\(\int (c+d x)^2 \cot (a+b x) \, dx\) [34]

Optimal result

Integrand size = 14, antiderivative size = 93 \[ \int (c+d x)^2 \cot (a+b x) \, dx=-\frac {i (c+d x)^3}{3 d}+\frac {(c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i d (c+d x) \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}+\frac {d^2 \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3} \] Output:

-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
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(356\) vs. \(2(93)=186\).

Time = 0.98 (sec) , antiderivative size = 356, normalized size of antiderivative = 3.83 \[ \int (c+d x)^2 \cot (a+b x) \, dx=\frac {3 i b^2 c d \pi x+i b^3 d^2 x^3-6 i b^2 c d x \arctan (\tan (a))+3 b^3 c d x^2 \cot (a)+3 b c d \pi \log \left (1+e^{-2 i b x}\right )+3 b^2 d^2 x^2 \log \left (1-e^{-i (a+b x)}\right )+3 b^2 d^2 x^2 \log \left (1+e^{-i (a+b x)}\right )+6 b^2 c d x \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )+6 b c d \arctan (\tan (a)) \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )-3 b c d \pi \log (\cos (b x))+3 b^2 c^2 \log (\sin (a+b x))-6 b c d \arctan (\tan (a)) \log (\sin (b x+\arctan (\tan (a))))+6 i b d^2 x \operatorname {PolyLog}\left (2,-e^{-i (a+b x)}\right )+6 i b d^2 x \operatorname {PolyLog}\left (2,e^{-i (a+b x)}\right )-3 i b c d \operatorname {PolyLog}\left (2,e^{2 i (b x+\arctan (\tan (a)))}\right )+6 d^2 \operatorname {PolyLog}\left (3,-e^{-i (a+b x)}\right )+6 d^2 \operatorname {PolyLog}\left (3,e^{-i (a+b x)}\right )-3 b^3 c d e^{i \arctan (\tan (a))} x^2 \cot (a) \sqrt {\sec ^2(a)}}{3 b^3} \] Input:

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

Output:

((3*I)*b^2*c*d*Pi*x + I*b^3*d^2*x^3 - (6*I)*b^2*c*d*x*ArcTan[Tan[a]] + 3*b 
^3*c*d*x^2*Cot[a] + 3*b*c*d*Pi*Log[1 + E^((-2*I)*b*x)] + 3*b^2*d^2*x^2*Log 
[1 - E^((-I)*(a + b*x))] + 3*b^2*d^2*x^2*Log[1 + E^((-I)*(a + b*x))] + 6*b 
^2*c*d*x*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))] + 6*b*c*d*ArcTan[Tan[a] 
]*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))] - 3*b*c*d*Pi*Log[Cos[b*x]] + 3 
*b^2*c^2*Log[Sin[a + b*x]] - 6*b*c*d*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[T 
an[a]]]] + (6*I)*b*d^2*x*PolyLog[2, -E^((-I)*(a + b*x))] + (6*I)*b*d^2*x*P 
olyLog[2, E^((-I)*(a + b*x))] - (3*I)*b*c*d*PolyLog[2, E^((2*I)*(b*x + Arc 
Tan[Tan[a]]))] + 6*d^2*PolyLog[3, -E^((-I)*(a + b*x))] + 6*d^2*PolyLog[3, 
E^((-I)*(a + b*x))] - 3*b^3*c*d*E^(I*ArcTan[Tan[a]])*x^2*Cot[a]*Sqrt[Sec[a 
]^2])/(3*b^3)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.34, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 25, 4202, 2620, 3011, 2720, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[(c + d*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)
 

Defintions of rubi rules used

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 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 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. 476 vs. \(2 (81 ) = 162\).

Time = 0.33 (sec) , antiderivative size = 477, normalized size of antiderivative = 5.13

method	result	size
risch	\(\frac {2 d c \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {2 d c \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {4 i d c a x}{b}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {2 i d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}-\frac {2 d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {4 i d^{2} a^{3}}{3 b^{3}}-i d c \,x^{2}+\frac {4 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}-\frac {2 i d c \,a^{2}}{b^{2}}-\frac {2 i d c \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i d c \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 i d^{2} a^{2} x}{b^{2}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}+\frac {i c^{3}}{3 d}+\frac {2 d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {i d^{2} x^{3}}{3}+i c^{2} x -\frac {2 c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}\)	\(477\)

Input:

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

Output:

-1/3*I*d^2*x^3+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*d*c*ln(1-exp(I*(b*x+a)))*a+4/b^2*c*d*a*ln(exp(I*(b*x+a)))-2/b^2*c 
*d*a*ln(exp(I*(b*x+a))-1)-2*I/b^2*d*c*a^2-2*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-4*I/ 
b*d*c*a*x+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^ 
2-1/b^3*d^2*ln(1-exp(I*(b*x+a)))*a^2-2/b^3*d^2*a^2*ln(exp(I*(b*x+a)))+1/b^ 
3*d^2*a^2*ln(exp(I*(b*x+a))-1)+4/3*I/b^3*d^2*a^3-I*d*c*x^2+I*c^2*x-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)+1/3*I/d*c^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
 

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. 350 vs. \(2 (78) = 156\).

Time = 0.09 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.76 \[ \int (c+d x)^2 \cot (a+b x) \, dx=\frac {d^{2} {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + d^{2} {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) - 2 \, {\left (i \, b d^{2} x + i \, b c d\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) - 2 \, {\left (-i \, b d^{2} x - i \, b c d\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right )}{4 \, b^{3}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((c + d*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. 411 vs. \(2 (78) = 156\).

Time = 0.13 (sec) , antiderivative size = 411, normalized size of antiderivative = 4.42 \[ \int (c+d x)^2 \cot (a+b x) \, dx=\frac {6 \, c^{2} \log \left (\sin \left (b x + a\right )\right ) - \frac {12 \, a c d \log \left (\sin \left (b x + a\right )\right )}{b} + \frac {6 \, a^{2} d^{2} \log \left (\sin \left (b x + a\right )\right )}{b^{2}} + \frac {-2 i \, {\left (b x + a\right )}^{3} d^{2} - 6 \, {\left (i \, b c d - i \, a d^{2}\right )} {\left (b x + a\right )}^{2} + 12 \, d^{2} {\rm Li}_{3}(-e^{\left (i \, b x + i \, a\right )}) + 12 \, d^{2} {\rm Li}_{3}(e^{\left (i \, b x + i \, a\right )}) - 6 \, {\left (-i \, {\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (-i \, b c d + i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - 6 \, {\left (i \, {\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (i \, b c d - i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) - 12 \, {\left (i \, b c d + i \, {\left (b x + a\right )} d^{2} - i \, a d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 12 \, {\left (i \, b c d + i \, {\left (b x + a\right )} d^{2} - i \, a d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + 3 \, {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) + 3 \, {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right )}{b^{2}}}{6 \, b} \] Input:

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

Output:

1/6*(6*c^2*log(sin(b*x + a)) - 12*a*c*d*log(sin(b*x + a))/b + 6*a^2*d^2*lo 
g(sin(b*x + a))/b^2 + (-2*I*(b*x + a)^3*d^2 - 6*(I*b*c*d - I*a*d^2)*(b*x + 
 a)^2 + 12*d^2*polylog(3, -e^(I*b*x + I*a)) + 12*d^2*polylog(3, e^(I*b*x + 
 I*a)) - 6*(-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) - 6*(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) - 12*(I*b*c*d + 
I*(b*x + a)*d^2 - I*a*d^2)*dilog(-e^(I*b*x + I*a)) - 12*(I*b*c*d + I*(b*x 
+ a)*d^2 - I*a*d^2)*dilog(e^(I*b*x + I*a)) + 3*((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) + 3*((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))/b^2)/b
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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