3.100 \(\int \frac{\coth ^{-1}(1+x)}{2+2 x} \, dx\)

Optimal. Leaf size=25 \[ \frac{1}{4} \text{PolyLog}\left (2,-\frac{1}{x+1}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{1}{x+1}\right ) \]

[Out]

PolyLog[2, -(1 + x)^(-1)]/4 - PolyLog[2, (1 + x)^(-1)]/4

________________________________________________________________________________________

Rubi [A]  time = 0.0227001, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6108, 12, 5913} \[ \frac{1}{4} \text{PolyLog}\left (2,-\frac{1}{x+1}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{1}{x+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[1 + x]/(2 + 2*x),x]

[Out]

PolyLog[2, -(1 + x)^(-1)]/4 - PolyLog[2, (1 + x)^(-1)]/4

Rule 6108

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((f*x)/d)^m*(a + b*ArcCoth[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f,
0] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5913

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Simp[(b*PolyLog[2, -(c*x)^(-1)
])/2, x] - Simp[(b*PolyLog[2, 1/(c*x)])/2, x]) /; FreeQ[{a, b, c}, x]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(1+x)}{2+2 x} \, dx &=\operatorname{Subst}\left (\int \frac{\coth ^{-1}(x)}{2 x} \, dx,x,1+x\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\coth ^{-1}(x)}{x} \, dx,x,1+x\right )\\ &=\frac{1}{4} \text{Li}_2\left (-\frac{1}{1+x}\right )-\frac{1}{4} \text{Li}_2\left (\frac{1}{1+x}\right )\\ \end{align*}

Mathematica [B]  time = 0.0131804, size = 117, normalized size = 4.68 \[ -\frac{1}{4} \text{PolyLog}(2,-x-1)+\frac{1}{4} \text{PolyLog}(2,x+1)+\frac{1}{8} \log ^2\left (-\frac{1}{x+1}\right )-\frac{1}{8} \log ^2\left (\frac{1}{x+1}\right )+\frac{1}{4} \log (x+2) \log \left (-\frac{1}{x+1}\right )-\frac{1}{4} \log \left (\frac{x+2}{x+1}\right ) \log \left (-\frac{1}{x+1}\right )-\frac{1}{4} \log (-x) \log \left (\frac{1}{x+1}\right )+\frac{1}{4} \log \left (\frac{1}{x+1}\right ) \log \left (\frac{x}{x+1}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[1 + x]/(2 + 2*x),x]

[Out]

Log[-(1 + x)^(-1)]^2/8 - (Log[-x]*Log[(1 + x)^(-1)])/4 - Log[(1 + x)^(-1)]^2/8 + (Log[(1 + x)^(-1)]*Log[x/(1 +
x)])/4 + (Log[-(1 + x)^(-1)]*Log[2 + x])/4 - (Log[-(1 + x)^(-1)]*Log[(2 + x)/(1 + x)])/4 - PolyLog[2, -1 - x]
/4 + PolyLog[2, 1 + x]/4

________________________________________________________________________________________

Maple [A]  time = 0.029, size = 34, normalized size = 1.4 \begin{align*}{\frac{\ln \left ( 1+x \right ){\rm arccoth} \left (1+x\right )}{2}}-{\frac{{\it dilog} \left ( 1+x \right ) }{4}}-{\frac{{\it dilog} \left ( x+2 \right ) }{4}}-{\frac{\ln \left ( 1+x \right ) \ln \left ( x+2 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(1+x)/(2+2*x),x)

[Out]

1/2*ln(1+x)*arccoth(1+x)-1/4*dilog(1+x)-1/4*dilog(x+2)-1/4*ln(1+x)*ln(x+2)

________________________________________________________________________________________

Maxima [B]  time = 0.978667, size = 78, normalized size = 3.12 \begin{align*} -\frac{1}{4} \,{\left (\log \left (x + 2\right ) - \log \left (x\right )\right )} \log \left (x + 1\right ) + \frac{1}{2} \, \operatorname{arcoth}\left (x + 1\right ) \log \left (x + 1\right ) - \frac{1}{4} \, \log \left (x + 1\right ) \log \left (x\right ) + \frac{1}{4} \, \log \left (x + 2\right ) \log \left (-x - 1\right ) - \frac{1}{4} \,{\rm Li}_2\left (-x\right ) + \frac{1}{4} \,{\rm Li}_2\left (x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1+x)/(2+2*x),x, algorithm="maxima")

[Out]

-1/4*(log(x + 2) - log(x))*log(x + 1) + 1/2*arccoth(x + 1)*log(x + 1) - 1/4*log(x + 1)*log(x) + 1/4*log(x + 2)
*log(-x - 1) - 1/4*dilog(-x) + 1/4*dilog(x + 2)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arcoth}\left (x + 1\right )}{2 \,{\left (x + 1\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1+x)/(2+2*x),x, algorithm="fricas")

[Out]

integral(1/2*arccoth(x + 1)/(x + 1), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{acoth}{\left (x + 1 \right )}}{x + 1}\, dx}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(1+x)/(2+2*x),x)

[Out]

Integral(acoth(x + 1)/(x + 1), x)/2

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (x + 1\right )}{2 \,{\left (x + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1+x)/(2+2*x),x, algorithm="giac")

[Out]

integrate(1/2*arccoth(x + 1)/(x + 1), x)