Integrand size = 13, antiderivative size = 162 \[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=\frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}+\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{2 f}-\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {f \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {f \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2} \]
1/2*(f*x+e)^2*arccoth(tan(b*x+a))/f+1/2*I*(f*x+e)^2*arctan(exp(2*I*(b*x+a) ))/f-1/4*I*(f*x+e)*polylog(2,-I*exp(2*I*(b*x+a)))/b+1/4*I*(f*x+e)*polylog( 2,I*exp(2*I*(b*x+a)))/b+1/8*f*polylog(3,-I*exp(2*I*(b*x+a)))/b^2-1/8*f*pol ylog(3,I*exp(2*I*(b*x+a)))/b^2
Time = 0.21 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.82 \[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=e x \coth ^{-1}(\tan (a+b x))+\frac {1}{2} f x^2 \coth ^{-1}(\tan (a+b x))-\frac {e \left ((-4 a+\pi -4 b x) \left (\log \left (1-i e^{-2 i (a+b x)}\right )-\log \left (1+i e^{-2 i (a+b x)}\right )\right )-(-4 a+\pi ) \log \left (\cot \left (a+\frac {\pi }{4}+b x\right )\right )+2 i \left (\operatorname {PolyLog}\left (2,-i e^{-2 i (a+b x)}\right )-\operatorname {PolyLog}\left (2,i e^{-2 i (a+b x)}\right )\right )\right )}{8 b}+\frac {f \left (4 i b^2 x^2 \arctan (\cos (2 (a+b x))+i \sin (2 (a+b x)))+2 i b x \operatorname {PolyLog}(2,i \cos (2 (a+b x))-\sin (2 (a+b x)))-2 i b x \operatorname {PolyLog}(2,-i \cos (2 (a+b x))+\sin (2 (a+b x)))-\operatorname {PolyLog}(3,i \cos (2 (a+b x))-\sin (2 (a+b x)))+\operatorname {PolyLog}(3,-i \cos (2 (a+b x))+\sin (2 (a+b x)))\right )}{8 b^2} \]
e*x*ArcCoth[Tan[a + b*x]] + (f*x^2*ArcCoth[Tan[a + b*x]])/2 - (e*((-4*a + Pi - 4*b*x)*(Log[1 - I/E^((2*I)*(a + b*x))] - Log[1 + I/E^((2*I)*(a + b*x) )]) - (-4*a + Pi)*Log[Cot[a + Pi/4 + b*x]] + (2*I)*(PolyLog[2, (-I)/E^((2* I)*(a + b*x))] - PolyLog[2, I/E^((2*I)*(a + b*x))])))/(8*b) + (f*((4*I)*b^ 2*x^2*ArcTan[Cos[2*(a + b*x)] + I*Sin[2*(a + b*x)]] + (2*I)*b*x*PolyLog[2, I*Cos[2*(a + b*x)] - Sin[2*(a + b*x)]] - (2*I)*b*x*PolyLog[2, (-I)*Cos[2* (a + b*x)] + Sin[2*(a + b*x)]] - PolyLog[3, I*Cos[2*(a + b*x)] - Sin[2*(a + b*x)]] + PolyLog[3, (-I)*Cos[2*(a + b*x)] + Sin[2*(a + b*x)]]))/(8*b^2)
Time = 0.60 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {6806, 3042, 4669, 3011, 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.
\(\displaystyle \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx\) |
\(\Big \downarrow \) 6806 |
\(\displaystyle \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}-\frac {b \int (e+f x)^2 \sec (2 a+2 b x)dx}{2 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}-\frac {b \int (e+f x)^2 \csc \left (2 a+2 b x+\frac {\pi }{2}\right )dx}{2 f}\) |
\(\Big \downarrow \) 4669 |
\(\displaystyle \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}-\frac {b \left (-\frac {f \int (e+f x) \log \left (1-i e^{2 i (a+b x)}\right )dx}{b}+\frac {f \int (e+f x) \log \left (1+i e^{2 i (a+b x)}\right )dx}{b}-\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{2 f}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}-\frac {b \left (\frac {f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \int \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )dx}{2 b}\right )}{b}-\frac {f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {i f \int \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )dx}{2 b}\right )}{b}-\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{2 f}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}-\frac {b \left (\frac {f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {f \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {f \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{2 f}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}-\frac {b \left (-\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{b}+\frac {f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {f \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{4 b^2}\right )}{b}-\frac {f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {f \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{4 b^2}\right )}{b}\right )}{2 f}\) |
((e + f*x)^2*ArcCoth[Tan[a + b*x]])/(2*f) - (b*(((-I)*(e + f*x)^2*ArcTan[E ^((2*I)*(a + b*x))])/b + (f*(((I/2)*(e + f*x)*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b - (f*PolyLog[3, (-I)*E^((2*I)*(a + b*x))])/(4*b^2)))/b - (f*(( (I/2)*(e + f*x)*PolyLog[2, I*E^((2*I)*(a + b*x))])/b - (f*PolyLog[3, I*E^( (2*I)*(a + b*x))])/(4*b^2)))/b))/(2*f)
3.3.33.3.1 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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.83 (sec) , antiderivative size = 1818, normalized size of antiderivative = 11.22
-1/2*e/b*a*ln(-exp(2*I*(b*x+a))+I)+1/2/b*f*a*ln(((-I)^(1/2)-exp(I*(b*x+a)) )/(-I)^(1/2))*x+1/2/b*f*a*ln(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))*x-1/2 *I/b^2*f*a*dilog(((-I)^(1/2)-exp(I*(b*x+a)))/(-I)^(1/2))-1/2*I/b^2*f*a*dil og(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))+1/2*f/b*ln(1+I*exp(2*I*(b*x+a)) )*a*x-1/4*I*f/b*polylog(2,-I*exp(2*I*(b*x+a)))*x-1/4*I*f/b^2*polylog(2,-I* exp(2*I*(b*x+a)))*a-1/4*f/b^2*ln(-I*exp(2*I*(b*x+a))+1)*a^2+1/4/b^2*f*a^2* ln(-exp(2*I*(b*x+a))+I)+1/4*f/b^2*ln(1+I*exp(2*I*(b*x+a)))*a^2+1/8*f*polyl og(3,-I*exp(2*I*(b*x+a)))/b^2-1/8*f*polylog(3,I*exp(2*I*(b*x+a)))/b^2+1/2* e/b*ln(1+exp(I*(b*x+a))*(-1)^(3/4))*a+1/2*e/b*ln(1-exp(I*(b*x+a))*(-1)^(3/ 4))*a-1/2*I*e/b*dilog(1+exp(I*(b*x+a))*(-1)^(3/4))-1/2*I*e/b*dilog(1-exp(I *(b*x+a))*(-1)^(3/4))-1/2/b^2*f*a^2*ln(1+exp(I*(b*x+a))*(-1)^(3/4))-1/2/b^ 2*f*a^2*ln(1-exp(I*(b*x+a))*(-1)^(3/4))+1/2/b^2*f*a^2*ln(((-I)^(1/2)-exp(I *(b*x+a)))/(-I)^(1/2))+1/2/b^2*f*a^2*ln(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^( 1/2))-1/2*e/b*ln(((-I)^(1/2)-exp(I*(b*x+a)))/(-I)^(1/2))*a-1/2*e/b*ln(((-I )^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))*a+1/2*I*e/b*dilog(((-I)^(1/2)-exp(I*(b *x+a)))/(-I)^(1/2))+1/2*I*e/b*dilog(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2) )-1/4/b^2*f*a^2*ln(exp(2*I*(b*x+a))+I)-1/4*ln(exp(2*I*(b*x+a))-I)*x^2*f-1/ 2*ln(exp(2*I*(b*x+a))-I)*e*x-1/2/b*f*a*ln(1+exp(I*(b*x+a))*(-1)^(3/4))*x-1 /2/b*f*a*ln(1-exp(I*(b*x+a))*(-1)^(3/4))*x+1/2*I/b^2*f*a*dilog(1+exp(I*(b* x+a))*(-1)^(3/4))+1/2*I/b^2*f*a*dilog(1-exp(I*(b*x+a))*(-1)^(3/4))-1/2*...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 834 vs. \(2 (130) = 260\).
Time = 0.29 (sec) , antiderivative size = 834, normalized size of antiderivative = 5.15 \[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=\text {Too large to display} \]
-1/16*(2*(-I*b*f*x - I*b*e)*dilog(-((I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a ) - I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2*(-I*b*f*x - I*b*e)*dilog(-((I + 1 )*tan(b*x + a)^2 - 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2*( I*b*f*x + I*b*e)*dilog(-(-(I - 1)*tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1) /(tan(b*x + a)^2 + 1) + 1) + 2*(I*b*f*x + I*b*e)*dilog(-(-(I - 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2*(b^2*f*x^2 + 2*b^2*e*x + 2*a*b*e - a^2*f)*log(((I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a ) - I + 1)/(tan(b*x + a)^2 + 1)) - 2*(2*a*b*e - a^2*f)*log(((I + 1)*tan(b* x + a)^2 + 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a)^2 + 1)) + 2*(2*a*b*e - a^2*f)*log(((I + 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a)^2 + 1)) - 2*(b^2*f*x^2 + 2*b^2*e*x + 2*a*b*e - a^2*f)*log(((I + 1)*tan( b*x + a)^2 - 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1)) + 2*(b^2*f*x^2 + 2*b^2*e*x + 2*a*b*e - a^2*f)*log((-(I - 1)*tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) - 2*(b^2*f*x^2 + 2*b^2*e*x + 2*a*b*e - a ^2*f)*log((-(I - 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a) ^2 + 1)) - 2*(2*a*b*e - a^2*f)*log(((I - 1)*tan(b*x + a)^2 + 2*I*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) + 2*(2*a*b*e - a^2*f)*log(((I - 1)*tan( b*x + a)^2 - 2*I*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) - 4*(b^2*f*x^ 2 + 2*b^2*e*x)*log((tan(b*x + a) + 1)/(tan(b*x + a) - 1)) - f*polylog(3, ( I*tan(b*x + a)^2 + 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) + f*polylo...
\[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=\int \left (e + f x\right ) \operatorname {acoth}{\left (\tan {\left (a + b x \right )} \right )}\, dx \]
\[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=\int { {\left (f x + e\right )} \operatorname {arcoth}\left (\tan \left (b x + a\right )\right ) \,d x } \]
1/8*(f*x^2 + 2*e*x)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x + 2*a)^2 + 4*si n(2*b*x + 2*a) + 2) - 1/8*(f*x^2 + 2*e*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(((b*f*x^2 + 2*b*e*x) *cos(4*b*x + 4*a)*cos(2*b*x + 2*a) + (b*f*x^2 + 2*b*e*x)*sin(4*b*x + 4*a)* sin(2*b*x + 2*a) + (b*f*x^2 + 2*b*e*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)
\[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=\int { {\left (f x + e\right )} \operatorname {arcoth}\left (\tan \left (b x + a\right )\right ) \,d x } \]
Timed out. \[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=\int \mathrm {acoth}\left (\mathrm {tan}\left (a+b\,x\right )\right )\,\left (e+f\,x\right ) \,d x \]