\(\int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx\) [338]

Optimal result

Integrand size = 20, antiderivative size = 168 \[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \text {arctanh}(1+i d+d \cot (a+b x))-\frac {1}{6} x^3 \log \left (1-(1+i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,(1+i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,(1+i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i \operatorname {PolyLog}\left (4,(1+i d) e^{2 i a+2 i b x}\right )}{8 b^3} \] Output:

1/12*I*b*x^4+1/3*x^3*arctanh(1+I*d+d*cot(b*x+a))-1/6*x^3*ln(1-(1+I*d)*exp( 
2*I*a+2*I*b*x))+1/4*I*x^2*polylog(2,(1+I*d)*exp(2*I*a+2*I*b*x))/b-1/4*x*po 
lylog(3,(1+I*d)*exp(2*I*a+2*I*b*x))/b^2-1/8*I*polylog(4,(1+I*d)*exp(2*I*a+ 
2*I*b*x))/b^3
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.92 \[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\frac {1}{3} x^3 \text {arctanh}(1+i d+d \cot (a+b x))-\frac {4 b^3 x^3 \log \left (1+\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+6 i b^2 x^2 \operatorname {PolyLog}\left (2,-\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+6 b x \operatorname {PolyLog}\left (3,-\frac {i e^{-2 i (a+b x)}}{-i+d}\right )-3 i \operatorname {PolyLog}\left (4,-\frac {i e^{-2 i (a+b x)}}{-i+d}\right )}{24 b^3} \] Input:

Integrate[x^2*ArcTanh[1 + I*d + d*Cot[a + b*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.30, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6819, 2615, 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.

\(\displaystyle \int x^2 \text {arctanh}(d \cot (a+b x)+i d+1) \, dx\)

\(\Big \downarrow \) 6819

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

\(\Big \downarrow \) 2615

\(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}+(1+i d) \int \frac {e^{2 i a+2 i b x} x^3}{1-(i d+1) e^{2 i a+2 i b x}}dx\right )+\frac {1}{3} x^3 \text {arctanh}(d \cot (a+b x)+i d+1)\)

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[x^2*ArcTanh[1 + I*d + d*Cot[a + b*x]],x]
 

Output:

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

Defintions of rubi rules used

rule 2615

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

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 6819

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/(f*(m + 1)))   Int[(e + f*x)^(m + 1)/(c - I*d - c*E^ 
(2*I*a + 2*I*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && 
 EqQ[(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 = 4.77 (sec) , antiderivative size = 2387, normalized size of antiderivative = 14.21

method	result	size
risch	\(\text {Expression too large to display}\)	\(2387\)

Input:

int(x^2*arctanh(1+I*d+d*cot(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/8/b^3/(I-d)*polylog(4,-I*exp(2*I*(b*x+a))*(I-d))+1/12*I*b*x^4-1/2/b^2*d/ 
(I-d)*ln(1+I*exp(2*I*(b*x+a))*(I-d))*a^2*x+1/2/b^2*d*a^2/(I-d)*ln(1-I*exp( 
I*(b*x+a))*(I*(I-d))^(1/2))*x+1/2/b^2*d*a^2/(I-d)*ln(1+I*exp(I*(b*x+a))*(I 
*(I-d))^(1/2))*x+1/2*I/b^2/(I-d)*ln(1+I*exp(2*I*(b*x+a))*(I-d))*a^2*x+1/6* 
x^3*ln(exp(2*I*(b*x+a))*d-I*exp(2*I*(b*x+a))+I)-1/3*x^3*ln(exp(I*(b*x+a))) 
-1/12*(-I*Pi*csgn(I/(-1+exp(2*I*(b*x+a)))*exp(2*I*(b*x+a)))^3-2*I*Pi+2*ln( 
d)-I*Pi*csgn(I/(-1+exp(2*I*(b*x+a))))*csgn(I*(exp(2*I*(b*x+a))*d-I*exp(2*I 
*(b*x+a))+I)/(-1+exp(2*I*(b*x+a))))^2-I*Pi*csgn(I/(-1+exp(2*I*(b*x+a)))*ex 
p(2*I*(b*x+a)))*csgn(I/(-1+exp(2*I*(b*x+a)))*exp(2*I*(b*x+a))*d)*csgn(I*d) 
-I*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I/(-1+exp(2*I*(b*x+a))))*csgn(I/(-1+ex 
p(2*I*(b*x+a)))*exp(2*I*(b*x+a)))-I*Pi*csgn(I*(exp(2*I*(b*x+a))*d-I*exp(2* 
I*(b*x+a))+I)/(-1+exp(2*I*(b*x+a))))*csgn((exp(2*I*(b*x+a))*d-I*exp(2*I*(b 
*x+a))+I)/(-1+exp(2*I*(b*x+a))))^2-I*Pi*csgn(I*(exp(2*I*(b*x+a))*d-I*exp(2 
*I*(b*x+a))+I))*csgn(I*(exp(2*I*(b*x+a))*d-I*exp(2*I*(b*x+a))+I)/(-1+exp(2 
*I*(b*x+a))))^2+I*Pi*csgn(I/(-1+exp(2*I*(b*x+a)))*exp(2*I*(b*x+a))*d)*csgn 
(1/(-1+exp(2*I*(b*x+a)))*exp(2*I*(b*x+a))*d)^2-I*Pi*csgn(I*exp(I*(b*x+a))) 
^2*csgn(I*exp(2*I*(b*x+a)))+I*Pi*csgn(I/(-1+exp(2*I*(b*x+a))))*csgn(I*(exp 
(2*I*(b*x+a))*d-I*exp(2*I*(b*x+a))+I))*csgn(I*(exp(2*I*(b*x+a))*d-I*exp(2* 
I*(b*x+a))+I)/(-1+exp(2*I*(b*x+a))))+I*Pi*csgn(1/(-1+exp(2*I*(b*x+a)))*exp 
(2*I*(b*x+a))*d)^2+I*Pi*csgn((exp(2*I*(b*x+a))*d-I*exp(2*I*(b*x+a))+I)/...
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.07 \[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\frac {2 i \, b^{4} x^{4} + 4 \, b^{3} x^{3} \log \left (-\frac {{\left ({\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (-{\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 2 i \, a^{4} + 4 \, a^{3} \log \left (\frac {{\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i}{d - i}\right ) - 6 \, b x {\rm polylog}\left (3, {\left (i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 4 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left ({\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) - 3 i \, {\rm polylog}\left (4, {\left (i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{24 \, b^{3}} \] Input:

integrate(x^2*arctanh(1+I*d+d*cot(b*x+a)),x, algorithm="fricas")
 

Output:

1/24*(2*I*b^4*x^4 + 4*b^3*x^3*log(-((d - I)*e^(2*I*b*x + 2*I*a) + I)*e^(-2 
*I*b*x - 2*I*a)/d) + 6*I*b^2*x^2*dilog(-(-I*d - 1)*e^(2*I*b*x + 2*I*a)) - 
2*I*a^4 + 4*a^3*log(((d - I)*e^(2*I*b*x + 2*I*a) + I)/(d - I)) - 6*b*x*pol 
ylog(3, (I*d + 1)*e^(2*I*b*x + 2*I*a)) - 4*(b^3*x^3 + a^3)*log((-I*d - 1)* 
e^(2*I*b*x + 2*I*a) + 1) - 3*I*polylog(4, (I*d + 1)*e^(2*I*b*x + 2*I*a)))/ 
b^3
 

Sympy [F]

\[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\int x^{2} \operatorname {atanh}{\left (d \cot {\left (a + b x \right )} + i d + 1 \right )}\, dx \] Input:

integrate(x**2*atanh(1+I*d+d*cot(b*x+a)),x)
 

Output:

Integral(x**2*atanh(d*cot(a + b*x) + I*d + 1), 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. 344 vs. \(2 (119) = 238\).

Time = 0.06 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.05 \[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\frac {\frac {12 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \operatorname {artanh}\left (d \cot \left (b x + a\right ) + i \, d + 1\right )}{b^{2}} - \frac {-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} - 2 \, {\left (4 i \, {\left (b x + a\right )}^{3} - 9 i \, {\left (b x + a\right )}^{2} a + 9 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), d \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \, {\left (4 i \, {\left (b x + a\right )}^{2} - 6 i \, {\left (b x + a\right )} a + 3 i \, a^{2}\right )} {\rm Li}_2\left ({\left (i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, d \sin \left (2 \, b x + 2 \, a\right ) - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}({\left (i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \, {\rm Li}_{4}({\left (i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b^{2}}}{36 \, b} \] Input:

integrate(x^2*arctanh(1+I*d+d*cot(b*x+a)),x, algorithm="maxima")
 

Output:

1/36*(12*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2)*arctanh(d*cot(b 
*x + a) + I*d + 1)/b^2 - (-3*I*(b*x + a)^4 + 12*I*(b*x + a)^3*a - 18*I*(b* 
x + a)^2*a^2 - 2*(4*I*(b*x + a)^3 - 9*I*(b*x + a)^2*a + 9*I*(b*x + a)*a^2) 
*arctan2(d*cos(2*b*x + 2*a) + sin(2*b*x + 2*a), d*sin(2*b*x + 2*a) - cos(2 
*b*x + 2*a) + 1) - 3*(4*I*(b*x + a)^2 - 6*I*(b*x + a)*a + 3*I*a^2)*dilog(( 
I*d + 1)*e^(2*I*b*x + 2*I*a)) + (4*(b*x + a)^3 - 9*(b*x + a)^2*a + 9*(b*x 
+ a)*a^2)*log((d^2 + 1)*cos(2*b*x + 2*a)^2 + (d^2 + 1)*sin(2*b*x + 2*a)^2 
+ 2*d*sin(2*b*x + 2*a) - 2*cos(2*b*x + 2*a) + 1) + 3*(4*b*x + a)*polylog(3 
, (I*d + 1)*e^(2*I*b*x + 2*I*a)) + 6*I*polylog(4, (I*d + 1)*e^(2*I*b*x + 2 
*I*a)))/b^2)/b
 

Giac [F]

\[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\int { x^{2} \operatorname {artanh}\left (d \cot \left (b x + a\right ) + i \, d + 1\right ) \,d x } \] Input:

integrate(x^2*arctanh(1+I*d+d*cot(b*x+a)),x, algorithm="giac")
 

Output:

integrate(x^2*arctanh(d*cot(b*x + a) + I*d + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\int x^2\,\mathrm {atanh}\left (d\,\mathrm {cot}\left (a+b\,x\right )+1+d\,1{}\mathrm {i}\right ) \,d x \] Input:

int(x^2*atanh(d*1i + d*cot(a + b*x) + 1),x)
 

Output:

int(x^2*atanh(d*1i + d*cot(a + b*x) + 1), x)
 

Reduce [F]

\[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\int \mathit {atanh} \left (\cot \left (b x +a \right ) d +d i +1\right ) x^{2}d x \] Input:

int(x^2*atanh(1+I*d+d*cot(b*x+a)),x)
 

Output:

int(atanh(cot(a + b*x)*d + d*i + 1)*x**2,x)