3.26 \(\int \frac{a+b \text{sech}^{-1}(c x)}{x} \, dx\)

Optimal. Leaf size=56 \[ \frac{1}{2} b \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(c x)}\right )-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{2 b}-\log \left (e^{-2 \text{sech}^{-1}(c x)}+1\right ) \left (a+b \text{sech}^{-1}(c x)\right ) \]

[Out]

-(a + b*ArcSech[c*x])^2/(2*b) - (a + b*ArcSech[c*x])*Log[1 + E^(-2*ArcSech[c*x])] + (b*PolyLog[2, -E^(-2*ArcSe
ch[c*x])])/2

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Rubi [A]  time = 0.0879521, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6281, 5660, 3718, 2190, 2279, 2391} \[ -\frac{1}{2} b \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(c x)}\right )+\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{2 b}-\log \left (e^{2 \text{sech}^{-1}(c x)}+1\right ) \left (a+b \text{sech}^{-1}(c x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcSech[c*x])/x,x]

[Out]

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

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

Mathematica [A]  time = 0.043811, size = 47, normalized size = 0.84 \[ \frac{1}{2} b \left (\text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(c x)}\right )-\text{sech}^{-1}(c x) \left (\text{sech}^{-1}(c x)+2 \log \left (e^{-2 \text{sech}^{-1}(c x)}+1\right )\right )\right )+a \log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSech[c*x])/x,x]

[Out]

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

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Maple [A]  time = 0.256, size = 100, normalized size = 1.8 \begin{align*} a\ln \left ( cx \right ) +{\frac{b \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{2}}-b{\rm arcsech} \left (cx\right )\ln \left ( 1+ \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) -{\frac{b}{2}{\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsech(c*x))/x,x)

[Out]

a*ln(c*x)+1/2*b*arcsech(c*x)^2-b*arcsech(c*x)*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)-1/2*b*polylog(2
,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))/x,x, algorithm="maxima")

[Out]

b*integrate(log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))/x, x) + a*log(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))/x,x, algorithm="fricas")

[Out]

integral((b*arcsech(c*x) + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asech(c*x))/x,x)

[Out]

Integral((a + b*asech(c*x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))/x,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)/x, x)