3.3.37 \(\int \frac {\tanh ^{-1}(a x^5)}{x} \, dx\) [237]

Optimal. Leaf size=24 \[ -\frac {1}{10} \text {PolyLog}\left (2,-a x^5\right )+\frac {1}{10} \text {PolyLog}\left (2,a x^5\right ) \]

[Out]

-1/10*polylog(2,-a*x^5)+1/10*polylog(2,a*x^5)

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Rubi [A]
time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6035, 6031} \begin {gather*} \frac {\text {Li}_2\left (a x^5\right )}{10}-\frac {1}{10} \text {Li}_2\left (-a x^5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x^5]/x,x]

[Out]

-1/10*PolyLog[2, -(a*x^5)] + PolyLog[2, a*x^5]/10

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 6035

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTanh[c*x])
^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}\left (a x^5\right )}{x} \, dx &=\frac {1}{5} \text {Subst}\left (\int \frac {\tanh ^{-1}(a x)}{x} \, dx,x,x^5\right )\\ &=-\frac {1}{10} \text {Li}_2\left (-a x^5\right )+\frac {\text {Li}_2\left (a x^5\right )}{10}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 22, normalized size = 0.92 \begin {gather*} \frac {1}{10} \left (-\text {PolyLog}\left (2,-a x^5\right )+\text {PolyLog}\left (2,a x^5\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x^5]/x,x]

[Out]

(-PolyLog[2, -(a*x^5)] + PolyLog[2, a*x^5])/10

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 95, normalized size = 3.96

method	result	size
meijerg	\(-\frac {i \left (\frac {2 i a \,x^{5} \polylog \left (2, \sqrt {a^{2} x^{10}}\right )}{\sqrt {a^{2} x^{10}}}-\frac {2 i a \,x^{5} \polylog \left (2, -\sqrt {a^{2} x^{10}}\right )}{\sqrt {a^{2} x^{10}}}\right )}{20}\)	\(61\)
default	\(\ln \left (x \right ) \arctanh \left (a \,x^{5}\right )-5 a \left (-\frac {\munderset {\textit {\_R1} =\RootOf \left (a \,\textit {\_Z}^{5}-1\right )}{\sum }\left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )}{10 a}+\frac {\munderset {\textit {\_R1} =\RootOf \left (a \,\textit {\_Z}^{5}+1\right )}{\sum }\left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )}{10 a}\right )\)	\(95\)
risch	\(\frac {\ln \left (x \right ) \ln \left (a \,x^{5}+1\right )}{2}-\frac {\left (\munderset {\textit {\_R1} =\RootOf \left (a \,\textit {\_Z}^{5}+1\right )}{\sum }\left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )\right )}{2}-\frac {\ln \left (x \right ) \ln \left (-a \,x^{5}+1\right )}{2}+\frac {\left (\munderset {\textit {\_R1} =\RootOf \left (a \,\textit {\_Z}^{5}-1\right )}{\sum }\left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )\right )}{2}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x^5)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)*arctanh(a*x^5)-5*a*(-1/10/a*sum(ln(x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R1),_R1=RootOf(_Z^5*a-1))+1/10/a*su
m(ln(x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R1),_R1=RootOf(_Z^5*a+1)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (18) = 36\).
time = 0.26, size = 104, normalized size = 4.33 \begin {gather*} -\frac {1}{2} \, a {\left (\frac {\log \left (a x^{5} + 1\right )}{a} - \frac {\log \left (a x^{5} - 1\right )}{a}\right )} \log \left (x\right ) - \frac {1}{10} \, a {\left (\frac {\log \left (a x^{5} - 1\right ) \log \left (a x^{5}\right ) + {\rm Li}_2\left (-a x^{5} + 1\right )}{a} - \frac {\log \left (a x^{5} + 1\right ) \log \left (-a x^{5}\right ) + {\rm Li}_2\left (a x^{5} + 1\right )}{a}\right )} + \operatorname {artanh}\left (a x^{5}\right ) \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x^5)/x,x, algorithm="maxima")

[Out]

-1/2*a*(log(a*x^5 + 1)/a - log(a*x^5 - 1)/a)*log(x) - 1/10*a*((log(a*x^5 - 1)*log(a*x^5) + dilog(-a*x^5 + 1))/
a - (log(a*x^5 + 1)*log(-a*x^5) + dilog(a*x^5 + 1))/a) + arctanh(a*x^5)*log(x)

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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(a*x^5)/x,x, algorithm="fricas")

[Out]

integral(arctanh(a*x^5)/x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {atanh}{\left (a x^{5} \right )}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x**5)/x,x)

[Out]

Integral(atanh(a*x**5)/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(a*x^5)/x,x, algorithm="giac")

[Out]

integrate(arctanh(a*x^5)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\mathrm {atanh}\left (a\,x^5\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x^5)/x,x)

[Out]

int(atanh(a*x^5)/x, x)

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