Integrand size = 15, antiderivative size = 234 \[ \int (e+f x)^2 \coth ^{-1}(\tan (a+b x)) \, dx=\frac {(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}+\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{3 f}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {f (e+f x) \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {f (e+f x) \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{4 b^2}+\frac {i f^2 \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {i f^2 \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3} \]
1/3*(f*x+e)^3*arccoth(tan(b*x+a))/f+1/3*I*(f*x+e)^3*arctan(exp(2*I*(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 = 0.14 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.75 \[ \int (e+f x)^2 \coth ^{-1}(\tan (a+b x)) \, dx=\frac {1}{3} x \left (3 e^2+3 e f x+f^2 x^2\right ) \coth ^{-1}(\tan (a+b x))+\frac {-12 b^3 e^2 x \log \left (1-i e^{2 i (a+b x)}\right )-12 b^3 e f x^2 \log \left (1-i e^{2 i (a+b x)}\right )-4 b^3 f^2 x^3 \log \left (1-i e^{2 i (a+b x)}\right )+12 b^3 e^2 x \log \left (1+i e^{2 i (a+b x)}\right )+12 b^3 e f x^2 \log \left (1+i e^{2 i (a+b x)}\right )+4 b^3 f^2 x^3 \log \left (1+i e^{2 i (a+b x)}\right )-6 i b^2 (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )+6 i b^2 (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )+6 b e f \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )+6 b f^2 x \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b e f \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )-6 b f^2 x \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )+3 i f^2 \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )-3 i f^2 \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{24 b^3} \]
(x*(3*e^2 + 3*e*f*x + f^2*x^2)*ArcCoth[Tan[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 = 0.86 (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 = {6806, 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.
\(\displaystyle \int (e+f x)^2 \coth ^{-1}(\tan (a+b x)) \, dx\) |
\(\Big \downarrow \) 6806 |
\(\displaystyle \frac {(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}-\frac {b \int (e+f x)^3 \sec (2 a+2 b x)dx}{3 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}-\frac {b \int (e+f x)^3 \csc \left (2 a+2 b x+\frac {\pi }{2}\right )dx}{3 f}\) |
\(\Big \downarrow \) 4669 |
\(\displaystyle \frac {(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}-\frac {b \left (-\frac {3 f \int (e+f x)^2 \log \left (1-i e^{2 i (a+b x)}\right )dx}{2 b}+\frac {3 f \int (e+f x)^2 \log \left (1+i e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{3 f}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}-\frac {b \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{3 f}\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}-\frac {b \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{3 f}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}-\frac {b \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {f \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {f \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{3 f}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}-\frac {b \left (-\frac {i (e+f x)^3 \arctan \left (e^{2 i (a+b x)}\right )}{b}+\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {f \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \left (\frac {f \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}\right )}{3 f}\) |
((e + f*x)^3*ArcCoth[Tan[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)
3.3.32.3.1 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[ArcCoth[Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcCoth[Tan[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 = 20.18 (sec) , antiderivative size = 2719, normalized size of antiderivative = 11.62
-1/3*f^2/b^3*ln(1+I*exp(2*I*(b*x+a)))*a^3+1/4*f^2/b^2*polylog(3,-I*exp(2*I *(b*x+a)))*x+1/2*f*e*ln(1+I*exp(2*I*(b*x+a)))*x^2+1/2*f^2/b^3*a^3*ln(1+exp (I*(b*x+a))*(-1)^(3/4))+1/2*f^2/b^3*a^3*ln(1-exp(I*(b*x+a))*(-1)^(3/4))-1/ 6*f^2/b^3*a^3*ln(-exp(2*I*(b*x+a))+I)+1/4*f*e/b^2*polylog(3,-I*exp(2*I*(b* x+a)))-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+1/12*I*Pi*(csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(exp( 2*I*(b*x+a))-I))*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))-csgn(I/ (exp(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))+I))*csgn(I*(exp(2*I*(b*x+a) )+I)/(exp(2*I*(b*x+a))+1))-csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b *x+a))-I)/(exp(2*I*(b*x+a))+1))^2+csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(exp (2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2-csgn(I*(exp(2*I*(b*x+a))-I))*csgn (I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^2+csgn(I*(exp(2*I*(b*x+a))+I ))*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2-csgn(I*(exp(2*I*(b* x+a))+I)/(exp(2*I*(b*x+a))+1))*csgn((1+I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b *x+a))+1))+csgn((1+I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2+csgn(I* (exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^3+csgn(I*(exp(2*I*(b*x+a))-I)/( exp(2*I*(b*x+a))+1))*csgn((1-I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1)) -csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))*csgn((1-I)*(exp(2*I*(b* x+a))-I)/(exp(2*I*(b*x+a))+1))^2-csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x +a))+1))^3+csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))*csgn((1+I)...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1282 vs. \(2 (180) = 360\).
Time = 0.29 (sec) , antiderivative size = 1282, normalized size of antiderivative = 5.48 \[ \int (e+f x)^2 \coth ^{-1}(\tan (a+b x)) \, dx=\text {Too large to display} \]
1/48*(3*I*f^2*polylog(4, (I*tan(b*x + a)^2 + 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) + 3*I*f^2*polylog(4, (I*tan(b*x + a)^2 - 2*tan(b*x + a) - I)/ (tan(b*x + a)^2 + 1)) - 3*I*f^2*polylog(4, (-I*tan(b*x + a)^2 + 2*tan(b*x + a) + I)/(tan(b*x + a)^2 + 1)) - 3*I*f^2*polylog(4, (-I*tan(b*x + a)^2 - 2*tan(b*x + a) + I)/(tan(b*x + a)^2 + 1)) - 6*(-I*b^2*f^2*x^2 - 2*I*b^2*e* f*x - I*b^2*e^2)*dilog(-((I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I + 1)/ (tan(b*x + a)^2 + 1) + 1) - 6*(-I*b^2*f^2*x^2 - 2*I*b^2*e*f*x - I*b^2*e^2) *dilog(-((I + 1)*tan(b*x + a)^2 - 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1) + 1) - 6*(I*b^2*f^2*x^2 + 2*I*b^2*e*f*x + I*b^2*e^2)*dilog(-(-(I - 1) *tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) - 6*(I *b^2*f^2*x^2 + 2*I*b^2*e*f*x + I*b^2*e^2)*dilog(-(-(I - 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 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 + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1)) + 4*(3*a *b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(((I + 1)*tan(b*x + a)^2 + 2*I*tan(b* x + a) + I - 1)/(tan(b*x + a)^2 + 1)) - 4*(3*a*b^2*e^2 - 3*a^2*b*e*f + a^3 *f^2)*log(((I + 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a )^2 + 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 + 1)*tan(b*x + a)^2 - 2*tan(b*x + a) - I + 1 )/(tan(b*x + a)^2 + 1)) - 4*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3*e^2*x ...
\[ \int (e+f x)^2 \coth ^{-1}(\tan (a+b x)) \, dx=\int \left (e + f x\right )^{2} \operatorname {acoth}{\left (\tan {\left (a + b x \right )} \right )}\, dx \]
\[ \int (e+f x)^2 \coth ^{-1}(\tan (a+b x)) \, dx=\int { {\left (f x + e\right )}^{2} \operatorname {arcoth}\left (\tan \left (b x + a\right )\right ) \,d x } \]
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)
\[ \int (e+f x)^2 \coth ^{-1}(\tan (a+b x)) \, dx=\int { {\left (f x + e\right )}^{2} \operatorname {arcoth}\left (\tan \left (b x + a\right )\right ) \,d x } \]
Timed out. \[ \int (e+f x)^2 \coth ^{-1}(\tan (a+b x)) \, dx=\int \mathrm {acoth}\left (\mathrm {tan}\left (a+b\,x\right )\right )\,{\left (e+f\,x\right )}^2 \,d x \]