\(\int x^2 \arctan (c+d \cot (a+b x)) \, dx\) [61]

Optimal result

Integrand size = 15, antiderivative size = 399 \[ \int x^2 \arctan (c+d \cot (a+b x)) \, dx=\frac {1}{3} x^3 \arctan (c+d \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}-\frac {\operatorname {PolyLog}\left (4,\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{8 b^3}+\frac {\operatorname {PolyLog}\left (4,\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{8 b^3} \] Output:

1/3*x^3*arctan(c+d*cot(b*x+a))+1/6*I*x^3*ln(1-(1+I*c-d)*exp(2*I*a+2*I*b*x) 
/(1+I*c+d))-1/6*I*x^3*ln(1-(c+I*(1+d))*exp(2*I*a+2*I*b*x)/(c+I*(1-d)))+1/4 
*x^2*polylog(2,(1+I*c-d)*exp(2*I*a+2*I*b*x)/(1+I*c+d))/b-1/4*x^2*polylog(2 
,(c+I*(1+d))*exp(2*I*a+2*I*b*x)/(c+I*(1-d)))/b+1/4*I*x*polylog(3,(1+I*c-d) 
*exp(2*I*a+2*I*b*x)/(1+I*c+d))/b^2-1/4*I*x*polylog(3,(c+I*(1+d))*exp(2*I*a 
+2*I*b*x)/(c+I*(1-d)))/b^2-1/8*polylog(4,(1+I*c-d)*exp(2*I*a+2*I*b*x)/(1+I 
*c+d))/b^3+1/8*polylog(4,(c+I*(1+d))*exp(2*I*a+2*I*b*x)/(c+I*(1-d)))/b^3
 

Mathematica [A] (verified)

Time = 2.65 (sec) , antiderivative size = 360, normalized size of antiderivative = 0.90 \[ \int x^2 \arctan (c+d \cot (a+b x)) \, dx=\frac {8 b^3 x^3 \arctan (c+d \cot (a+b x))+4 i b^3 x^3 \log \left (1+\frac {(-c+i (1+d)) e^{-2 i (a+b x)}}{c+i (-1+d)}\right )-4 i b^3 x^3 \log \left (1+\frac {(-c+i (-1+d)) e^{-2 i (a+b x)}}{c+i (1+d)}\right )-6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(c-i (1+d)) e^{-2 i (a+b x)}}{c+i (-1+d)}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(i+c-i d) e^{-2 i (a+b x)}}{c+i (1+d)}\right )+6 i b x \operatorname {PolyLog}\left (3,\frac {(c-i (1+d)) e^{-2 i (a+b x)}}{c+i (-1+d)}\right )-6 i b x \operatorname {PolyLog}\left (3,\frac {(i+c-i d) e^{-2 i (a+b x)}}{c+i (1+d)}\right )+3 \operatorname {PolyLog}\left (4,\frac {(c-i (1+d)) e^{-2 i (a+b x)}}{c+i (-1+d)}\right )-3 \operatorname {PolyLog}\left (4,\frac {(i+c-i d) e^{-2 i (a+b x)}}{c+i (1+d)}\right )}{24 b^3} \] Input:

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

Output:

(8*b^3*x^3*ArcTan[c + d*Cot[a + b*x]] + (4*I)*b^3*x^3*Log[1 + (-c + I*(1 + 
 d))/((c + I*(-1 + d))*E^((2*I)*(a + b*x)))] - (4*I)*b^3*x^3*Log[1 + (-c + 
 I*(-1 + d))/((c + I*(1 + d))*E^((2*I)*(a + b*x)))] - 6*b^2*x^2*PolyLog[2, 
 (c - I*(1 + d))/((c + I*(-1 + d))*E^((2*I)*(a + b*x)))] + 6*b^2*x^2*PolyL 
og[2, (I + c - I*d)/((c + I*(1 + d))*E^((2*I)*(a + b*x)))] + (6*I)*b*x*Pol 
yLog[3, (c - I*(1 + d))/((c + I*(-1 + d))*E^((2*I)*(a + b*x)))] - (6*I)*b* 
x*PolyLog[3, (I + c - I*d)/((c + I*(1 + d))*E^((2*I)*(a + b*x)))] + 3*Poly 
Log[4, (c - I*(1 + d))/((c + I*(-1 + d))*E^((2*I)*(a + b*x)))] - 3*PolyLog 
[4, (I + c - I*d)/((c + I*(1 + d))*E^((2*I)*(a + b*x)))])/(24*b^3)
 

Rubi [A] (verified)

Time = 1.55 (sec) , antiderivative size = 515, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5700, 2620, 3011, 7163, 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.

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

\(\Big \downarrow \) 5700

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[x^2*ArcTan[c + d*Cot[a + b*x]],x]
 

Output:

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

Defintions of rubi rules used

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 5700

Int[ArcTan[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_. 
), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTan[c + d*Cot[a + b*x]]/(f*(m + 
1))), x] + (Simp[b*((1 + I*c - d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*(E^( 
2*I*a + 2*I*b*x)/(1 + I*c + d - (1 + I*c - d)*E^(2*I*a + 2*I*b*x))), x], x] 
 - Simp[b*((1 - I*c + d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*(E^(2*I*a + 2 
*I*b*x)/(1 - I*c - d - (1 - I*c + d)*E^(2*I*a + 2*I*b*x))), x], x]) /; Free 
Q[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NeQ[(c - I*d)^2, -1]
 

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]
 

rule 7163

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 60.35 (sec) , antiderivative size = 7869, normalized size of antiderivative = 19.72

method	result	size
risch	\(\text {Expression too large to display}\)	\(7869\)

Input:

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

Output:

result too large to display
 

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. 1589 vs. \(2 (283) = 566\).

Time = 0.27 (sec) , antiderivative size = 1589, normalized size of antiderivative = 3.98 \[ \int x^2 \arctan (c+d \cot (a+b x)) \, dx=\text {Too large to display} \] Input:

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

Output:

1/48*(16*b^3*x^3*arctan(d*cot(b*x + a) + c) + 6*b^2*x^2*dilog(-(c^2 + d^2 
- (c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 + 2*c*d + I*d^2 - I 
)*sin(2*b*x + 2*a) + 2*d + 1)/(c^2 + d^2 + 2*d + 1) + 1) + 6*b^2*x^2*dilog 
(-(c^2 + d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d 
 - I*d^2 + I)*sin(2*b*x + 2*a) + 2*d + 1)/(c^2 + d^2 + 2*d + 1) + 1) - 6*b 
^2*x^2*dilog(-(c^2 + d^2 - (c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (- 
I*c^2 + 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a) - 2*d + 1)/(c^2 + d^2 - 2*d + 
1) + 1) - 6*b^2*x^2*dilog(-(c^2 + d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b* 
x + 2*a) + (I*c^2 + 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a) - 2*d + 1)/(c^2 + 
d^2 - 2*d + 1) + 1) - 4*I*a^3*log(1/2*c^2 + I*c*d - 1/2*d^2 - 1/2*(c^2 + d 
^2 + 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 + 2*I*d + I)*sin(2*b*x 
 + 2*a) + 1/2) + 4*I*a^3*log(1/2*c^2 + I*c*d - 1/2*d^2 - 1/2*(c^2 + d^2 - 
2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 - 2*I*d + I)*sin(2*b*x + 2* 
a) + 1/2) + 4*I*a^3*log(-1/2*c^2 + I*c*d + 1/2*d^2 + 1/2*(c^2 + d^2 + 2*d 
+ 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 + 2*I*d + I)*sin(2*b*x + 2*a) - 
 1/2) - 4*I*a^3*log(-1/2*c^2 + I*c*d + 1/2*d^2 + 1/2*(c^2 + d^2 - 2*d + 1) 
*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 - 2*I*d + I)*sin(2*b*x + 2*a) - 1/2 
) + 6*I*b*x*polylog(3, ((c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^ 
2 - 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a))/(c^2 + d^2 + 2*d + 1)) - 6*I*b*x* 
polylog(3, ((c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 - 2*c*d...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

-1/6*x^3*arctan2(-c*cos(2*b*x + 2*a) + (d + 1)*sin(2*b*x + 2*a) + c, (d + 
1)*cos(2*b*x + 2*a) + c*sin(2*b*x + 2*a) + d - 1) - 1/6*x^3*arctan2(-c*cos 
(2*b*x + 2*a) + (d - 1)*sin(2*b*x + 2*a) + c, -(d - 1)*cos(2*b*x + 2*a) - 
c*sin(2*b*x + 2*a) - d - 1) + 4*b*d*integrate(1/3*(2*(c^2 + d^2 + 1)*x^3*c 
os(2*b*x + 2*a)^2 + 2*c*d*x^3*sin(2*b*x + 2*a) + 2*(c^2 + d^2 + 1)*x^3*sin 
(2*b*x + 2*a)^2 - (c^2 - d^2 + 1)*x^3*cos(2*b*x + 2*a) - (2*c*d*x^3*sin(2* 
b*x + 2*a) + (c^2 - d^2 + 1)*x^3*cos(2*b*x + 2*a))*cos(4*b*x + 4*a) + (2*c 
*d*x^3*cos(2*b*x + 2*a) - (c^2 - d^2 + 1)*x^3*sin(2*b*x + 2*a))*sin(4*b*x 
+ 4*a))/(c^4 + d^4 + 2*(c^2 - 1)*d^2 + (c^4 + d^4 + 2*(c^2 - 1)*d^2 + 2*c^ 
2 + 1)*cos(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 + 1)*d^2 + 2*c^2 + 1)*co 
s(2*b*x + 2*a)^2 + (c^4 + d^4 + 2*(c^2 - 1)*d^2 + 2*c^2 + 1)*sin(4*b*x + 4 
*a)^2 + 4*(c^4 + d^4 + 2*(c^2 + 1)*d^2 + 2*c^2 + 1)*sin(2*b*x + 2*a)^2 + 2 
*c^2 + 2*(c^4 + d^4 - 2*(3*c^2 + 1)*d^2 + 2*c^2 - 2*(c^4 - d^4 + 2*c^2 + 1 
)*cos(2*b*x + 2*a) - 4*(c*d^3 + (c^3 + c)*d)*sin(2*b*x + 2*a) + 1)*cos(4*b 
*x + 4*a) - 4*(c^4 - d^4 + 2*c^2 + 1)*cos(2*b*x + 2*a) + 4*(2*c*d^3 - 2*(c 
^3 + c)*d + 2*(c*d^3 + (c^3 + c)*d)*cos(2*b*x + 2*a) - (c^4 - d^4 + 2*c^2 
+ 1)*sin(2*b*x + 2*a))*sin(4*b*x + 4*a) + 8*(c*d^3 + (c^3 + c)*d)*sin(2*b* 
x + 2*a) + 1), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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