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

Optimal. Leaf size=66 \[ a \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-\frac{\sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2}{x}-a \sinh ^{-1}(a x)^2+2 a \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right ) \]

[Out]

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

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Rubi [A]  time = 0.162083, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {5723, 5659, 3716, 2190, 2279, 2391} \[ a \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-\frac{\sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2}{x}-a \sinh ^{-1}(a x)^2+2 a \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5723

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] - Dist[(b*c*n*d^IntPart[p]*(d + e
*x^2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Arc
Sinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 5659

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tanh[x], x], x, ArcSinh
[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && 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{\sinh ^{-1}(a x)^2}{x^2 \sqrt{1+a^2 x^2}} \, dx &=-\frac{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{x}+(2 a) \int \frac{\sinh ^{-1}(a x)}{x} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{x}+(2 a) \operatorname{Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-a \sinh ^{-1}(a x)^2-\frac{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{x}-(4 a) \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-a \sinh ^{-1}(a x)^2-\frac{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{x}+2 a \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )-(2 a) \operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-a \sinh ^{-1}(a x)^2-\frac{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{x}+2 a \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )-a \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a x)}\right )\\ &=-a \sinh ^{-1}(a x)^2-\frac{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{x}+2 a \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+a \text{Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.320649, size = 65, normalized size = 0.98 \[ a \left (\sinh ^{-1}(a x) \left (-\frac{\sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{a x}+\sinh ^{-1}(a x)+2 \log \left (1-e^{-2 \sinh ^{-1}(a x)}\right )\right )-\text{PolyLog}\left (2,e^{-2 \sinh ^{-1}(a x)}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*(ArcSinh[a*x]*(ArcSinh[a*x] - (Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(a*x) + 2*Log[1 - E^(-2*ArcSinh[a*x])]) - Pol
yLog[2, E^(-2*ArcSinh[a*x])])

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Maple [A]  time = 0.078, size = 132, normalized size = 2. \begin{align*}{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}{x} \left ( ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) }-2\,a \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}+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/(a^2*x^2+1)^(1/2),x)

[Out]

(a*x-(a^2*x^2+1)^(1/2))/x*arcsinh(a*x)^2-2*a*arcsinh(a*x)^2+2*a*arcsinh(a*x)*ln(1+a*x+(a^2*x^2+1)^(1/2))+2*a*p
olylog(2,-a*x-(a^2*x^2+1)^(1/2))+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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(a^2*x^2 + 1)*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))/(sqrt(a^2*x^2 + 1)*a*x^2 + (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 )^{2}}{a^{2} x^{4} + x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a^2*x^2 + 1)*arcsinh(a*x)^2/(a^2*x^4 + 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} \sqrt{a^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x)**2/(x**2*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 )^{2}}{\sqrt{a^{2} x^{2} + 1} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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