\(\int (c+d x)^2 \cos ^2(a+b x) \cot (a+b x) \, dx\) [166]

Optimal result

Integrand size = 22, antiderivative size = 172 \[ \int (c+d x)^2 \cos ^2(a+b x) \cot (a+b x) \, dx=\frac {(c+d x)^2}{4 b}-\frac {i (c+d x)^3}{3 d}+\frac {(c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i d (c+d x) \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}+\frac {d^2 \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac {d (c+d x) \cos (a+b x) \sin (a+b x)}{2 b^2}+\frac {d^2 \sin ^2(a+b x)}{4 b^3}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b} \] Output:

1/4*(d*x+c)^2/b-1/3*I*(d*x+c)^3/d+(d*x+c)^2*ln(1-exp(2*I*(b*x+a)))/b-I*d*( 
d*x+c)*polylog(2,exp(2*I*(b*x+a)))/b^2+1/2*d^2*polylog(3,exp(2*I*(b*x+a))) 
/b^3-1/2*d*(d*x+c)*cos(b*x+a)*sin(b*x+a)/b^2+1/4*d^2*sin(b*x+a)^2/b^3-1/2* 
(d*x+c)^2*sin(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. \(564\) vs. \(2(172)=344\).

Time = 1.99 (sec) , antiderivative size = 564, normalized size of antiderivative = 3.28 \[ \int (c+d x)^2 \cos ^2(a+b x) \cot (a+b x) \, dx=\frac {48 i b^2 c d \pi x+16 i b^3 d^2 x^3-96 i b^2 c d x \arctan (\tan (a))+48 b^3 c d x^2 \cot (a)-6 b c d \cos (a+2 b x) \csc (a)-6 b d^2 x \cos (a+2 b x) \csc (a)+6 b c d \cos (3 a+2 b x) \csc (a)+6 b d^2 x \cos (3 a+2 b x) \csc (a)+48 b c d \pi \log \left (1+e^{-2 i b x}\right )+48 b^2 d^2 x^2 \log \left (1-e^{-i (a+b x)}\right )+48 b^2 d^2 x^2 \log \left (1+e^{-i (a+b x)}\right )+96 b^2 c d x \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )+96 b c d \arctan (\tan (a)) \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )-48 b c d \pi \log (\cos (b x))+48 b^2 c^2 \log (\sin (a+b x))-96 b c d \arctan (\tan (a)) \log (\sin (b x+\arctan (\tan (a))))+96 i b d^2 x \operatorname {PolyLog}\left (2,-e^{-i (a+b x)}\right )+96 i b d^2 x \operatorname {PolyLog}\left (2,e^{-i (a+b x)}\right )-48 i b c d \operatorname {PolyLog}\left (2,e^{2 i (b x+\arctan (\tan (a)))}\right )+96 d^2 \operatorname {PolyLog}\left (3,-e^{-i (a+b x)}\right )+96 d^2 \operatorname {PolyLog}\left (3,e^{-i (a+b x)}\right )-48 b^3 c d e^{i \arctan (\tan (a))} x^2 \cot (a) \sqrt {\sec ^2(a)}-6 b^2 c^2 \csc (a) \sin (a+2 b x)+3 d^2 \csc (a) \sin (a+2 b x)-12 b^2 c d x \csc (a) \sin (a+2 b x)-6 b^2 d^2 x^2 \csc (a) \sin (a+2 b x)+6 b^2 c^2 \csc (a) \sin (3 a+2 b x)-3 d^2 \csc (a) \sin (3 a+2 b x)+12 b^2 c d x \csc (a) \sin (3 a+2 b x)+6 b^2 d^2 x^2 \csc (a) \sin (3 a+2 b x)}{48 b^3} \] Input:

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

Output:

((48*I)*b^2*c*d*Pi*x + (16*I)*b^3*d^2*x^3 - (96*I)*b^2*c*d*x*ArcTan[Tan[a] 
] + 48*b^3*c*d*x^2*Cot[a] - 6*b*c*d*Cos[a + 2*b*x]*Csc[a] - 6*b*d^2*x*Cos[ 
a + 2*b*x]*Csc[a] + 6*b*c*d*Cos[3*a + 2*b*x]*Csc[a] + 6*b*d^2*x*Cos[3*a + 
2*b*x]*Csc[a] + 48*b*c*d*Pi*Log[1 + E^((-2*I)*b*x)] + 48*b^2*d^2*x^2*Log[1 
 - E^((-I)*(a + b*x))] + 48*b^2*d^2*x^2*Log[1 + E^((-I)*(a + b*x))] + 96*b 
^2*c*d*x*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))] + 96*b*c*d*ArcTan[Tan[a 
]]*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))] - 48*b*c*d*Pi*Log[Cos[b*x]] + 
 48*b^2*c^2*Log[Sin[a + b*x]] - 96*b*c*d*ArcTan[Tan[a]]*Log[Sin[b*x + ArcT 
an[Tan[a]]]] + (96*I)*b*d^2*x*PolyLog[2, -E^((-I)*(a + b*x))] + (96*I)*b*d 
^2*x*PolyLog[2, E^((-I)*(a + b*x))] - (48*I)*b*c*d*PolyLog[2, E^((2*I)*(b* 
x + ArcTan[Tan[a]]))] + 96*d^2*PolyLog[3, -E^((-I)*(a + b*x))] + 96*d^2*Po 
lyLog[3, E^((-I)*(a + b*x))] - 48*b^3*c*d*E^(I*ArcTan[Tan[a]])*x^2*Cot[a]* 
Sqrt[Sec[a]^2] - 6*b^2*c^2*Csc[a]*Sin[a + 2*b*x] + 3*d^2*Csc[a]*Sin[a + 2* 
b*x] - 12*b^2*c*d*x*Csc[a]*Sin[a + 2*b*x] - 6*b^2*d^2*x^2*Csc[a]*Sin[a + 2 
*b*x] + 6*b^2*c^2*Csc[a]*Sin[3*a + 2*b*x] - 3*d^2*Csc[a]*Sin[3*a + 2*b*x] 
+ 12*b^2*c*d*x*Csc[a]*Sin[3*a + 2*b*x] + 6*b^2*d^2*x^2*Csc[a]*Sin[3*a + 2* 
b*x])/(48*b^3)
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.20, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {4908, 3042, 25, 4202, 2620, 3011, 2720, 4904, 3042, 3791, 17, 7143}

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.

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

\(\Big \downarrow \) 4908

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

\(\displaystyle 2 i \int \frac {e^{i (2 a+2 b x+\pi )} (c+d x)^2}{1+e^{i (2 a+2 b x+\pi )}}dx-\int (c+d x)^2 \cos (a+b x) \sin (a+b x)dx-\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 2620

\(\displaystyle 2 i \left (\frac {i d \int (c+d x) \log \left (1+e^{i (2 a+2 b x+\pi )}\right )dx}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )-\int (c+d x)^2 \cos (a+b x) \sin (a+b x)dx-\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 3011

\(\displaystyle 2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {i d \int \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )dx}{2 b}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )-\int (c+d x)^2 \cos (a+b x) \sin (a+b x)dx-\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 2720

\(\displaystyle 2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {d \int e^{-i (2 a+2 b x+\pi )} \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )de^{i (2 a+2 b x+\pi )}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )-\int (c+d x)^2 \cos (a+b x) \sin (a+b x)dx-\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 4904

\(\displaystyle 2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {d \int e^{-i (2 a+2 b x+\pi )} \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )de^{i (2 a+2 b x+\pi )}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )+\frac {d \int (c+d x) \sin ^2(a+b x)dx}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {d \int e^{-i (2 a+2 b x+\pi )} \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )de^{i (2 a+2 b x+\pi )}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )+\frac {d \int (c+d x) \sin (a+b x)^2dx}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 3791

\(\displaystyle 2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {d \int e^{-i (2 a+2 b x+\pi )} \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )de^{i (2 a+2 b x+\pi )}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )+\frac {d \left (\frac {1}{2} \int (c+d x)dx+\frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}\right )}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 17

\(\displaystyle 2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {d \int e^{-i (2 a+2 b x+\pi )} \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )de^{i (2 a+2 b x+\pi )}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )+\frac {d \left (\frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^2}{4 d}\right )}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 7143

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

Input:

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

Output:

((-1/3*I)*(c + d*x)^3)/d + (2*I)*(((-1/2*I)*(c + d*x)^2*Log[1 + E^(I*(2*a 
+ Pi + 2*b*x))])/b + (I*d*(((I/2)*(c + d*x)*PolyLog[2, -E^(I*(2*a + Pi + 2 
*b*x))])/b - (d*PolyLog[3, -E^(I*(2*a + Pi + 2*b*x))])/(4*b^2)))/b) - ((c 
+ d*x)^2*Sin[a + b*x]^2)/(2*b) + (d*((c + d*x)^2/(4*d) - ((c + d*x)*Cos[a 
+ b*x]*Sin[a + b*x])/(2*b) + (d*Sin[a + b*x]^2)/(4*b^2)))/b
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 25

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

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

rule 3791

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> 
 Simp[d*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Simp[b*(c + d*x)*Cos[e + f*x 
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n)   Int[(c + d* 
x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 
 1]
 

rule 4202

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

rule 4904

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x 
_)]^(n_.), x_Symbol] :> Simp[(c + d*x)^m*(Sin[a + b*x]^(n + 1)/(b*(n + 1))) 
, x] - Simp[d*(m/(b*(n + 1)))   Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n + 1), 
 x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]
 

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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 543 vs. \(2 (152 ) = 304\).

Time = 1.27 (sec) , antiderivative size = 544, normalized size of antiderivative = 3.16

method	result	size
risch	\(\frac {2 d c \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {2 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}+\frac {4 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {4 i d c a x}{b}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}+\frac {d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}-\frac {2 d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {4 i d^{2} a^{3}}{3 b^{3}}-i d c \,x^{2}-\frac {2 i d c \,a^{2}}{b^{2}}+\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}-d^{2}\right ) \cos \left (2 b x +2 a \right )}{8 b^{3}}+\frac {2 d c \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {2 i d c \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i d c \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 i d^{2} a^{2} x}{b^{2}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}+i c^{2} x +\frac {i c^{3}}{3 d}-\frac {i d^{2} x^{3}}{3}+\frac {2 d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {d \left (d x +c \right ) \sin \left (2 b x +2 a \right )}{4 b^{2}}\)	\(544\)

Input:

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

Output:

-4*I/b*d*c*a*x-2*I/b^2*d*c*polylog(2,-exp(I*(b*x+a)))-2*I/b^2*d*c*polylog( 
2,exp(I*(b*x+a)))+2*I/b^2*d^2*a^2*x-2*I/b^2*d^2*polylog(2,-exp(I*(b*x+a))) 
*x-2*I/b^2*d^2*polylog(2,exp(I*(b*x+a)))*x+1/b^3*d^2*a^2*ln(exp(I*(b*x+a)) 
-1)-2/b^3*d^2*a^2*ln(exp(I*(b*x+a)))-1/b^3*d^2*ln(1-exp(I*(b*x+a)))*a^2+1/ 
b*d^2*ln(exp(I*(b*x+a))+1)*x^2+1/b*d^2*ln(1-exp(I*(b*x+a)))*x^2+4/3*I/b^3* 
d^2*a^3+2/b^2*d*c*ln(1-exp(I*(b*x+a)))*a+2/b*d*c*ln(exp(I*(b*x+a))+1)*x+2/ 
b*d*c*ln(1-exp(I*(b*x+a)))*x-2/b^2*c*d*a*ln(exp(I*(b*x+a))-1)+4/b^2*c*d*a* 
ln(exp(I*(b*x+a)))-2*I/b^2*d*c*a^2-I*d*c*x^2+1/8*(2*b^2*d^2*x^2+4*b^2*c*d* 
x+2*b^2*c^2-d^2)/b^3*cos(2*b*x+2*a)-2/b*c^2*ln(exp(I*(b*x+a)))+1/b*c^2*ln( 
exp(I*(b*x+a))-1)+1/b*c^2*ln(exp(I*(b*x+a))+1)+I*c^2*x+1/3*I/d*c^3-1/3*I*d 
^2*x^3+2*d^2*polylog(3,-exp(I*(b*x+a)))/b^3+2*d^2*polylog(3,exp(I*(b*x+a)) 
)/b^3-1/4*d*(d*x+c)*sin(2*b*x+2*a)/b^2
 

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. 594 vs. \(2 (149) = 298\).

Time = 0.12 (sec) , antiderivative size = 594, normalized size of antiderivative = 3.45 \[ \int (c+d x)^2 \cos ^2(a+b x) \cot (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/4*(b^2*d^2*x^2 + 2*b^2*c*d*x - (2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 
 - d^2)*cos(b*x + a)^2 - 4*d^2*polylog(3, cos(b*x + a) + I*sin(b*x + a)) - 
 4*d^2*polylog(3, cos(b*x + a) - I*sin(b*x + a)) - 4*d^2*polylog(3, -cos(b 
*x + a) + I*sin(b*x + a)) - 4*d^2*polylog(3, -cos(b*x + a) - I*sin(b*x + a 
)) + 2*(b*d^2*x + b*c*d)*cos(b*x + a)*sin(b*x + a) + 4*(I*b*d^2*x + I*b*c* 
d)*dilog(cos(b*x + a) + I*sin(b*x + a)) + 4*(-I*b*d^2*x - I*b*c*d)*dilog(c 
os(b*x + a) - I*sin(b*x + a)) + 4*(-I*b*d^2*x - I*b*c*d)*dilog(-cos(b*x + 
a) + I*sin(b*x + a)) + 4*(I*b*d^2*x + I*b*c*d)*dilog(-cos(b*x + a) - I*sin 
(b*x + a)) - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(cos(b*x + a) + I* 
sin(b*x + a) + 1) - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(cos(b*x + 
a) - I*sin(b*x + a) + 1) - 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-1/2*cos( 
b*x + a) + 1/2*I*sin(b*x + a) + 1/2) - 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*l 
og(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2) - 2*(b^2*d^2*x^2 + 2*b^2* 
c*d*x + 2*a*b*c*d - a^2*d^2)*log(-cos(b*x + a) + I*sin(b*x + a) + 1) - 2*( 
b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(-cos(b*x + a) - I*sin 
(b*x + a) + 1))/b^3
 

Sympy [F]

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

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

Output:

Integral((c + d*x)**2*cos(a + b*x)**2*cot(a + b*x), 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. 529 vs. \(2 (149) = 298\).

Time = 0.14 (sec) , antiderivative size = 529, normalized size of antiderivative = 3.08 \[ \int (c+d x)^2 \cos ^2(a+b x) \cot (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/24*(12*(sin(b*x + a)^2 - log(sin(b*x + a)^2))*c^2 - 24*(sin(b*x + a)^2 
- log(sin(b*x + a)^2))*a*c*d/b + 12*(sin(b*x + a)^2 - log(sin(b*x + a)^2)) 
*a^2*d^2/b^2 - (-8*I*(b*x + a)^3*d^2 - 24*(I*b*c*d - I*a*d^2)*(b*x + a)^2 
+ 48*d^2*polylog(3, -e^(I*b*x + I*a)) + 48*d^2*polylog(3, e^(I*b*x + I*a)) 
 - 24*(-I*(b*x + a)^2*d^2 + 2*(-I*b*c*d + I*a*d^2)*(b*x + a))*arctan2(sin( 
b*x + a), cos(b*x + a) + 1) - 24*(I*(b*x + a)^2*d^2 + 2*(I*b*c*d - I*a*d^2 
)*(b*x + a))*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 3*(2*(b*x + a)^2*d 
^2 + 4*(b*c*d - a*d^2)*(b*x + a) - d^2)*cos(2*b*x + 2*a) - 48*(I*b*c*d + I 
*(b*x + a)*d^2 - I*a*d^2)*dilog(-e^(I*b*x + I*a)) - 48*(I*b*c*d + I*(b*x + 
 a)*d^2 - I*a*d^2)*dilog(e^(I*b*x + I*a)) + 12*((b*x + a)^2*d^2 + 2*(b*c*d 
 - a*d^2)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) 
+ 1) + 12*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*log(cos(b*x + a) 
^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - 6*(b*c*d + (b*x + a)*d^2 - a*d 
^2)*sin(2*b*x + 2*a))/b^2)/b
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*int(cos(a + b*x)**2*cot(a + b*x)*x**2,x)*b*d**2 + 4*int(cos(a + b*x)**2 
*cot(a + b*x)*x,x)*b*c*d - 2*log(tan((a + b*x)/2)**2 + 1)*c**2 + 2*log(tan 
((a + b*x)/2))*c**2 - sin(a + b*x)**2*c**2)/(2*b)