\(\int x^2 \text {arctanh}(1+i d-d \tan (a+b x)) \, dx\) [325]

Optimal result

Integrand size = 21, antiderivative size = 171 \[ \int x^2 \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \text {arctanh}(1+i d-d \tan (a+b x))-\frac {1}{6} x^3 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,-\left ((1+i d) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,-\left ((1+i d) e^{2 i a+2 i b x}\right )\right )}{4 b^2}-\frac {i \operatorname {PolyLog}\left (4,-\left ((1+i d) e^{2 i a+2 i b x}\right )\right )}{8 b^3} \] Output:

1/12*I*b*x^4-1/3*x^3*arctanh(-1-I*d+d*tan(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* 
polylog(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 = 0.23 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.91 \[ \int x^2 \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=\frac {1}{3} x^3 \text {arctanh}(1+i d-d \tan (a+b x))-\frac {4 b^3 x^3 \log \left (1-\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+6 i b^2 x^2 \operatorname {PolyLog}\left (2,\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+6 b x \operatorname {PolyLog}\left (3,\frac {i e^{-2 i (a+b x)}}{-i+d}\right )-3 i \operatorname {PolyLog}\left (4,\frac {i e^{-2 i (a+b x)}}{-i+d}\right )}{24 b^3} \] Input:

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

Output:

(x^3*ArcTanh[1 + I*d - d*Tan[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)*Pol 
yLog[4, I/((-I + d)*E^((2*I)*(a + b*x)))])/(24*b^3)
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.30, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6817, 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.

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

\(\Big \downarrow \) 6817

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

(x^3*ArcTanh[1 + I*d - d*Tan[a + b*x]])/3 + (I/3)*b*(x^4/4 - (1 + I*d)*((x 
^3*Log[1 + (1 + I*d)*E^((2*I)*a + (2*I)*b*x)])/(2*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 6817

Int[ArcTanh[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Tan[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 = 4.48 (sec) , antiderivative size = 2387, normalized size of antiderivative = 13.96

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

Input:

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

Output:

1/12*I*b*x^4+1/6*x^3*ln(exp(2*I*(b*x+a))*d-I*exp(2*I*(b*x+a))-I)-1/3*x^3*l 
n(exp(I*(b*x+a)))-1/12*(-2*I*Pi+2*ln(d)+I*Pi*csgn((exp(2*I*(b*x+a))*d-I*ex 
p(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^2-I*Pi*csgn((exp(2*I*(b*x+a))*d-I* 
exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^3+I*Pi*csgn(1/(exp(2*I*(b*x+a))+ 
1)*exp(2*I*(b*x+a))*d)^2+I*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I/(exp(2*I 
*(b*x+a))+1)*exp(2*I*(b*x+a)))^2-I*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I* 
(exp(2*I*(b*x+a))*d-I*exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^2-I*Pi*csg 
n(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)/(exp(2*I*(b*x+a))+1))^2-I*Pi*csgn(1/(exp(2*I*(b*x+a) 
)+1)*exp(2*I*(b*x+a))*d)^3+I*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*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)/(exp(2*I*(b*x+a))+1))-I*Pi*csgn(I*exp(I*(b*x+a)))^2*csgn(I*exp 
(2*I*(b*x+a)))-I*Pi*csgn(I*exp(2*I*(b*x+a)))^3-I*Pi*csgn(I/(exp(2*I*(b*x+a 
))+1)*exp(2*I*(b*x+a)))^3-I*Pi*csgn(I/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a) 
)*d)^3-I*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*exp(2*I*(b*x+a)))*csgn(I/( 
exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))+2*I*Pi*csgn(I*exp(I*(b*x+a)))*csgn(I 
*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) 
/(exp(2*I*(b*x+a))+1))^3-I*Pi*csgn(I/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)) 
)*csgn(I/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a))*d)*csgn(I*d)+I*Pi*csgn(I*(e 
xp(2*I*(b*x+a))*d-I*exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))*csgn((exp...
 

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. 345 vs. \(2 (118) = 236\).

Time = 0.10 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.02 \[ \int x^2 \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=\frac {i \, b^{4} x^{4} - 2 \, b^{3} x^{3} \log \left (-\frac {d e^{\left (2 i \, b x + 2 i \, a\right )}}{{\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - i \, a^{4} + 2 \, a^{3} \log \left (\frac {2 \, {\left (d - i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {-4 i \, d - 4}}{2 \, {\left (d - i\right )}}\right ) + 2 \, a^{3} \log \left (\frac {2 \, {\left (d - i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {-4 i \, d - 4}}{2 \, {\left (d - i\right )}}\right ) - 12 \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 12 \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 12 i \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 12 i \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{3}} \] Input:

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

Output:

1/12*(I*b^4*x^4 - 2*b^3*x^3*log(-d*e^(2*I*b*x + 2*I*a)/((d - I)*e^(2*I*b*x 
 + 2*I*a) - I)) + 6*I*b^2*x^2*dilog(1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a)) 
+ 6*I*b^2*x^2*dilog(-1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a)) - I*a^4 + 2*a^3 
*log(1/2*(2*(d - I)*e^(I*b*x + I*a) + I*sqrt(-4*I*d - 4))/(d - I)) + 2*a^3 
*log(1/2*(2*(d - I)*e^(I*b*x + I*a) - I*sqrt(-4*I*d - 4))/(d - I)) - 12*b* 
x*polylog(3, 1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a)) - 12*b*x*polylog(3, -1/ 
2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a)) - 2*(b^3*x^3 + a^3)*log(1/2*sqrt(-4*I* 
d - 4)*e^(I*b*x + I*a) + 1) - 2*(b^3*x^3 + a^3)*log(-1/2*sqrt(-4*I*d - 4)* 
e^(I*b*x + I*a) + 1) - 12*I*polylog(4, 1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a 
)) - 12*I*polylog(4, -1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a)))/b^3
 

Sympy [F]

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

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

Output:

Integral(x**2*atanh(-d*tan(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. 342 vs. \(2 (118) = 236\).

Time = 0.06 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.00 \[ \int x^2 \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=-\frac {\frac {12 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \operatorname {artanh}\left (d \tan \left (b x + a\right ) - i \, d - 1\right )}{b^{2}} + \frac {-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} - 2 \, {\left (-4 i \, {\left (b x + a\right )}^{3} + 9 i \, {\left (b x + a\right )}^{2} a - 9 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), -d \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \, {\left (4 i \, {\left (b x + a\right )}^{2} - 6 i \, {\left (b x + a\right )} a + 3 i \, a^{2}\right )} {\rm Li}_2\left ({\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, d \sin \left (2 \, b x + 2 \, a\right ) + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}({\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \, {\rm Li}_{4}({\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b^{2}}}{36 \, b} \] Input:

integrate(-x^2*arctanh(-1-I*d+d*tan(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*tan( 
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) + co 
s(2*b*x + 2*a) + 1) - 3*(4*I*(b*x + a)^2 - 6*I*(b*x + a)*a + 3*I*a^2)*dilo 
g((-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)*polyl 
og(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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