3.4.46 \(\int \tanh ^{-1}(e^x) \, dx\) [346]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2320, 6031} \begin {gather*} \frac {\text {Li}_2\left (e^x\right )}{2}-\frac {\text {Li}_2\left (-e^x\right )}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[E^x],x]

[Out]

-1/2*PolyLog[2, -E^x] + PolyLog[2, E^x]/2

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[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]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(46\) vs. \(2(21)=42\).
time = 0.18, size = 46, normalized size = 2.19 \begin {gather*} x \tanh ^{-1}\left (e^x\right )+\frac {1}{2} \left (-x \left (-\log \left (1-e^x\right )+\log \left (1+e^x\right )\right )-\text {PolyLog}\left (2,-e^x\right )+\text {PolyLog}\left (2,e^x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[E^x],x]

[Out]

x*ArcTanh[E^x] + (-(x*(-Log[1 - E^x] + Log[1 + E^x])) - PolyLog[2, -E^x] + PolyLog[2, E^x])/2

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Maple [A]
time = 0.05, size = 31, normalized size = 1.48

method	result	size
risch	\(-\frac {\dilog \left ({\mathrm e}^{x}+1\right )}{2}+\frac {\dilog \left (1-{\mathrm e}^{x}\right )}{2}\)	\(18\)
derivativedivides	\(\ln \left ({\mathrm e}^{x}\right ) \arctanh \left ({\mathrm e}^{x}\right )-\frac {\dilog \left ({\mathrm e}^{x}\right )}{2}-\frac {\dilog \left ({\mathrm e}^{x}+1\right )}{2}-\frac {\ln \left ({\mathrm e}^{x}\right ) \ln \left ({\mathrm e}^{x}+1\right )}{2}\)	\(31\)
default	\(\ln \left ({\mathrm e}^{x}\right ) \arctanh \left ({\mathrm e}^{x}\right )-\frac {\dilog \left ({\mathrm e}^{x}\right )}{2}-\frac {\dilog \left ({\mathrm e}^{x}+1\right )}{2}-\frac {\ln \left ({\mathrm e}^{x}\right ) \ln \left ({\mathrm e}^{x}+1\right )}{2}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (13) = 26\).
time = 0.26, size = 58, normalized size = 2.76 \begin {gather*} -\frac {1}{2} \, x {\left (\log \left (e^{x} + 1\right ) - \log \left (e^{x} - 1\right )\right )} + x \operatorname {artanh}\left (e^{x}\right ) + \frac {1}{2} \, \log \left (-e^{x}\right ) \log \left (e^{x} + 1\right ) - \frac {1}{2} \, x \log \left (e^{x} - 1\right ) + \frac {1}{2} \, {\rm Li}_2\left (e^{x} + 1\right ) - \frac {1}{2} \, {\rm Li}_2\left (-e^{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(exp(x)),x, algorithm="maxima")

[Out]

-1/2*x*(log(e^x + 1) - log(e^x - 1)) + x*arctanh(e^x) + 1/2*log(-e^x)*log(e^x + 1) - 1/2*x*log(e^x - 1) + 1/2*
dilog(e^x + 1) - 1/2*dilog(-e^x + 1)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (13) = 26\).
time = 0.45, size = 65, normalized size = 3.10 \begin {gather*} \frac {1}{2} \, x \log \left (-\frac {\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) + \sinh \left (x\right ) - 1}\right ) - \frac {1}{2} \, x \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \frac {1}{2} \, x \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) + \frac {1}{2} \, {\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - \frac {1}{2} \, {\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(exp(x)),x, algorithm="fricas")

[Out]

1/2*x*log(-(cosh(x) + sinh(x) + 1)/(cosh(x) + sinh(x) - 1)) - 1/2*x*log(cosh(x) + sinh(x) + 1) + 1/2*x*log(-co
sh(x) - sinh(x) + 1) + 1/2*dilog(cosh(x) + sinh(x)) - 1/2*dilog(-cosh(x) - sinh(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(exp(x)),x)

[Out]

Integral(atanh(exp(x)), x)

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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(exp(x)),x, algorithm="giac")

[Out]

integrate(arctanh(e^x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(exp(x)),x)

[Out]

int(atanh(exp(x)), x)

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