\(\int x^2 \text {arctanh}(c+d \cot (a+b x)) \, dx\) [334]

Optimal result

Integrand size = 15, antiderivative size = 391 \[ \int x^2 \text {arctanh}(c+d \cot (a+b x)) \, dx=\frac {1}{3} x^3 \text {arctanh}(c+d \cot (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b}+\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b}+\frac {x \operatorname {PolyLog}\left (3,\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (3,\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b^2}+\frac {i \operatorname {PolyLog}\left (4,\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{8 b^3}-\frac {i \operatorname {PolyLog}\left (4,\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{8 b^3} \] Output:

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

Mathematica [A] (verified)

Time = 3.33 (sec) , antiderivative size = 341, normalized size of antiderivative = 0.87 \[ \int x^2 \text {arctanh}(c+d \cot (a+b x)) \, dx=\frac {8 b^3 x^3 \text {arctanh}(c+d \cot (a+b x))+4 b^3 x^3 \log \left (1+\frac {(1-c+i d) e^{-2 i (a+b x)}}{-1+c+i d}\right )-4 b^3 x^3 \log \left (1+\frac {(-1-c+i d) e^{-2 i (a+b x)}}{1+c+i d}\right )+6 i b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(-1+c-i d) e^{-2 i (a+b x)}}{-1+c+i d}\right )-6 i b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(1+c-i d) e^{-2 i (a+b x)}}{1+c+i d}\right )+6 b x \operatorname {PolyLog}\left (3,\frac {(-1+c-i d) e^{-2 i (a+b x)}}{-1+c+i d}\right )-6 b x \operatorname {PolyLog}\left (3,\frac {(1+c-i d) e^{-2 i (a+b x)}}{1+c+i d}\right )-3 i \operatorname {PolyLog}\left (4,\frac {(-1+c-i d) e^{-2 i (a+b x)}}{-1+c+i d}\right )+3 i \operatorname {PolyLog}\left (4,\frac {(1+c-i d) e^{-2 i (a+b x)}}{1+c+i d}\right )}{24 b^3} \] Input:

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

Output:

(8*b^3*x^3*ArcTanh[c + d*Cot[a + b*x]] + 4*b^3*x^3*Log[1 + (1 - c + I*d)/( 
(-1 + c + I*d)*E^((2*I)*(a + b*x)))] - 4*b^3*x^3*Log[1 + (-1 - c + I*d)/(( 
1 + c + I*d)*E^((2*I)*(a + b*x)))] + (6*I)*b^2*x^2*PolyLog[2, (-1 + c - I* 
d)/((-1 + c + I*d)*E^((2*I)*(a + b*x)))] - (6*I)*b^2*x^2*PolyLog[2, (1 + c 
 - I*d)/((1 + c + I*d)*E^((2*I)*(a + b*x)))] + 6*b*x*PolyLog[3, (-1 + c - 
I*d)/((-1 + c + I*d)*E^((2*I)*(a + b*x)))] - 6*b*x*PolyLog[3, (1 + c - I*d 
)/((1 + c + I*d)*E^((2*I)*(a + b*x)))] - (3*I)*PolyLog[4, (-1 + c - I*d)/( 
(-1 + c + I*d)*E^((2*I)*(a + b*x)))] + (3*I)*PolyLog[4, (1 + c - I*d)/((1 
+ c + I*d)*E^((2*I)*(a + b*x)))])/(24*b^3)
 

Rubi [A] (verified)

Time = 1.50 (sec) , antiderivative size = 514, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6823, 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.

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

\(\Big \downarrow \) 6823

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

(x^3*ArcTanh[c + d*Cot[a + b*x]])/3 - (b*(I - I*c + d)*(-1/2*(x^3*Log[1 - 
((1 - c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c + I*d)])/(b*(I*(1 - c) + d) 
) + (3*(((I/2)*x^2*PolyLog[2, ((1 - c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - 
 c + I*d)])/b - (I*(((-1/2*I)*x*PolyLog[3, ((1 - c - I*d)*E^((2*I)*a + (2* 
I)*b*x))/(1 - c + I*d)])/b + PolyLog[4, ((1 - c - I*d)*E^((2*I)*a + (2*I)* 
b*x))/(1 - c + I*d)]/(4*b^2)))/b))/(2*b*(I*(1 - c) + d))))/3 + (b*(I*(1 + 
c) - d)*(-1/2*(x^3*Log[1 - ((1 + c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c 
- I*d)])/(I*(b + b*c) - b*d) + (3*(((I/2)*x^2*PolyLog[2, ((1 + c + I*d)*E^ 
((2*I)*a + (2*I)*b*x))/(1 + c - I*d)])/b - (I*(((-1/2*I)*x*PolyLog[3, ((1 
+ c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c - I*d)])/b + PolyLog[4, ((1 + c 
 + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c - I*d)]/(4*b^2)))/b))/(2*(I*(b + b 
*c) - b*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 6823

Int[ArcTanh[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Cot[a + b*x]]/(f*(m 
+ 1))), x] + (-Simp[I*b*((1 - c - I*d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1) 
*(E^(2*I*a + 2*I*b*x)/(1 - c + I*d - (1 - c - I*d)*E^(2*I*a + 2*I*b*x))), x 
], x] + Simp[I*b*((1 + c + I*d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*(E^(2* 
I*a + 2*I*b*x)/(1 + c - I*d - (1 + c + I*d)*E^(2*I*a + 2*I*b*x))), x], x]) 
/; FreeQ[{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 = 49.45 (sec) , antiderivative size = 6725, normalized size of antiderivative = 17.20

method	result	size
risch	\(\text {Expression too large to display}\)	\(6725\)

Input:

int(x^2*arctanh(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. 1799 vs. \(2 (275) = 550\).

Time = 0.27 (sec) , antiderivative size = 1799, normalized size of antiderivative = 4.60 \[ \int x^2 \text {arctanh}(c+d \cot (a+b x)) \, dx=\text {Too large to display} \] Input:

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

Output:

1/48*(8*b^3*x^3*log(-(d*cos(2*b*x + 2*a) + (c + 1)*sin(2*b*x + 2*a) + d)/( 
d*cos(2*b*x + 2*a) + (c - 1)*sin(2*b*x + 2*a) + d)) + 6*I*b^2*x^2*dilog(-( 
c^2 + d^2 - (c^2 + 2*I*(c + 1)*d - d^2 + 2*c + 1)*cos(2*b*x + 2*a) + (-I*c 
^2 + 2*(c + 1)*d + I*d^2 - 2*I*c - I)*sin(2*b*x + 2*a) + 2*c + 1)/(c^2 + d 
^2 + 2*c + 1) + 1) - 6*I*b^2*x^2*dilog(-(c^2 + d^2 - (c^2 - 2*I*(c + 1)*d 
- d^2 + 2*c + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*(c + 1)*d - I*d^2 + 2*I*c + 
 I)*sin(2*b*x + 2*a) + 2*c + 1)/(c^2 + d^2 + 2*c + 1) + 1) - 6*I*b^2*x^2*d 
ilog(-(c^2 + d^2 - (c^2 + 2*I*(c - 1)*d - d^2 - 2*c + 1)*cos(2*b*x + 2*a) 
+ (-I*c^2 + 2*(c - 1)*d + I*d^2 + 2*I*c - I)*sin(2*b*x + 2*a) - 2*c + 1)/( 
c^2 + d^2 - 2*c + 1) + 1) + 6*I*b^2*x^2*dilog(-(c^2 + d^2 - (c^2 - 2*I*(c 
- 1)*d - d^2 - 2*c + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*(c - 1)*d - I*d^2 - 
2*I*c + I)*sin(2*b*x + 2*a) - 2*c + 1)/(c^2 + d^2 - 2*c + 1) + 1) + 4*a^3* 
log(1/2*c^2 + I*(c + 1)*d - 1/2*d^2 - 1/2*(c^2 + d^2 + 2*c + 1)*cos(2*b*x 
+ 2*a) + 1/2*(I*c^2 + I*d^2 + 2*I*c + I)*sin(2*b*x + 2*a) + c + 1/2) - 4*a 
^3*log(1/2*c^2 + I*(c - 1)*d - 1/2*d^2 - 1/2*(c^2 + d^2 - 2*c + 1)*cos(2*b 
*x + 2*a) + 1/2*(I*c^2 + I*d^2 - 2*I*c + I)*sin(2*b*x + 2*a) - c + 1/2) + 
4*a^3*log(-1/2*c^2 + I*(c + 1)*d + 1/2*d^2 + 1/2*(c^2 + d^2 + 2*c + 1)*cos 
(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 + 2*I*c + I)*sin(2*b*x + 2*a) - c - 1/2 
) - 4*a^3*log(-1/2*c^2 + I*(c - 1)*d + 1/2*d^2 + 1/2*(c^2 + d^2 - 2*c + 1) 
*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 - 2*I*c + I)*sin(2*b*x + 2*a) + ...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/12*x^3*log((c^2 + d^2 + 2*c + 1)*cos(2*b*x + 2*a)^2 + 4*(c + 1)*d*sin(2* 
b*x + 2*a) + (c^2 + d^2 + 2*c + 1)*sin(2*b*x + 2*a)^2 + c^2 + d^2 - 2*(c^2 
 - d^2 + 2*c + 1)*cos(2*b*x + 2*a) + 2*c + 1) - 1/12*x^3*log((c^2 + d^2 - 
2*c + 1)*cos(2*b*x + 2*a)^2 + 4*(c - 1)*d*sin(2*b*x + 2*a) + (c^2 + d^2 - 
2*c + 1)*sin(2*b*x + 2*a)^2 + c^2 + d^2 - 2*(c^2 - d^2 - 2*c + 1)*cos(2*b* 
x + 2*a) - 2*c + 1) - 4*b*d*integrate(1/3*(2*(c^2 + d^2 - 1)*x^3*cos(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*c 
os(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)*c 
os(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 - 1)*d^2 - 2*c^2 + 1)*cos(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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