3.305 \(\int \frac{\sinh ^{-1}(\frac{a}{x})}{x^3} \, dx\)

Optimal. Leaf size=50 \[ \frac{\sqrt{\frac{a^2}{x^2}+1}}{4 a x}-\frac{\text{csch}^{-1}\left (\frac{x}{a}\right )}{4 a^2}-\frac{\text{csch}^{-1}\left (\frac{x}{a}\right )}{2 x^2} \]

[Out]

Sqrt[1 + a^2/x^2]/(4*a*x) - ArcCsch[x/a]/(4*a^2) - ArcCsch[x/a]/(2*x^2)

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Rubi [A]  time = 0.034073, antiderivative size = 50, 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 = {5892, 6284, 335, 321, 215} \[ \frac{\sqrt{\frac{a^2}{x^2}+1}}{4 a x}-\frac{\text{csch}^{-1}\left (\frac{x}{a}\right )}{4 a^2}-\frac{\text{csch}^{-1}\left (\frac{x}{a}\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a/x]/x^3,x]

[Out]

Sqrt[1 + a^2/x^2]/(4*a*x) - ArcCsch[x/a]/(4*a^2) - ArcCsch[x/a]/(2*x^2)

Rule 5892

Int[ArcSinh[(c_.)/((a_.) + (b_.)*(x_)^(n_.))]^(m_.)*(u_.), x_Symbol] :> Int[u*ArcCsch[a/c + (b*x^n)/c]^m, x] /
; FreeQ[{a, b, c, n, m}, x]

Rule 6284

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

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \frac{\sinh ^{-1}\left (\frac{a}{x}\right )}{x^3} \, dx &=\int \frac{\text{csch}^{-1}\left (\frac{x}{a}\right )}{x^3} \, dx\\ &=-\frac{\text{csch}^{-1}\left (\frac{x}{a}\right )}{2 x^2}-\frac{1}{2} a \int \frac{1}{\sqrt{1+\frac{a^2}{x^2}} x^4} \, dx\\ &=-\frac{\text{csch}^{-1}\left (\frac{x}{a}\right )}{2 x^2}+\frac{1}{2} a \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+a^2 x^2}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{\sqrt{1+\frac{a^2}{x^2}}}{4 a x}-\frac{\text{csch}^{-1}\left (\frac{x}{a}\right )}{2 x^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1+a^2 x^2}} \, dx,x,\frac{1}{x}\right )}{4 a}\\ &=\frac{\sqrt{1+\frac{a^2}{x^2}}}{4 a x}-\frac{\text{csch}^{-1}\left (\frac{x}{a}\right )}{4 a^2}-\frac{\text{csch}^{-1}\left (\frac{x}{a}\right )}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0224858, size = 44, normalized size = 0.88 \[ \frac{a x \sqrt{\frac{a^2}{x^2}+1}-\left (2 a^2+x^2\right ) \sinh ^{-1}\left (\frac{a}{x}\right )}{4 a^2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a/x]/x^3,x]

[Out]

(a*Sqrt[1 + a^2/x^2]*x - (2*a^2 + x^2)*ArcSinh[a/x])/(4*a^2*x^2)

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Maple [A]  time = 0.004, size = 46, normalized size = 0.9 \begin{align*} -{\frac{1}{{a}^{2}} \left ({\frac{{a}^{2}}{2\,{x}^{2}}{\it Arcsinh} \left ({\frac{a}{x}} \right ) }-{\frac{a}{4\,x}\sqrt{1+{\frac{{a}^{2}}{{x}^{2}}}}}+{\frac{1}{4}{\it Arcsinh} \left ({\frac{a}{x}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a/x)/x^3,x)

[Out]

-1/a^2*(1/2*a^2/x^2*arcsinh(a/x)-1/4*a/x*(1+a^2/x^2)^(1/2)+1/4*arcsinh(a/x))

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Maxima [B]  time = 1.10012, size = 131, normalized size = 2.62 \begin{align*} \frac{1}{8} \, a{\left (\frac{2 \, x \sqrt{\frac{a^{2}}{x^{2}} + 1}}{a^{2} x^{2}{\left (\frac{a^{2}}{x^{2}} + 1\right )} - a^{4}} - \frac{\log \left (x \sqrt{\frac{a^{2}}{x^{2}} + 1} + a\right )}{a^{3}} + \frac{\log \left (x \sqrt{\frac{a^{2}}{x^{2}} + 1} - a\right )}{a^{3}}\right )} - \frac{\operatorname{arsinh}\left (\frac{a}{x}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a/x)/x^3,x, algorithm="maxima")

[Out]

1/8*a*(2*x*sqrt(a^2/x^2 + 1)/(a^2*x^2*(a^2/x^2 + 1) - a^4) - log(x*sqrt(a^2/x^2 + 1) + a)/a^3 + log(x*sqrt(a^2
/x^2 + 1) - a)/a^3) - 1/2*arcsinh(a/x)/x^2

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Fricas [A]  time = 2.68232, size = 130, normalized size = 2.6 \begin{align*} \frac{a x \sqrt{\frac{a^{2} + x^{2}}{x^{2}}} -{\left (2 \, a^{2} + x^{2}\right )} \log \left (\frac{x \sqrt{\frac{a^{2} + x^{2}}{x^{2}}} + a}{x}\right )}{4 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a/x)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asinh}{\left (\frac{a}{x} \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a/x)/x**3,x)

[Out]

Integral(asinh(a/x)/x**3, x)

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Giac [B]  time = 1.41869, size = 124, normalized size = 2.48 \begin{align*} -\frac{1}{8} \, a{\left (\frac{\log \left (a + \sqrt{a^{2} + x^{2}}\right )}{a^{3} \mathrm{sgn}\left (x\right )} - \frac{\log \left (-a + \sqrt{a^{2} + x^{2}}\right )}{a^{3} \mathrm{sgn}\left (x\right )} - \frac{2 \, \sqrt{a^{2} + x^{2}}}{a^{2} x^{2} \mathrm{sgn}\left (x\right )}\right )} - \frac{\log \left (\sqrt{\frac{a^{2}}{x^{2}} + 1} + \frac{a}{x}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a/x)/x^3,x, algorithm="giac")

[Out]

-1/8*a*(log(a + sqrt(a^2 + x^2))/(a^3*sgn(x)) - log(-a + sqrt(a^2 + x^2))/(a^3*sgn(x)) - 2*sqrt(a^2 + x^2)/(a^
2*x^2*sgn(x))) - 1/2*log(sqrt(a^2/x^2 + 1) + a/x)/x^2