3.182 \(\int \frac{\sinh ^{-1}(a x)}{x \sqrt{1+a^2 x^2}} \, dx\)

Optimal. Leaf size=34 \[ -\text{PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+\text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )-2 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \]

[Out]

-2*ArcSinh[a*x]*ArcTanh[E^ArcSinh[a*x]] - PolyLog[2, -E^ArcSinh[a*x]] + PolyLog[2, E^ArcSinh[a*x]]

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Rubi [A]  time = 0.0854683, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {5760, 4182, 2279, 2391} \[ -\text{PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+\text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )-2 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]/(x*Sqrt[1 + a^2*x^2]),x]

[Out]

-2*ArcSinh[a*x]*ArcTanh[E^ArcSinh[a*x]] - PolyLog[2, -E^ArcSinh[a*x]] + PolyLog[2, E^ArcSinh[a*x]]

Rule 5760

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[
e, c^2*d] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, 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{\sinh ^{-1}(a x)}{x \sqrt{1+a^2 x^2}} \, dx &=\operatorname{Subst}\left (\int x \text{csch}(x) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-2 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-\operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )+\operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-2 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )+\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )\\ &=-2 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-\text{Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+\text{Li}_2\left (e^{\sinh ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.041103, size = 57, normalized size = 1.68 \[ \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(a x)}\right )-\text{PolyLog}\left (2,e^{-\sinh ^{-1}(a x)}\right )+\sinh ^{-1}(a x) \left (\log \left (1-e^{-\sinh ^{-1}(a x)}\right )-\log \left (e^{-\sinh ^{-1}(a x)}+1\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSinh[a*x]/(x*Sqrt[1 + a^2*x^2]),x]

[Out]

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

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Maple [A]  time = 0., size = 42, normalized size = 1.2 \begin{align*} 2\,{\it dilog} \left ( \left ( ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) ^{-1} \right ) -{\frac{1}{2}{\it dilog} \left ( \left ( ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) ^{-2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)/x/(a^2*x^2+1)^(1/2),x)

[Out]

2*dilog(1/(a*x+(a^2*x^2+1)^(1/2)))-1/2*dilog(1/(a*x+(a^2*x^2+1)^(1/2))^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(arcsinh(a*x)/(sqrt(a^2*x^2 + 1)*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a^{2} x^{2} + 1} \operatorname{arsinh}\left (a x\right )}{a^{2} x^{3} + x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*x^2 + 1)*arcsinh(a*x)/(a^2*x^3 + x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asinh}{\left (a x \right )}}{x \sqrt{a^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)/x/(a**2*x**2+1)**(1/2),x)

[Out]

Integral(asinh(a*x)/(x*sqrt(a**2*x**2 + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(arcsinh(a*x)/(sqrt(a^2*x^2 + 1)*x), x)