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

Optimal. Leaf size=25 \[ \frac {1}{4} \text {Li}_2\left (-\frac {1}{x+1}\right )-\frac {1}{4} \text {Li}_2\left (\frac {1}{x+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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]

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*}

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Mathematica [B]  time = 0.02, size = 117, normalized size = 4.68 \[ -\frac {\text {Li}_2(-x-1)}{4}+\frac {\text {Li}_2(x+1)}{4}+\frac {1}{8} \log ^2\left (-\frac {1}{x+1}\right )-\frac {1}{8} \log ^2\left (\frac {1}{x+1}\right )+\frac {1}{4} \log (x+2) \log \left (-\frac {1}{x+1}\right )-\frac {1}{4} \log \left (\frac {x+2}{x+1}\right ) \log \left (-\frac {1}{x+1}\right )-\frac {1}{4} \log (-x) \log \left (\frac {1}{x+1}\right )+\frac {1}{4} \log \left (\frac {1}{x+1}\right ) \log \left (\frac {x}{x+1}\right ) \]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcoth}\left (x + 1\right )}{2 \, {\left (x + 1\right )}}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (x + 1\right )}{2 \, {\left (x + 1\right )}}\,{d x} \]

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)

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maple [A]  time = 0.05, size = 34, normalized size = 1.36 \[ \frac {\ln \left (1+x \right ) \mathrm {arccoth}\left (1+x \right )}{2}-\frac {\dilog \left (1+x \right )}{4}-\frac {\dilog \left (x +2\right )}{4}-\frac {\ln \left (1+x \right ) \ln \left (x +2\right )}{4} \]

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)

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maxima [B]  time = 0.31, size = 58, normalized size = 2.32 \[ -\frac {1}{4} \, {\left (\log \left (x + 2\right ) - \log \relax (x)\right )} \log \left (x + 1\right ) + \frac {1}{2} \, \operatorname {arcoth}\left (x + 1\right ) \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x + 1\right ) \log \relax (x) + \frac {1}{4} \, \log \left (x + 2\right ) \log \left (-x - 1\right ) - \frac {1}{4} \, {\rm Li}_2\left (-x\right ) + \frac {1}{4} \, {\rm Li}_2\left (x + 2\right ) \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {\mathrm {acoth}\left (x+1\right )}{2\,x+2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(x + 1)/(2*x + 2),x)

[Out]

int(acoth(x + 1)/(2*x + 2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {acoth}{\left (x + 1 \right )}}{x + 1}\, dx}{2} \]

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

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