3.1.29 \(\int \frac {\tanh ^{-1}(1+x)}{2+2 x} \, dx\) [29]

Optimal. Leaf size=21 \[ -\frac {1}{4} \text {PolyLog}(2,-1-x)+\frac {1}{4} \text {PolyLog}(2,1+x) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 21, 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 = {6242, 12, 6031} \begin {gather*} \frac {\text {Li}_2(x+1)}{4}-\frac {\text {Li}_2(-x-1)}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[1 + x]/(2 + 2*x),x]

[Out]

-1/4*PolyLog[2, -1 - x] + PolyLog[2, 1 + x]/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 6031

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

Rule 6242

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[(f*(x/d))^m*(a + b*ArcTanh[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 {\tanh ^{-1}(1+x)}{2+2 x} \, dx &=\text {Subst}\left (\int \frac {\tanh ^{-1}(x)}{2 x} \, dx,x,1+x\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {\tanh ^{-1}(x)}{x} \, dx,x,1+x\right )\\ &=-\frac {1}{4} \text {Li}_2(-1-x)+\frac {\text {Li}_2(1+x)}{4}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 31, normalized size = 1.48 \begin {gather*} -\frac {1}{4} \text {PolyLog}\left (2,\frac {1}{2} (-2-2 x)\right )+\frac {1}{4} \text {PolyLog}\left (2,\frac {1}{2} (2+2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[1 + x]/(2 + 2*x),x]

[Out]

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

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Maple [A]
time = 0.93, size = 34, normalized size = 1.62

method	result	size
risch	\(\frac {\dilog \left (-x \right )}{4}-\frac {\dilog \left (x +2\right )}{4}\)	\(14\)
derivativedivides	\(\frac {\ln \left (1+x \right ) \arctanh \left (1+x \right )}{2}-\frac {\dilog \left (x +2\right )}{4}-\frac {\ln \left (1+x \right ) \ln \left (x +2\right )}{4}-\frac {\dilog \left (1+x \right )}{4}\)	\(34\)
default	\(\frac {\ln \left (1+x \right ) \arctanh \left (1+x \right )}{2}-\frac {\dilog \left (x +2\right )}{4}-\frac {\ln \left (1+x \right ) \ln \left (x +2\right )}{4}-\frac {\dilog \left (1+x \right )}{4}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (15) = 30\).
time = 0.25, size = 58, normalized size = 2.76 \begin {gather*} -\frac {1}{4} \, {\left (\log \left (x + 2\right ) - \log \left (x\right )\right )} \log \left (x + 1\right ) + \frac {1}{2} \, \operatorname {artanh}\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(log(x + 2) - log(x))*log(x + 1) + 1/2*arctanh(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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\operatorname {atanh}{\left (x + 1 \right )}}{x + 1}\, dx}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atanh(x + 1)/(x + 1), x)/2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {\mathrm {atanh}\left (x+1\right )}{2\,x+2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(x + 1)/(2*x + 2),x)

[Out]

int(atanh(x + 1)/(2*x + 2), x)

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