Integrand size = 15, antiderivative size = 234 \[ \int (e+f x)^2 \text {arctanh}(\cot (a+b x)) \, dx=\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{3 f}+\frac {(e+f x)^3 \text {arctanh}(\cot (a+b x))}{3 f}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {f (e+f x) \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {f (e+f x) \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{4 b^2}+\frac {i f^2 \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {i f^2 \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3} \]
1/3*I*(f*x+e)^3*arctan(exp(2*I*(b*x+a)))/f+1/3*(f*x+e)^3*arctanh(cot(b*x+a ))/f-1/4*I*(f*x+e)^2*polylog(2,-I*exp(2*I*(b*x+a)))/b+1/4*I*(f*x+e)^2*poly log(2,I*exp(2*I*(b*x+a)))/b+1/4*f*(f*x+e)*polylog(3,-I*exp(2*I*(b*x+a)))/b ^2-1/4*f*(f*x+e)*polylog(3,I*exp(2*I*(b*x+a)))/b^2+1/8*I*f^2*polylog(4,-I* exp(2*I*(b*x+a)))/b^3-1/8*I*f^2*polylog(4,I*exp(2*I*(b*x+a)))/b^3
Time = 0.11 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.75 \[ \int (e+f x)^2 \text {arctanh}(\cot (a+b x)) \, dx=\frac {1}{3} x \left (3 e^2+3 e f x+f^2 x^2\right ) \text {arctanh}(\cot (a+b x))+\frac {-12 b^3 e^2 x \log \left (1-i e^{2 i (a+b x)}\right )-12 b^3 e f x^2 \log \left (1-i e^{2 i (a+b x)}\right )-4 b^3 f^2 x^3 \log \left (1-i e^{2 i (a+b x)}\right )+12 b^3 e^2 x \log \left (1+i e^{2 i (a+b x)}\right )+12 b^3 e f x^2 \log \left (1+i e^{2 i (a+b x)}\right )+4 b^3 f^2 x^3 \log \left (1+i e^{2 i (a+b x)}\right )-6 i b^2 (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )+6 i b^2 (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )+6 b e f \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )+6 b f^2 x \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b e f \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )-6 b f^2 x \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )+3 i f^2 \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )-3 i f^2 \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{24 b^3} \]
(x*(3*e^2 + 3*e*f*x + f^2*x^2)*ArcTanh[Cot[a + b*x]])/3 + (-12*b^3*e^2*x*L og[1 - I*E^((2*I)*(a + b*x))] - 12*b^3*e*f*x^2*Log[1 - I*E^((2*I)*(a + b*x ))] - 4*b^3*f^2*x^3*Log[1 - I*E^((2*I)*(a + b*x))] + 12*b^3*e^2*x*Log[1 + I*E^((2*I)*(a + b*x))] + 12*b^3*e*f*x^2*Log[1 + I*E^((2*I)*(a + b*x))] + 4 *b^3*f^2*x^3*Log[1 + I*E^((2*I)*(a + b*x))] - (6*I)*b^2*(e + f*x)^2*PolyLo g[2, (-I)*E^((2*I)*(a + b*x))] + (6*I)*b^2*(e + f*x)^2*PolyLog[2, I*E^((2* I)*(a + b*x))] + 6*b*e*f*PolyLog[3, (-I)*E^((2*I)*(a + b*x))] + 6*b*f^2*x* PolyLog[3, (-I)*E^((2*I)*(a + b*x))] - 6*b*e*f*PolyLog[3, I*E^((2*I)*(a + b*x))] - 6*b*f^2*x*PolyLog[3, I*E^((2*I)*(a + b*x))] + (3*I)*f^2*PolyLog[4 , (-I)*E^((2*I)*(a + b*x))] - (3*I)*f^2*PolyLog[4, I*E^((2*I)*(a + b*x))]) /(24*b^3)
Time = 0.85 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {6807, 3042, 4669, 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.
| \(\displaystyle \int (e+f x)^2 \text {arctanh}(\cot (a+b x)) \, dx\) |
| \(\Big \downarrow \) 6807 |
| \(\displaystyle \frac {(e+f x)^3 \text {arctanh}(\cot (a+b x))}{3 f}-\frac {b \int (e+f x)^3 \sec (2 a+2 b x)dx}{3 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(e+f x)^3 \text {arctanh}(\cot (a+b x))}{3 f}-\frac {b \int (e+f x)^3 \csc \left (2 a+2 b x+\frac {\pi }{2}\right )dx}{3 f}\) |
| \(\Big \downarrow \) 4669 |
| \(\displaystyle \frac {(e+f x)^3 \text {arctanh}(\cot (a+b x))}{3 f}-\frac {b \left (-\frac {3 f \int (e+f x)^2 \log \left (1-i e^{2 i (a+b x)}\right )dx}{2 b}+\frac {3 f \int (e+f x)^2 \log \left (1+i e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{3 f}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {(e+f x)^3 \text {arctanh}(\cot (a+b x))}{3 f}-\frac {b \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{3 f}\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {(e+f x)^3 \text {arctanh}(\cot (a+b x))}{3 f}-\frac {b \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{3 f}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {(e+f x)^3 \text {arctanh}(\cot (a+b x))}{3 f}-\frac {b \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {f \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {f \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{3 f}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {(e+f x)^3 \text {arctanh}(\cot (a+b x))}{3 f}-\frac {b \left (-\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{b}+\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {f \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {f \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}\right )}{3 f}\) |
((e + f*x)^3*ArcTanh[Cot[a + b*x]])/(3*f) - (b*(((-I)*(e + f*x)^3*ArcTan[E ^((2*I)*(a + b*x))])/b + (3*f*(((I/2)*(e + f*x)^2*PolyLog[2, (-I)*E^((2*I) *(a + b*x))])/b - (I*f*(((-1/2*I)*(e + f*x)*PolyLog[3, (-I)*E^((2*I)*(a + b*x))])/b + (f*PolyLog[4, (-I)*E^((2*I)*(a + b*x))])/(4*b^2)))/b))/(2*b) - (3*f*(((I/2)*(e + f*x)^2*PolyLog[2, I*E^((2*I)*(a + b*x))])/b - (I*f*(((- 1/2*I)*(e + f*x)*PolyLog[3, I*E^((2*I)*(a + b*x))])/b + (f*PolyLog[4, I*E^ ((2*I)*(a + b*x))])/(4*b^2)))/b))/(2*b)))/(3*f)
3.4.30.3.1 Defintions of rubi rules used
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[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol ] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si mp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x ))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Int[ArcTanh[Cot[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[Cot[a + b*x]]/(f*(m + 1))), x] - Simp[b/ (f*(m + 1)) Int[(e + f*x)^(m + 1)*Sec[2*a + 2*b*x], x], x] /; FreeQ[{a, b , e, f}, x] && IGtQ[m, 0]
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 = 10.21 (sec) , antiderivative size = 2719, normalized size of antiderivative = 11.62
-1/6*f^2*ln(-I*exp(2*I*(b*x+a))+1)*x^3+1/6*(f*x+e)^3/f*ln(exp(2*I*(b*x+a)) +I)-1/8*I*f^2*polylog(4,I*exp(2*I*(b*x+a)))/b^3+1/8*I*f^2*polylog(4,-I*exp (2*I*(b*x+a)))/b^3-f*e/b*ln(-I*exp(2*I*(b*x+a))+1)*a*x+1/2*I*f*e/b*polylog (2,I*exp(2*I*(b*x+a)))*x+1/2*I*f*e/b^2*polylog(2,I*exp(2*I*(b*x+a)))*a+f*e /b*ln(1+I*exp(2*I*(b*x+a)))*a*x-1/2*I*f*e/b*polylog(2,-I*exp(2*I*(b*x+a))) *x-1/2*I*f*e/b^2*polylog(2,-I*exp(2*I*(b*x+a)))*a-I*f/b^2*a*e*dilog(((-I)^ (1/2)-exp(I*(b*x+a)))/(-I)^(1/2))-I*f/b^2*a*e*dilog(((-I)^(1/2)+exp(I*(b*x +a)))/(-I)^(1/2))+f/b*a*e*ln(((-I)^(1/2)-exp(I*(b*x+a)))/(-I)^(1/2))*x+f/b *a*e*ln(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))*x-f/b*a*e*ln(1+exp(I*(b*x+ a))*(-1)^(3/4))*x-f/b*a*e*ln(1-exp(I*(b*x+a))*(-1)^(3/4))*x+I*f/b^2*a*e*di log(1+exp(I*(b*x+a))*(-1)^(3/4))+I*f/b^2*a*e*dilog(1-exp(I*(b*x+a))*(-1)^( 3/4))+1/6/f*e^3*ln(-exp(2*I*(b*x+a))+I)+1/6*f^2*ln(1+I*exp(2*I*(b*x+a)))*x ^3-1/2*f*ln(exp(2*I*(b*x+a))-I)*x^2*e-1/2*f^2/b^2*ln(1+I*exp(2*I*(b*x+a))) *a^2*x+1/2*f/b^2*a^2*e*ln(-exp(2*I*(b*x+a))+I)+1/4*I*f^2*a^2/b^3*polylog(2 ,-I*exp(2*I*(b*x+a)))-1/4*I*f^2/b*polylog(2,-I*exp(2*I*(b*x+a)))*x^2-1/3*f ^2/b^3*ln(1+I*exp(2*I*(b*x+a)))*a^3+1/2*f*e*ln(1+I*exp(2*I*(b*x+a)))*x^2+1 /4*f*e/b^2*polylog(3,-I*exp(2*I*(b*x+a)))-1/6*f^2/b^3*a^3*ln(-exp(2*I*(b*x +a))+I)+1/4*f^2/b^2*polylog(3,-I*exp(2*I*(b*x+a)))*x-1/2/b*a*e^2*ln(-exp(2 *I*(b*x+a))+I)-1/2*f^2*ln(((-I)^(1/2)-exp(I*(b*x+a)))/(-I)^(1/2))/b^3*a^3- 1/2*f^2*ln(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))/b^3*a^3-1/2*ln(((-I)...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1085 vs. \(2 (180) = 360\).
Time = 0.34 (sec) , antiderivative size = 1085, normalized size of antiderivative = 4.64 \[ \int (e+f x)^2 \text {arctanh}(\cot (a+b x)) \, dx=\text {Too large to display} \]
1/48*(-3*I*f^2*polylog(4, I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - 3*I*f^2 *polylog(4, I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)) + 3*I*f^2*polylog(4, -I *cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) + 3*I*f^2*polylog(4, -I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)) - 6*(-I*b^2*f^2*x^2 - 2*I*b^2*e*f*x - I*b^2*e^2)* dilog(I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - 6*(-I*b^2*f^2*x^2 - 2*I*b^2 *e*f*x - I*b^2*e^2)*dilog(I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)) - 6*(I*b^ 2*f^2*x^2 + 2*I*b^2*e*f*x + I*b^2*e^2)*dilog(-I*cos(2*b*x + 2*a) + sin(2*b *x + 2*a)) - 6*(I*b^2*f^2*x^2 + 2*I*b^2*e*f*x + I*b^2*e^2)*dilog(-I*cos(2* b*x + 2*a) - sin(2*b*x + 2*a)) + 8*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3*e^ 2*x)*log(-(cos(2*b*x + 2*a) + sin(2*b*x + 2*a) + 1)/(cos(2*b*x + 2*a) - si n(2*b*x + 2*a) + 1)) + 4*(3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(cos(2*b *x + 2*a) + I*sin(2*b*x + 2*a) + I) - 4*(3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f ^2)*log(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + I) - 4*(b^3*f^2*x^3 + 3*b^ 3*e*f*x^2 + 3*b^3*e^2*x + 3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(I*cos(2 *b*x + 2*a) + sin(2*b*x + 2*a) + 1) + 4*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b ^3*e^2*x + 3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(I*cos(2*b*x + 2*a) - s in(2*b*x + 2*a) + 1) - 4*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3*e^2*x + 3*a* b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(-I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a ) + 1) + 4*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3*e^2*x + 3*a*b^2*e^2 - 3*a^ 2*b*e*f + a^3*f^2)*log(-I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a) + 1) + 4*...
\[ \int (e+f x)^2 \text {arctanh}(\cot (a+b x)) \, dx=\int \left (e + f x\right )^{2} \operatorname {atanh}{\left (\cot {\left (a + b x \right )} \right )}\, dx \]
\[ \int (e+f x)^2 \text {arctanh}(\cot (a+b x)) \, dx=\int { {\left (f x + e\right )}^{2} \operatorname {artanh}\left (\cot \left (b x + a\right )\right ) \,d x } \]
1/12*(f^2*x^3 + 3*e*f*x^2 + 3*e^2*x)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b* x + 2*a)^2 + 4*sin(2*b*x + 2*a) + 2) - 1/12*(f^2*x^3 + 3*e*f*x^2 + 3*e^2*x )*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x + 2*a)^2 - 4*sin(2*b*x + 2*a) + 2 ) - integrate(2/3*((b*f^2*x^3 + 3*b*e*f*x^2 + 3*b*e^2*x)*cos(4*b*x + 4*a)* cos(2*b*x + 2*a) + (b*f^2*x^3 + 3*b*e*f*x^2 + 3*b*e^2*x)*sin(4*b*x + 4*a)* sin(2*b*x + 2*a) + (b*f^2*x^3 + 3*b*e*f*x^2 + 3*b*e^2*x)*cos(2*b*x + 2*a)) /(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1), x)
\[ \int (e+f x)^2 \text {arctanh}(\cot (a+b x)) \, dx=\int { {\left (f x + e\right )}^{2} \operatorname {artanh}\left (\cot \left (b x + a\right )\right ) \,d x } \]
Timed out. \[ \int (e+f x)^2 \text {arctanh}(\cot (a+b x)) \, dx=\int \mathrm {atanh}\left (\mathrm {cot}\left (a+b\,x\right )\right )\,{\left (e+f\,x\right )}^2 \,d x \]