3.251 \(\int \coth ^{-1}(\cot (a+b x)) \, dx\)

Optimal. Leaf size=79 \[ -\frac{i \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+i x \tan ^{-1}\left (e^{2 i (a+b x)}\right )+x \coth ^{-1}(\cot (a+b x)) \]

[Out]

x*ArcCoth[Cot[a + b*x]] + I*x*ArcTan[E^((2*I)*(a + b*x))] - ((I/4)*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b + (
(I/4)*PolyLog[2, I*E^((2*I)*(a + b*x))])/b

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Rubi [A]  time = 0.048343, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {6250, 4181, 2279, 2391} \[ -\frac{i \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+i x \tan ^{-1}\left (e^{2 i (a+b x)}\right )+x \coth ^{-1}(\cot (a+b x)) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Cot[a + b*x]],x]

[Out]

x*ArcCoth[Cot[a + b*x]] + I*x*ArcTan[E^((2*I)*(a + b*x))] - ((I/4)*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b + (
(I/4)*PolyLog[2, I*E^((2*I)*(a + b*x))])/b

Rule 6250

Int[ArcCoth[Cot[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcCoth[Cot[a + b*x]], x] - Dist[b, Int[x*Sec[2*a +
2*b*x], x], x] /; FreeQ[{a, b}, x]

Rule 4181

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] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(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]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \coth ^{-1}(\cot (a+b x)) \, dx &=x \coth ^{-1}(\cot (a+b x))-b \int x \sec (2 a+2 b x) \, dx\\ &=x \coth ^{-1}(\cot (a+b x))+i x \tan ^{-1}\left (e^{2 i (a+b x)}\right )+\frac{1}{2} \int \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac{1}{2} \int \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx\\ &=x \coth ^{-1}(\cot (a+b x))+i x \tan ^{-1}\left (e^{2 i (a+b x)}\right )-\frac{i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{4 b}+\frac{i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{4 b}\\ &=x \coth ^{-1}(\cot (a+b x))+i x \tan ^{-1}\left (e^{2 i (a+b x)}\right )-\frac{i \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0317604, size = 78, normalized size = 0.99 \[ \frac{-i \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )+i \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )+4 b x \left (\coth ^{-1}(\cot (a+b x))+i \tan ^{-1}\left (e^{2 i (a+b x)}\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Cot[a + b*x]],x]

[Out]

(4*b*x*(ArcCoth[Cot[a + b*x]] + I*ArcTan[E^((2*I)*(a + b*x))]) - I*PolyLog[2, (-I)*E^((2*I)*(a + b*x))] + I*Po
lyLog[2, I*E^((2*I)*(a + b*x))])/(4*b)

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Maple [B]  time = 0.161, size = 265, normalized size = 3.4 \begin{align*} -{\frac{{\rm arccoth} \left (\cot \left ( bx+a \right ) \right )\pi }{2\,b}}+{\frac{{\rm arccoth} \left (\cot \left ( bx+a \right ) \right ){\rm arccot} \left (\cot \left ( bx+a \right ) \right )}{b}}-{\frac{\pi }{4\,b}\ln \left ( 1+{\frac{i \left ( 1+i\cot \left ( bx+a \right ) \right ) ^{2}}{ \left ( \cot \left ( bx+a \right ) \right ) ^{2}+1}} \right ) }+{\frac{{\rm arccot} \left (\cot \left ( bx+a \right ) \right )}{2\,b}\ln \left ( 1+{\frac{i \left ( 1+i\cot \left ( bx+a \right ) \right ) ^{2}}{ \left ( \cot \left ( bx+a \right ) \right ) ^{2}+1}} \right ) }+{\frac{{\frac{i}{4}}}{b}{\it polylog} \left ( 2,{\frac{-i \left ( 1+i\cot \left ( bx+a \right ) \right ) ^{2}}{ \left ( \cot \left ( bx+a \right ) \right ) ^{2}+1}} \right ) }+{\frac{\pi }{4\,b}\ln \left ( 1-{\frac{i \left ( 1+i\cot \left ( bx+a \right ) \right ) ^{2}}{ \left ( \cot \left ( bx+a \right ) \right ) ^{2}+1}} \right ) }-{\frac{{\rm arccot} \left (\cot \left ( bx+a \right ) \right )}{2\,b}\ln \left ( 1-{\frac{i \left ( 1+i\cot \left ( bx+a \right ) \right ) ^{2}}{ \left ( \cot \left ( bx+a \right ) \right ) ^{2}+1}} \right ) }-{\frac{{\frac{i}{4}}}{b}{\it polylog} \left ( 2,{\frac{i \left ( 1+i\cot \left ( bx+a \right ) \right ) ^{2}}{ \left ( \cot \left ( bx+a \right ) \right ) ^{2}+1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(cot(b*x+a)),x)

[Out]

-1/2/b*arccoth(cot(b*x+a))*Pi+1/b*arccoth(cot(b*x+a))*arccot(cot(b*x+a))-1/4/b*ln(1+I*(1+I*cot(b*x+a))^2/(cot(
b*x+a)^2+1))*Pi+1/2/b*ln(1+I*(1+I*cot(b*x+a))^2/(cot(b*x+a)^2+1))*arccot(cot(b*x+a))+1/4*I/b*polylog(2,-I*(1+I
*cot(b*x+a))^2/(cot(b*x+a)^2+1))+1/4/b*ln(1-I*(1+I*cot(b*x+a))^2/(cot(b*x+a)^2+1))*Pi-1/2/b*ln(1-I*(1+I*cot(b*
x+a))^2/(cot(b*x+a)^2+1))*arccot(cot(b*x+a))-1/4*I/b*polylog(2,I*(1+I*cot(b*x+a))^2/(cot(b*x+a)^2+1))

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Maxima [B]  time = 1.68979, size = 248, normalized size = 3.14 \begin{align*} \frac{4 \,{\left (b x + a\right )} \operatorname{arcoth}\left (\frac{1}{\tan \left (b x + a\right )}\right ) +{\left (\arctan \left (\frac{1}{2} \, \tan \left (b x + a\right ) + \frac{1}{2}, \frac{1}{2} \, \tan \left (b x + a\right ) + \frac{1}{2}\right ) - \arctan \left (\frac{1}{2} \, \tan \left (b x + a\right ) - \frac{1}{2}, -\frac{1}{2} \, \tan \left (b x + a\right ) + \frac{1}{2}\right )\right )} \log \left (\tan \left (b x + a\right )^{2} + 1\right ) -{\left (b x + a\right )} \log \left (\frac{1}{2} \, \tan \left (b x + a\right )^{2} + \tan \left (b x + a\right ) + \frac{1}{2}\right ) +{\left (b x + a\right )} \log \left (\frac{1}{2} \, \tan \left (b x + a\right )^{2} - \tan \left (b x + a\right ) + \frac{1}{2}\right ) - i \,{\rm Li}_2\left (\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \tan \left (b x + a\right ) - \frac{1}{2} i + \frac{1}{2}\right ) + i \,{\rm Li}_2\left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \tan \left (b x + a\right ) + \frac{1}{2} i + \frac{1}{2}\right ) + i \,{\rm Li}_2\left (\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \tan \left (b x + a\right ) + \frac{1}{2} i + \frac{1}{2}\right ) - i \,{\rm Li}_2\left (-\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \tan \left (b x + a\right ) - \frac{1}{2} i + \frac{1}{2}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(cot(b*x+a)),x, algorithm="maxima")

[Out]

1/4*(4*(b*x + a)*arccoth(1/tan(b*x + a)) + (arctan2(1/2*tan(b*x + a) + 1/2, 1/2*tan(b*x + a) + 1/2) - arctan2(
1/2*tan(b*x + a) - 1/2, -1/2*tan(b*x + a) + 1/2))*log(tan(b*x + a)^2 + 1) - (b*x + a)*log(1/2*tan(b*x + a)^2 +
tan(b*x + a) + 1/2) + (b*x + a)*log(1/2*tan(b*x + a)^2 - tan(b*x + a) + 1/2) - I*dilog((1/2*I + 1/2)*tan(b*x
+ a) - 1/2*I + 1/2) + I*dilog(-(1/2*I - 1/2)*tan(b*x + a) + 1/2*I + 1/2) + I*dilog((1/2*I - 1/2)*tan(b*x + a)
+ 1/2*I + 1/2) - I*dilog(-(1/2*I + 1/2)*tan(b*x + a) - 1/2*I + 1/2))/b

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Fricas [B]  time = 1.83244, size = 1029, normalized size = 13.03 \begin{align*} \frac{4 \, b x \log \left (\frac{\cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ) + 1}{\cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right ) + 1}\right ) + 2 \, a \log \left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + i\right ) - 2 \, a \log \left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + i\right ) - 2 \,{\left (b x + a\right )} \log \left (i \, \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \,{\left (b x + a\right )} \log \left (i \, \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) - 2 \,{\left (b x + a\right )} \log \left (-i \, \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \,{\left (b x + a\right )} \log \left (-i \, \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, a \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + i\right ) - 2 \, a \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + i\right ) + i \,{\rm Li}_2\left (i \, \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right )\right ) + i \,{\rm Li}_2\left (i \, \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right ) - i \,{\rm Li}_2\left (-i \, \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right )\right ) - i \,{\rm Li}_2\left (-i \, \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(cot(b*x+a)),x, algorithm="fricas")

[Out]

1/8*(4*b*x*log((cos(2*b*x + 2*a) + sin(2*b*x + 2*a) + 1)/(cos(2*b*x + 2*a) - sin(2*b*x + 2*a) + 1)) + 2*a*log(
cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + I) - 2*a*log(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + I) - 2*(b*x + a)*
log(I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a) + 1) + 2*(b*x + a)*log(I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a) + 1) -
2*(b*x + a)*log(-I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a) + 1) + 2*(b*x + a)*log(-I*cos(2*b*x + 2*a) - sin(2*b*x
+ 2*a) + 1) + 2*a*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + I) - 2*a*log(-cos(2*b*x + 2*a) - I*sin(2*b*x +
2*a) + I) + I*dilog(I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) + I*dilog(I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)) -
I*dilog(-I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - I*dilog(-I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)))/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{acoth}{\left (\cot{\left (a + b x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(cot(b*x+a)),x)

[Out]

Integral(acoth(cot(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{arcoth}\left (\cot \left (b x + a\right )\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(cot(b*x+a)),x, algorithm="giac")

[Out]

integrate(arccoth(cot(b*x + a)), x)