Integrand size = 20, antiderivative size = 170 \[ \int x^2 \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac {1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b^2}-\frac {i \operatorname {PolyLog}\left (4,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{8 b^3} \]
1/12*I*b*x^4+1/3*x^3*arccoth(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*p olylog(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
Time = 0.16 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.91 \[ \int x^2 \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx=\frac {1}{3} x^3 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac {4 b^3 x^3 \log \left (1+\frac {i e^{-2 i (a+b x)}}{i+d}\right )+6 i b^2 x^2 \operatorname {PolyLog}\left (2,-\frac {i e^{-2 i (a+b x)}}{i+d}\right )+6 b x \operatorname {PolyLog}\left (3,-\frac {i e^{-2 i (a+b x)}}{i+d}\right )-3 i \operatorname {PolyLog}\left (4,-\frac {i e^{-2 i (a+b x)}}{i+d}\right )}{24 b^3} \]
(x^3*ArcCoth[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)* PolyLog[4, (-I)/((I + d)*E^((2*I)*(a + b*x)))])/(24*b^3)
Time = 0.90 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.28, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6818, 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.
| \(\displaystyle \int x^2 \coth ^{-1}(d \tan (a+b x)-i d+1) \, dx\) |
| \(\Big \downarrow \) 6818 |
| \(\displaystyle \frac {1}{3} i b \int \frac {x^3}{e^{2 i a+2 i b x} (1-i d)+1}dx+\frac {1}{3} x^3 \coth ^{-1}(d \tan (a+b x)-i d+1)\) |
| \(\Big \downarrow \) 2615 |
| \(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}-(1-i d) \int \frac {e^{2 i a+2 i b x} x^3}{e^{2 i a+2 i b x} (1-i d)+1}dx\right )+\frac {1}{3} x^3 \coth ^{-1}(d \tan (a+b x)-i d+1)\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}-(1-i d) \left (\frac {x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )}{2 b (d+i)}-\frac {3 \int x^2 \log \left (e^{2 i a+2 i b x} (1-i d)+1\right )dx}{2 b (d+i)}\right )\right )+\frac {1}{3} x^3 \coth ^{-1}(d \tan (a+b x)-i d+1)\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}-(1-i d) \left (\frac {x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )}{2 b (d+i)}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )dx}{b}\right )}{2 b (d+i)}\right )\right )+\frac {1}{3} x^3 \coth ^{-1}(d \tan (a+b x)-i d+1)\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}-(1-i d) \left (\frac {x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )}{2 b (d+i)}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{2 b}-\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (3,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )dx}{2 b}-\frac {i x \operatorname {PolyLog}\left (3,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{2 b}\right )}{b}\right )}{2 b (d+i)}\right )\right )+\frac {1}{3} x^3 \coth ^{-1}(d \tan (a+b x)-i d+1)\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}-(1-i d) \left (\frac {x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )}{2 b (d+i)}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{2 b}-\frac {i \left (\frac {\int e^{-2 i a-2 i b x} \operatorname {PolyLog}\left (3,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )de^{2 i a+2 i b x}}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{2 b}\right )}{b}\right )}{2 b (d+i)}\right )\right )+\frac {1}{3} x^3 \coth ^{-1}(d \tan (a+b x)-i d+1)\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}-(1-i d) \left (\frac {x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )}{2 b (d+i)}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{2 b}-\frac {i \left (\frac {\operatorname {PolyLog}\left (4,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{2 b}\right )}{b}\right )}{2 b (d+i)}\right )\right )+\frac {1}{3} x^3 \coth ^{-1}(d \tan (a+b x)-i d+1)\) |
(x^3*ArcCoth[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))))
3.3.40.3.1 Defintions of rubi rules used
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]
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]
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]]]
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]
Int[ArcCoth[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcCoth[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]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.31 (sec) , antiderivative size = 2273, normalized size of antiderivative = 13.37
1/2/b^2*d/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*a^2*x-1/2/b^2*d*a^2/(I+d)*l n(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+1/4*I/b*d/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b* x+a)))*x^2+1/2*I/b^3*d*a^2/(I+d)*dilog(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2) )+1/2*I/b^3*d*a^2/(I+d)*dilog(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/2*I/b ^2*a^2/(I+d)*ln(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))*x-1/2*I/b^2*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*(I+d)*ex p(2*I*(b*x+a)))*a^2*x-1/4*I/b^3*a^2*d/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b*x +a)))+1/12*I*b*x^4-1/6*d/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*x^3-1/2/b^3* a^2/(I+d)*dilog(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/4/b/(I+d)*polylog(2 ,I*(I+d)*exp(2*I*(b*x+a)))*x^2+1/4/b^3/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b* x+a)))*a^2-1/2/b^3*a^2/(I+d)*dilog(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/ 6*I/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*x^3-1/3*x^3*ln(exp(I*(b*x+a)))+1/ 3/b^3*d/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*a^3-1/4/b^2*d/(I+d)*polylog(3 ,I*(I+d)*exp(2*I*(b*x+a)))*x-1/2/b^3*d*a^3/(I+d)*ln(1-I*exp(I*(b*x+a))*(-I *(I+d))^(1/2))-1/2/b^3*d*a^3/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2)) +1/6/b^3*a^3*d/(I+d)*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)+1/3*I/b^3 /(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*a^3-1/4*I/b^2/(I+d)*polylog(3,I*(I+d )*exp(2*I*(b*x+a)))*x-1/8*I/b^3*d/(I+d)*polylog(4,I*(I+d)*exp(2*I*(b*x+a)) )+1/6*I/b^3*a^3/(I+d)*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)-1/2*I...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (118) = 236\).
Time = 0.28 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.02 \[ \int x^2 \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx=\frac {i \, b^{4} x^{4} + 2 \, b^{3} x^{3} \log \left (\frac {{\left ({\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - i \, a^{4} + 2 \, a^{3} \log \left (\frac {2 \, {\left (d + i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, d - 4}}{2 \, {\left (d + i\right )}}\right ) + 2 \, a^{3} \log \left (\frac {2 \, {\left (d + i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, d - 4}}{2 \, {\left (d + i\right )}}\right ) - 12 \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 12 \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 12 i \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 12 i \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{3}} \]
1/12*(I*b^4*x^4 + 2*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(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*l og(1/2*(2*(d + I)*e^(I*b*x + I*a) + I*sqrt(4*I*d - 4))/(d + I)) + 2*a^3*lo g(1/2*(2*(d + I)*e^(I*b*x + I*a) - I*sqrt(4*I*d - 4))/(d + I)) - 12*b*x*po lylog(3, 1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) - 12*b*x*polylog(3, -1/2*sqr t(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
\[ \int x^2 \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx=\int x^{2} \operatorname {acoth}{\left (d \tan {\left (a + b x \right )} - i d + 1 \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (118) = 236\).
Time = 0.24 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.02 \[ \int x^2 \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx=\frac {\frac {12 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \operatorname {arcoth}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )}{b^{2}} - \frac {-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} - 2 \, {\left (-4 i \, {\left (b x + a\right )}^{3} + 9 i \, {\left (b x + a\right )}^{2} a - 9 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (-d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), d \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \, {\left (4 i \, {\left (b x + a\right )}^{2} - 6 i \, {\left (b x + a\right )} a + 3 i \, a^{2}\right )} {\rm Li}_2\left ({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, d \sin \left (2 \, b x + 2 \, a\right ) + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \, {\rm Li}_{4}({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b^{2}}}{36 \, b} \]
1/36*(12*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2)*arccoth(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) + 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
\[ \int x^2 \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx=\int { x^{2} \operatorname {arcoth}\left (d \tan \left (b x + a\right ) - i \, d + 1\right ) \,d x } \]
Timed out. \[ \int x^2 \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx=\int x^2\,\mathrm {acoth}\left (d\,\mathrm {tan}\left (a+b\,x\right )+1-d\,1{}\mathrm {i}\right ) \,d x \]