3.30 \(\int \frac{\text{sech}^{-1}(a x^5)}{x} \, dx\)

Optimal. Leaf size=54 \[ -\frac{1}{10} \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}\left (a x^5\right )}\right )+\frac{1}{10} \text{sech}^{-1}\left (a x^5\right )^2-\frac{1}{5} \text{sech}^{-1}\left (a x^5\right ) \log \left (e^{2 \text{sech}^{-1}\left (a x^5\right )}+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.107238, antiderivative size = 54, 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} \[ -\frac{1}{10} \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}\left (a x^5\right )}\right )+\frac{1}{10} \text{sech}^{-1}\left (a x^5\right )^2-\frac{1}{5} \text{sech}^{-1}\left (a x^5\right ) \log \left (e^{2 \text{sech}^{-1}\left (a x^5\right )}+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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^5\right )}{x} \, dx &=\frac{1}{5} \operatorname{Subst}\left (\int \frac{\text{sech}^{-1}(a x)}{x} \, dx,x,x^5\right )\\ &=-\left (\frac{1}{5} \operatorname{Subst}\left (\int \frac{\cosh ^{-1}\left (\frac{x}{a}\right )}{x} \, dx,x,\frac{1}{x^5}\right )\right )\\ &=-\left (\frac{1}{5} \operatorname{Subst}\left (\int x \tanh (x) \, dx,x,\text{sech}^{-1}\left (a x^5\right )\right )\right )\\ &=\frac{1}{10} \text{sech}^{-1}\left (a x^5\right )^2-\frac{2}{5} \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1+e^{2 x}} \, dx,x,\text{sech}^{-1}\left (a x^5\right )\right )\\ &=\frac{1}{10} \text{sech}^{-1}\left (a x^5\right )^2-\frac{1}{5} \text{sech}^{-1}\left (a x^5\right ) \log \left (1+e^{2 \text{sech}^{-1}\left (a x^5\right )}\right )+\frac{1}{5} \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\text{sech}^{-1}\left (a x^5\right )\right )\\ &=\frac{1}{10} \text{sech}^{-1}\left (a x^5\right )^2-\frac{1}{5} \text{sech}^{-1}\left (a x^5\right ) \log \left (1+e^{2 \text{sech}^{-1}\left (a x^5\right )}\right )+\frac{1}{10} \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \text{sech}^{-1}\left (a x^5\right )}\right )\\ &=\frac{1}{10} \text{sech}^{-1}\left (a x^5\right )^2-\frac{1}{5} \text{sech}^{-1}\left (a x^5\right ) \log \left (1+e^{2 \text{sech}^{-1}\left (a x^5\right )}\right )-\frac{1}{10} \text{Li}_2\left (-e^{2 \text{sech}^{-1}\left (a x^5\right )}\right )\\ \end{align*}

Mathematica [A]  time = 0.0460431, size = 49, normalized size = 0.91 \[ \frac{1}{10} \left (\text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}\left (a x^5\right )}\right )-\text{sech}^{-1}\left (a x^5\right ) \left (\text{sech}^{-1}\left (a x^5\right )+2 \log \left (e^{-2 \text{sech}^{-1}\left (a x^5\right )}+1\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(ArcSech[a*x^5]*(ArcSech[a*x^5] + 2*Log[1 + E^(-2*ArcSech[a*x^5])])) + PolyLog[2, -E^(-2*ArcSech[a*x^5])])/1
0

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Maple [F]  time = 0.284, size = 0, normalized size = 0. \begin{align*} \int{\frac{{\rm arcsech} \left (a{x}^{5}\right )}{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsech}\left (a x^{5}\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(a*x^5)/x, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arsech}\left (a x^{5}\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(a*x^5)/x,x, algorithm="fricas")

[Out]

integral(arcsech(a*x^5)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a*x**5)/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsech}\left (a x^{5}\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(a*x^5)/x,x, algorithm="giac")

[Out]

integrate(arcsech(a*x^5)/x, x)