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

Optimal. Leaf size=50 \[ -2 a \text{PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+2 a \text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )-\frac{\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \]

[Out]

-(ArcSinh[a*x]^2/x) - 4*a*ArcSinh[a*x]*ArcTanh[E^ArcSinh[a*x]] - 2*a*PolyLog[2, -E^ArcSinh[a*x]] + 2*a*PolyLog
[2, E^ArcSinh[a*x]]

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Rubi [A]  time = 0.0997795, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5661, 5760, 4182, 2279, 2391} \[ -2 a \text{PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+2 a \text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )-\frac{\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^2/x^2,x]

[Out]

-(ArcSinh[a*x]^2/x) - 4*a*ArcSinh[a*x]*ArcTanh[E^ArcSinh[a*x]] - 2*a*PolyLog[2, -E^ArcSinh[a*x]] + 2*a*PolyLog
[2, E^ArcSinh[a*x]]

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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)^2}{x^2} \, dx &=-\frac{\sinh ^{-1}(a x)^2}{x}+(2 a) \int \frac{\sinh ^{-1}(a x)}{x \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{\sinh ^{-1}(a x)^2}{x}+(2 a) \operatorname{Subst}\left (\int x \text{csch}(x) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac{\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-(2 a) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )+(2 a) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac{\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-(2 a) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )+(2 a) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )\\ &=-\frac{\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-2 a \text{Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+2 a \text{Li}_2\left (e^{\sinh ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.191734, size = 75, normalized size = 1.5 \[ a \left (2 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(a x)}\right )-2 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(a x)}\right )-\sinh ^{-1}(a x) \left (\frac{\sinh ^{-1}(a x)}{a x}-2 \log \left (1-e^{-\sinh ^{-1}(a x)}\right )+2 \log \left (e^{-\sinh ^{-1}(a x)}+1\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSinh[a*x]^2/x^2,x]

[Out]

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

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Maple [A]  time = 0.049, size = 107, normalized size = 2.1 \begin{align*} -{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}{x}}-2\,a{\it Arcsinh} \left ( ax \right ) \ln \left ( 1+ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) -2\,a{\it polylog} \left ( 2,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) +2\,a{\it Arcsinh} \left ( ax \right ) \ln \left ( 1-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) +2\,a{\it polylog} \left ( 2,ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsinh(a*x)^2/x-2*a*arcsinh(a*x)*ln(1+a*x+(a^2*x^2+1)^(1/2))-2*a*polylog(2,-a*x-(a^2*x^2+1)^(1/2))+2*a*arcsi
nh(a*x)*ln(1-a*x-(a^2*x^2+1)^(1/2))+2*a*polylog(2,a*x+(a^2*x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{\log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{x} + \int \frac{2 \,{\left (a^{3} x^{2} + \sqrt{a^{2} x^{2} + 1} a^{2} x + a\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )}{a^{3} x^{4} + a x^{2} +{\left (a^{2} x^{3} + x\right )} \sqrt{a^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(a*x + sqrt(a^2*x^2 + 1))^2/x + integrate(2*(a^3*x^2 + sqrt(a^2*x^2 + 1)*a^2*x + a)*log(a*x + sqrt(a^2*x^2
+ 1))/(a^3*x^4 + a*x^2 + (a^2*x^3 + x)*sqrt(a^2*x^2 + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(a*x)^2/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x)**2/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x)^2/x^2, x)