\(\int \frac {\coth ^{-1}(1+x)}{2+2 x} \, dx\) [100]

Optimal result

Integrand size = 12, antiderivative size = 25 \[ \int \frac {\coth ^{-1}(1+x)}{2+2 x} \, dx=\frac {1}{4} \operatorname {PolyLog}\left (2,-\frac {1}{1+x}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1}{1+x}\right ) \]

[Out]

1/4*polylog(2,-1/(1+x))-1/4*polylog(2,1/(1+x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6243, 12, 6032} \[ \int \frac {\coth ^{-1}(1+x)}{2+2 x} \, dx=\frac {1}{4} \operatorname {PolyLog}\left (2,-\frac {1}{x+1}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1}{x+1}\right ) \]

[In]

Int[ArcCoth[1 + x]/(2 + 2*x),x]

[Out]

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

Rule 12

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

Rule 6032

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

Rule 6243

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]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(117\) vs. \(2(25)=50\).

Time = 0.01 (sec) , antiderivative size = 117, normalized size of antiderivative = 4.68 \[ \int \frac {\coth ^{-1}(1+x)}{2+2 x} \, dx=\frac {1}{8} \log ^2\left (-\frac {1}{1+x}\right )-\frac {1}{4} \log (-x) \log \left (\frac {1}{1+x}\right )-\frac {1}{8} \log ^2\left (\frac {1}{1+x}\right )+\frac {1}{4} \log \left (\frac {1}{1+x}\right ) \log \left (\frac {x}{1+x}\right )+\frac {1}{4} \log \left (-\frac {1}{1+x}\right ) \log (2+x)-\frac {1}{4} \log \left (-\frac {1}{1+x}\right ) \log \left (\frac {2+x}{1+x}\right )-\frac {\operatorname {PolyLog}(2,-1-x)}{4}+\frac {\operatorname {PolyLog}(2,1+x)}{4} \]

[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] (verified)

Time = 0.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {\operatorname {dilog}\left (1+x \right )}{4}-\frac {\ln \left (x \right ) \ln \left (1+x \right )}{4}-\frac {\operatorname {dilog}\left (x +2\right )}{4}\)	\(22\)
derivativedivides	\(\frac {\ln \left (1+x \right ) \operatorname {arccoth}\left (1+x \right )}{2}-\frac {\operatorname {dilog}\left (x +2\right )}{4}-\frac {\ln \left (1+x \right ) \ln \left (x +2\right )}{4}-\frac {\operatorname {dilog}\left (1+x \right )}{4}\)	\(34\)
default	\(\frac {\ln \left (1+x \right ) \operatorname {arccoth}\left (1+x \right )}{2}-\frac {\operatorname {dilog}\left (x +2\right )}{4}-\frac {\ln \left (1+x \right ) \ln \left (x +2\right )}{4}-\frac {\operatorname {dilog}\left (1+x \right )}{4}\)	\(34\)
parts	\(\frac {\ln \left (1+x \right ) \operatorname {arccoth}\left (1+x \right )}{2}-\frac {\operatorname {dilog}\left (x +2\right )}{4}-\frac {\ln \left (1+x \right ) \ln \left (x +2\right )}{4}-\frac {\operatorname {dilog}\left (1+x \right )}{4}\)	\(34\)

[In]

int(arccoth(1+x)/(2+2*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {\coth ^{-1}(1+x)}{2+2 x} \, dx=\int { \frac {\operatorname {arcoth}\left (x + 1\right )}{2 \, {\left (x + 1\right )}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\coth ^{-1}(1+x)}{2+2 x} \, dx=\frac {\int \frac {\operatorname {acoth}{\left (x + 1 \right )}}{x + 1}\, dx}{2} \]

[In]

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

[Out]

Integral(acoth(x + 1)/(x + 1), x)/2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (19) = 38\).

Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.32 \[ \int \frac {\coth ^{-1}(1+x)}{2+2 x} \, dx=-\frac {1}{4} \, {\left (\log \left (x + 2\right ) - \log \left (x\right )\right )} \log \left (x + 1\right ) + \frac {1}{2} \, \operatorname {arcoth}\left (x + 1\right ) \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x + 1\right ) \log \left (x\right ) + \frac {1}{4} \, \log \left (x + 2\right ) \log \left (-x - 1\right ) - \frac {1}{4} \, {\rm Li}_2\left (-x\right ) + \frac {1}{4} \, {\rm Li}_2\left (x + 2\right ) \]

[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)

Giac [F]

\[ \int \frac {\coth ^{-1}(1+x)}{2+2 x} \, dx=\int { \frac {\operatorname {arcoth}\left (x + 1\right )}{2 \, {\left (x + 1\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\coth ^{-1}(1+x)}{2+2 x} \, dx=\int \frac {\mathrm {acoth}\left (x+1\right )}{2\,x+2} \,d x \]

[In]

int(acoth(x + 1)/(2*x + 2),x)

[Out]

int(acoth(x + 1)/(2*x + 2), x)