3.6 \(\int \frac{\cosh ^{-1}(a x)}{x} \, dx\)

Optimal. Leaf size=43 \[ \frac{1}{2} \text{PolyLog}\left (2,-e^{2 \cosh ^{-1}(a x)}\right )-\frac{1}{2} \cosh ^{-1}(a x)^2+\cosh ^{-1}(a x) \log \left (e^{2 \cosh ^{-1}(a x)}+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.058621, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {5660, 3718, 2190, 2279, 2391} \[ \frac{1}{2} \text{PolyLog}\left (2,-e^{2 \cosh ^{-1}(a x)}\right )-\frac{1}{2} \cosh ^{-1}(a x)^2+\cosh ^{-1}(a x) \log \left (e^{2 \cosh ^{-1}(a x)}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a*x]/x,x]

[Out]

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

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

Mathematica [A]  time = 0.0340923, size = 42, normalized size = 0.98 \[ \frac{1}{2} \left (\cosh ^{-1}(a x) \left (\cosh ^{-1}(a x)+2 \log \left (e^{-2 \cosh ^{-1}(a x)}+1\right )\right )-\text{PolyLog}\left (2,-e^{-2 \cosh ^{-1}(a x)}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCosh[a*x]/x,x]

[Out]

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

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Maple [A]  time = 0.034, size = 66, normalized size = 1.5 \begin{align*} -{\frac{ \left ({\rm arccosh} \left (ax\right ) \right ) ^{2}}{2}}+{\rm arccosh} \left (ax\right )\ln \left ( 1+ \left ( ax+\sqrt{ax-1}\sqrt{ax+1} \right ) ^{2} \right ) +{\frac{1}{2}{\it polylog} \left ( 2,- \left ( ax+\sqrt{ax-1}\sqrt{ax+1} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh(a*x)/x,x)

[Out]

-1/2*arccosh(a*x)^2+arccosh(a*x)*ln(1+(a*x+(a*x-1)^(1/2)*(a*x+1)^(1/2))^2)+1/2*polylog(2,-(a*x+(a*x-1)^(1/2)*(
a*x+1)^(1/2))^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x,x, algorithm="maxima")

[Out]

integrate(arccosh(a*x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x,x, algorithm="fricas")

[Out]

integral(arccosh(a*x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acosh(a*x)/x,x)

[Out]

Integral(acosh(a*x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x,x, algorithm="giac")

[Out]

integrate(arccosh(a*x)/x, x)