3.11 \(\int \frac{\sinh ^{-1}(a x)}{x^6} \, dx\)

Optimal. Leaf size=77 \[ \frac{3 a^3 \sqrt{a^2 x^2+1}}{40 x^2}-\frac{a \sqrt{a^2 x^2+1}}{20 x^4}-\frac{3}{40} a^5 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right )-\frac{\sinh ^{-1}(a x)}{5 x^5} \]

[Out]

-(a*Sqrt[1 + a^2*x^2])/(20*x^4) + (3*a^3*Sqrt[1 + a^2*x^2])/(40*x^2) - ArcSinh[a*x]/(5*x^5) - (3*a^5*ArcTanh[S
qrt[1 + a^2*x^2]])/40

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Rubi [A]  time = 0.0449983, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {5661, 266, 51, 63, 208} \[ \frac{3 a^3 \sqrt{a^2 x^2+1}}{40 x^2}-\frac{a \sqrt{a^2 x^2+1}}{20 x^4}-\frac{3}{40} a^5 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right )-\frac{\sinh ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]/x^6,x]

[Out]

-(a*Sqrt[1 + a^2*x^2])/(20*x^4) + (3*a^3*Sqrt[1 + a^2*x^2])/(40*x^2) - ArcSinh[a*x]/(5*x^5) - (3*a^5*ArcTanh[S
qrt[1 + a^2*x^2]])/40

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 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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sinh ^{-1}(a x)}{x^6} \, dx &=-\frac{\sinh ^{-1}(a x)}{5 x^5}+\frac{1}{5} a \int \frac{1}{x^5 \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{\sinh ^{-1}(a x)}{5 x^5}+\frac{1}{10} a \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{a \sqrt{1+a^2 x^2}}{20 x^4}-\frac{\sinh ^{-1}(a x)}{5 x^5}-\frac{1}{40} \left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{a \sqrt{1+a^2 x^2}}{20 x^4}+\frac{3 a^3 \sqrt{1+a^2 x^2}}{40 x^2}-\frac{\sinh ^{-1}(a x)}{5 x^5}+\frac{1}{80} \left (3 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{a \sqrt{1+a^2 x^2}}{20 x^4}+\frac{3 a^3 \sqrt{1+a^2 x^2}}{40 x^2}-\frac{\sinh ^{-1}(a x)}{5 x^5}+\frac{1}{40} \left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{a^2}+\frac{x^2}{a^2}} \, dx,x,\sqrt{1+a^2 x^2}\right )\\ &=-\frac{a \sqrt{1+a^2 x^2}}{20 x^4}+\frac{3 a^3 \sqrt{1+a^2 x^2}}{40 x^2}-\frac{\sinh ^{-1}(a x)}{5 x^5}-\frac{3}{40} a^5 \tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )\\ \end{align*}

Mathematica [C]  time = 0.0111742, size = 49, normalized size = 0.64 \[ -\frac{1}{5} a^5 \sqrt{a^2 x^2+1} \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},a^2 x^2+1\right )-\frac{\sinh ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x]/x^6,x]

[Out]

-ArcSinh[a*x]/(5*x^5) - (a^5*Sqrt[1 + a^2*x^2]*Hypergeometric2F1[1/2, 3, 3/2, 1 + a^2*x^2])/5

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Maple [A]  time = 0.006, size = 70, normalized size = 0.9 \begin{align*}{a}^{5} \left ( -{\frac{{\it Arcsinh} \left ( ax \right ) }{5\,{a}^{5}{x}^{5}}}-{\frac{1}{20\,{a}^{4}{x}^{4}}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{3}{40\,{a}^{2}{x}^{2}}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{3}{40}{\it Artanh} \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)/x^6,x)

[Out]

a^5*(-1/5*arcsinh(a*x)/a^5/x^5-1/20/a^4/x^4*(a^2*x^2+1)^(1/2)+3/40/a^2/x^2*(a^2*x^2+1)^(1/2)-3/40*arctanh(1/(a
^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.16405, size = 88, normalized size = 1.14 \begin{align*} -\frac{1}{40} \,{\left (3 \, a^{4} \operatorname{arsinh}\left (\frac{1}{\sqrt{a^{2}}{\left | x \right |}}\right ) - \frac{3 \, \sqrt{a^{2} x^{2} + 1} a^{2}}{x^{2}} + \frac{2 \, \sqrt{a^{2} x^{2} + 1}}{x^{4}}\right )} a - \frac{\operatorname{arsinh}\left (a x\right )}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x^6,x, algorithm="maxima")

[Out]

-1/40*(3*a^4*arcsinh(1/(sqrt(a^2)*abs(x))) - 3*sqrt(a^2*x^2 + 1)*a^2/x^2 + 2*sqrt(a^2*x^2 + 1)/x^4)*a - 1/5*ar
csinh(a*x)/x^5

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Fricas [B]  time = 1.95841, size = 302, normalized size = 3.92 \begin{align*} -\frac{3 \, a^{5} x^{5} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} + 1\right ) - 3 \, a^{5} x^{5} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} - 1\right ) - 8 \, x^{5} \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right ) - 8 \,{\left (x^{5} - 1\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) -{\left (3 \, a^{3} x^{3} - 2 \, a x\right )} \sqrt{a^{2} x^{2} + 1}}{40 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x^6,x, algorithm="fricas")

[Out]

-1/40*(3*a^5*x^5*log(-a*x + sqrt(a^2*x^2 + 1) + 1) - 3*a^5*x^5*log(-a*x + sqrt(a^2*x^2 + 1) - 1) - 8*x^5*log(-
a*x + sqrt(a^2*x^2 + 1)) - 8*(x^5 - 1)*log(a*x + sqrt(a^2*x^2 + 1)) - (3*a^3*x^3 - 2*a*x)*sqrt(a^2*x^2 + 1))/x
^5

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asinh}{\left (a x \right )}}{x^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)/x**6,x)

[Out]

Integral(asinh(a*x)/x**6, x)

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Giac [A]  time = 1.42769, size = 128, normalized size = 1.66 \begin{align*} \frac{1}{80} \, a^{5}{\left (\frac{2 \,{\left (3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 5 \, \sqrt{a^{2} x^{2} + 1}\right )}}{a^{4} x^{4}} - 3 \, \log \left (\sqrt{a^{2} x^{2} + 1} + 1\right ) + 3 \, \log \left (\sqrt{a^{2} x^{2} + 1} - 1\right )\right )} - \frac{\log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x^6,x, algorithm="giac")

[Out]

1/80*a^5*(2*(3*(a^2*x^2 + 1)^(3/2) - 5*sqrt(a^2*x^2 + 1))/(a^4*x^4) - 3*log(sqrt(a^2*x^2 + 1) + 1) + 3*log(sqr
t(a^2*x^2 + 1) - 1)) - 1/5*log(a*x + sqrt(a^2*x^2 + 1))/x^5