3.29 \(\int \frac{\text{sech}^{-1}(a x^n)}{x} \, dx\)

Optimal. Leaf size=61 \[ -\frac{\text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}\left (a x^n\right )}\right )}{2 n}+\frac{\text{sech}^{-1}\left (a x^n\right )^2}{2 n}-\frac{\text{sech}^{-1}\left (a x^n\right ) \log \left (e^{2 \text{sech}^{-1}\left (a x^n\right )}+1\right )}{n} \]

[Out]

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

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Rubi [A]  time = 0.105367, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6281, 5660, 3718, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}\left (a x^n\right )}\right )}{2 n}+\frac{\text{sech}^{-1}\left (a x^n\right )^2}{2 n}-\frac{\text{sech}^{-1}\left (a x^n\right ) \log \left (e^{2 \text{sech}^{-1}\left (a x^n\right )}+1\right )}{n} \]

Antiderivative was successfully verified.

[In]

Int[ArcSech[a*x^n]/x,x]

[Out]

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

Rule 6281

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCosh[x/c])/x, x], x, 1/x] /; F
reeQ[{a, b, c}, x]

Rule 5660

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

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [B]  time = 0.956552, size = 219, normalized size = 3.59 \[ \frac{\sqrt{\frac{1-a x^n}{a x^n+1}} \left (\sqrt{1-a^2 x^{2 n}} \left (-4 \text{PolyLog}\left (2,\frac{1}{2}-\frac{1}{2} \sqrt{1-a^2 x^{2 n}}\right )+\log ^2\left (a^2 x^{2 n}\right )+2 \log ^2\left (\frac{1}{2} \left (\sqrt{1-a^2 x^{2 n}}+1\right )\right )-4 \log \left (\frac{1}{2} \left (\sqrt{1-a^2 x^{2 n}}+1\right )\right ) \log \left (a^2 x^{2 n}\right )\right )+4 \sqrt{a^2 x^{2 n}-1} \left (2 n \log (x)-\log \left (a^2 x^{2 n}\right )\right ) \tan ^{-1}\left (\sqrt{a^2 x^{2 n}-1}\right )\right )}{8 \left (n-a n x^n\right )}+\log (x) \text{sech}^{-1}\left (a x^n\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSech[a*x^n]/x,x]

[Out]

ArcSech[a*x^n]*Log[x] + (Sqrt[(1 - a*x^n)/(1 + a*x^n)]*(4*Sqrt[-1 + a^2*x^(2*n)]*ArcTan[Sqrt[-1 + a^2*x^(2*n)]
]*(2*n*Log[x] - Log[a^2*x^(2*n)]) + Sqrt[1 - a^2*x^(2*n)]*(Log[a^2*x^(2*n)]^2 - 4*Log[a^2*x^(2*n)]*Log[(1 + Sq
rt[1 - a^2*x^(2*n)])/2] + 2*Log[(1 + Sqrt[1 - a^2*x^(2*n)])/2]^2 - 4*PolyLog[2, 1/2 - Sqrt[1 - a^2*x^(2*n)]/2]
)))/(8*(n - a*n*x^n))

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Maple [A]  time = 0.333, size = 116, normalized size = 1.9 \begin{align*}{\frac{ \left ({\rm arcsech} \left (a{x}^{n}\right ) \right ) ^{2}}{2\,n}}-{\frac{{\rm arcsech} \left (a{x}^{n}\right )}{n}\ln \left ( 1+ \left ({\frac{1}{a{x}^{n}}}+\sqrt{{\frac{1}{a{x}^{n}}}-1}\sqrt{{\frac{1}{a{x}^{n}}}+1} \right ) ^{2} \right ) }-{\frac{1}{2\,n}{\it polylog} \left ( 2,- \left ({\frac{1}{a{x}^{n}}}+\sqrt{{\frac{1}{a{x}^{n}}}-1}\sqrt{{\frac{1}{a{x}^{n}}}+1} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(a*x^n)/x,x)

[Out]

1/2*arcsech(a*x^n)^2/n-arcsech(a*x^n)*ln(1+(1/a/(x^n)+(1/a/(x^n)-1)^(1/2)*(1/a/(x^n)+1)^(1/2))^2)/n-1/2*polylo
g(2,-(1/a/(x^n)+(1/a/(x^n)-1)^(1/2)*(1/a/(x^n)+1)^(1/2))^2)/n

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} n \int \frac{x^{2 \, n} \log \left (x\right )}{a^{2} x x^{2 \, n} +{\left (a^{2} x x^{2 \, n} - x\right )} \sqrt{a x^{n} + 1} \sqrt{-a x^{n} + 1} - x}\,{d x} + n \int \frac{\log \left (x\right )}{2 \,{\left (a x x^{n} + x\right )}}\,{d x} - n \int \frac{\log \left (x\right )}{2 \,{\left (a x x^{n} - x\right )}}\,{d x} + \log \left (\sqrt{a x^{n} + 1} \sqrt{-a x^{n} + 1} + 1\right ) \log \left (x\right ) - \log \left (a\right ) \log \left (x\right ) - \log \left (x\right ) \log \left (x^{n}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(a*x^n)/x,x, algorithm="maxima")

[Out]

a^2*n*integrate(x^(2*n)*log(x)/(a^2*x*x^(2*n) + (a^2*x*x^(2*n) - x)*sqrt(a*x^n + 1)*sqrt(-a*x^n + 1) - x), x)
+ n*integrate(1/2*log(x)/(a*x*x^n + x), x) - n*integrate(1/2*log(x)/(a*x*x^n - x), x) + log(sqrt(a*x^n + 1)*sq
rt(-a*x^n + 1) + 1)*log(x) - log(a)*log(x) - log(x)*log(x^n)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(a*x^n)/x,x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asech}{\left (a x^{n} \right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asech(a*x**n)/x,x)

[Out]

Integral(asech(a*x**n)/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsech}\left (a x^{n}\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(a*x^n)/x,x, algorithm="giac")

[Out]

integrate(arcsech(a*x^n)/x, x)