(b*x+a)^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((d*x+c)^2*cos(b*x+a)*cot(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate((d*x + c)^2*cos(b*x + a)*cot(b*x + a)^2, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^2 \cos (a+b x) \cot ^2(a+b x) \, dx=\int \cos \left (a+b\,x\right )\,{\mathrm {cot}\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^2 \,d x \] Input:

int(cos(a + b*x)*cot(a + b*x)^2*(c + d*x)^2,x)
 

Output:

int(cos(a + b*x)*cot(a + b*x)^2*(c + d*x)^2, x)
 

Reduce [F]

\[ \int (c+d x)^2 \cos (a+b x) \cot ^2(a+b x) \, dx=\frac {-16 \cos \left (b x +a \right ) \sin \left (b x +a \right ) b c d -18 \cos \left (b x +a \right ) \sin \left (b x +a \right ) b \,d^{2} x +4 \cos \left (b x +a \right ) b^{2} d^{2} x^{2}+2 \left (\int \frac {x^{2}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}d x \right ) \sin \left (b x +a \right ) b^{3} d^{2}+8 \left (\int \frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3} x}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}d x \right ) \sin \left (b x +a \right ) b^{2} d^{2}+16 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right ) b c d -8 \sin \left (b x +a \right )^{2} b^{2} c^{2}-16 \sin \left (b x +a \right )^{2} b^{2} c d x -7 \sin \left (b x +a \right )^{2} b^{2} d^{2} x^{2}+18 \sin \left (b x +a \right )^{2} d^{2}+\sin \left (b x +a \right ) b^{3} d^{2} x^{3}+16 \sin \left (b x +a \right ) b c d -8 b^{2} c^{2}-16 b^{2} c d x -4 b^{2} d^{2} x^{2}}{8 \sin \left (b x +a \right ) b^{3}} \] Input:

int((d*x+c)^2*cos(b*x+a)*cot(b*x+a)^2,x)
 

Output:

( - 16*cos(a + b*x)*sin(a + b*x)*b*c*d - 18*cos(a + b*x)*sin(a + b*x)*b*d* 
*2*x + 4*cos(a + b*x)*b**2*d**2*x**2 + 2*int(x**2/(tan((a + b*x)/2)**4 + t 
an((a + b*x)/2)**2),x)*sin(a + b*x)*b**3*d**2 + 8*int((tan((a + b*x)/2)**3 
*x)/(tan((a + b*x)/2)**2 + 1),x)*sin(a + b*x)*b**2*d**2 + 16*log(tan((a + 
b*x)/2))*sin(a + b*x)*b*c*d - 8*sin(a + b*x)**2*b**2*c**2 - 16*sin(a + b*x 
)**2*b**2*c*d*x - 7*sin(a + b*x)**2*b**2*d**2*x**2 + 18*sin(a + b*x)**2*d* 
*2 + sin(a + b*x)*b**3*d**2*x**3 + 16*sin(a + b*x)*b*c*d - 8*b**2*c**2 - 1 
6*b**2*c*d*x - 4*b**2*d**2*x**2)/(8*sin(a + b*x)*b**3)