3.306 \(\int \frac {\sinh ^{-1}(\frac {a}{x})}{x^4} \, dx\)

Optimal. Leaf size=54 \[ \frac {\left (\frac {a^2}{x^2}+1\right )^{3/2}}{9 a^3}-\frac {\sqrt {\frac {a^2}{x^2}+1}}{3 a^3}-\frac {\text {csch}^{-1}\left (\frac {x}{a}\right )}{3 x^3} \]

[Out]

1/9*(1+a^2/x^2)^(3/2)/a^3-1/3*arccsch(x/a)/x^3-1/3*(1+a^2/x^2)^(1/2)/a^3

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Rubi [A]  time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5892, 6284, 266, 43} \[ \frac {\left (\frac {a^2}{x^2}+1\right )^{3/2}}{9 a^3}-\frac {\sqrt {\frac {a^2}{x^2}+1}}{3 a^3}-\frac {\text {csch}^{-1}\left (\frac {x}{a}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a/x]/x^4,x]

[Out]

-Sqrt[1 + a^2/x^2]/(3*a^3) + (1 + a^2/x^2)^(3/2)/(9*a^3) - ArcCsch[x/a]/(3*x^3)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

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]

Rubi steps

\begin {align*} \int \frac {\sinh ^{-1}\left (\frac {a}{x}\right )}{x^4} \, dx &=\int \frac {\text {csch}^{-1}\left (\frac {x}{a}\right )}{x^4} \, dx\\ &=-\frac {\text {csch}^{-1}\left (\frac {x}{a}\right )}{3 x^3}-\frac {1}{3} a \int \frac {1}{\sqrt {1+\frac {a^2}{x^2}} x^5} \, dx\\ &=-\frac {\text {csch}^{-1}\left (\frac {x}{a}\right )}{3 x^3}+\frac {1}{6} a \operatorname {Subst}\left (\int \frac {x}{\sqrt {1+a^2 x}} \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {\text {csch}^{-1}\left (\frac {x}{a}\right )}{3 x^3}+\frac {1}{6} a \operatorname {Subst}\left (\int \left (-\frac {1}{a^2 \sqrt {1+a^2 x}}+\frac {\sqrt {1+a^2 x}}{a^2}\right ) \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {\sqrt {1+\frac {a^2}{x^2}}}{3 a^3}+\frac {\left (1+\frac {a^2}{x^2}\right )^{3/2}}{9 a^3}-\frac {\text {csch}^{-1}\left (\frac {x}{a}\right )}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 48, normalized size = 0.89 \[ \left (\frac {1}{9 a x^2}-\frac {2}{9 a^3}\right ) \sqrt {\frac {a^2+x^2}{x^2}}-\frac {\sinh ^{-1}\left (\frac {a}{x}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a/x]/x^4,x]

[Out]

(-2/(9*a^3) + 1/(9*a*x^2))*Sqrt[(a^2 + x^2)/x^2] - ArcSinh[a/x]/(3*x^3)

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fricas [A]  time = 0.81, size = 62, normalized size = 1.15 \[ -\frac {3 \, a^{3} \log \left (\frac {x \sqrt {\frac {a^{2} + x^{2}}{x^{2}}} + a}{x}\right ) - {\left (a^{2} x - 2 \, x^{3}\right )} \sqrt {\frac {a^{2} + x^{2}}{x^{2}}}}{9 \, a^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a/x)/x^4,x, algorithm="fricas")

[Out]

-1/9*(3*a^3*log((x*sqrt((a^2 + x^2)/x^2) + a)/x) - (a^2*x - 2*x^3)*sqrt((a^2 + x^2)/x^2))/(a^3*x^3)

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giac [A]  time = 0.67, size = 75, normalized size = 1.39 \[ -\frac {\log \left (\sqrt {\frac {a^{2}}{x^{2}} + 1} + \frac {a}{x}\right )}{3 \, x^{3}} - \frac {4 \, {\left (a^{2} - 3 \, {\left (x - \sqrt {a^{2} + x^{2}}\right )}^{2}\right )} a}{9 \, {\left (a^{2} - {\left (x - \sqrt {a^{2} + x^{2}}\right )}^{2}\right )}^{3} \mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a/x)/x^4,x, algorithm="giac")

[Out]

-1/3*log(sqrt(a^2/x^2 + 1) + a/x)/x^3 - 4/9*(a^2 - 3*(x - sqrt(a^2 + x^2))^2)*a/((a^2 - (x - sqrt(a^2 + x^2))^
2)^3*sgn(x))

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maple [A]  time = 0.01, size = 53, normalized size = 0.98 \[ -\frac {\frac {a^{3} \arcsinh \left (\frac {a}{x}\right )}{3 x^{3}}-\frac {a^{2} \sqrt {1+\frac {a^{2}}{x^{2}}}}{9 x^{2}}+\frac {2 \sqrt {1+\frac {a^{2}}{x^{2}}}}{9}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a/x)/x^4,x)

[Out]

-1/a^3*(1/3*a^3/x^3*arcsinh(a/x)-1/9*a^2/x^2*(1+a^2/x^2)^(1/2)+2/9*(1+a^2/x^2)^(1/2))

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maxima [A]  time = 0.46, size = 47, normalized size = 0.87 \[ \frac {1}{9} \, a {\left (\frac {{\left (\frac {a^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}}}{a^{4}} - \frac {3 \, \sqrt {\frac {a^{2}}{x^{2}} + 1}}{a^{4}}\right )} - \frac {\operatorname {arsinh}\left (\frac {a}{x}\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a/x)/x^4,x, algorithm="maxima")

[Out]

1/9*a*((a^2/x^2 + 1)^(3/2)/a^4 - 3*sqrt(a^2/x^2 + 1)/a^4) - 1/3*arcsinh(a/x)/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asinh}\left (\frac {a}{x}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a/x)/x^4,x)

[Out]

int(asinh(a/x)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (\frac {a}{x} \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a/x)/x**4,x)

[Out]

Integral(asinh(a/x)/x**4, x)

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