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

Optimal. Leaf size=54 \[ \frac{1}{4} \text{PolyLog}\left (2,e^{2 \sinh ^{-1}\left (a x^2\right )}\right )-\frac{1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac{1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right ) \]

[Out]

-ArcSinh[a*x^2]^2/4 + (ArcSinh[a*x^2]*Log[1 - E^(2*ArcSinh[a*x^2])])/2 + PolyLog[2, E^(2*ArcSinh[a*x^2])]/4

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Rubi [A]  time = 0.0622429, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5890, 3716, 2190, 2279, 2391} \[ \frac{1}{4} \text{PolyLog}\left (2,e^{2 \sinh ^{-1}\left (a x^2\right )}\right )-\frac{1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac{1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcSinh[a*x^2]^2/4 + (ArcSinh[a*x^2]*Log[1 - E^(2*ArcSinh[a*x^2])])/2 + PolyLog[2, E^(2*ArcSinh[a*x^2])]/4

Rule 5890

Int[ArcSinh[(a_.)*(x_)^(p_)]^(n_.)/(x_), x_Symbol] :> Dist[1/p, Subst[Int[x^n*Coth[x], x], x, ArcSinh[a*x^p]],
x] /; FreeQ[{a, p}, 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}\left (a x^2\right )}{x} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}\left (a x^2\right )\right )\\ &=-\frac{1}{4} \sinh ^{-1}\left (a x^2\right )^2-\operatorname{Subst}\left (\int \frac{e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}\left (a x^2\right )\right )\\ &=-\frac{1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac{1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}\left (a x^2\right )\right )\\ &=-\frac{1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac{1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}\left (a x^2\right )}\right )\\ &=-\frac{1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac{1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right )+\frac{1}{4} \text{Li}_2\left (e^{2 \sinh ^{-1}\left (a x^2\right )}\right )\\ \end{align*}

Mathematica [A]  time = 0.0070841, size = 54, normalized size = 1. \[ \frac{1}{4} \text{PolyLog}\left (2,e^{2 \sinh ^{-1}\left (a x^2\right )}\right )-\frac{1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac{1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcSinh[a*x^2]^2/4 + (ArcSinh[a*x^2]*Log[1 - E^(2*ArcSinh[a*x^2])])/2 + PolyLog[2, E^(2*ArcSinh[a*x^2])]/4

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Maple [F]  time = 0.048, size = 0, normalized size = 0. \begin{align*} \int{\frac{{\it Arcsinh} \left ( a{x}^{2} \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x^2)/x, 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^{2}\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(a*x^2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x**2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x^2)/x, x)