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

Optimal. Leaf size=27 \[ -a \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right )-\frac {\sinh ^{-1}(a x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5661, 266, 63, 208} \[ -a \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right )-\frac {\sinh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 27, normalized size = 1.00 \[ -a \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right )-\frac {\sinh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.47, size = 90, normalized size = 3.33 \[ -\frac {a x \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - a x \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) - {\left (x - 1\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) - x \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.13, size = 56, normalized size = 2.07 \[ -\frac {1}{2} \, a {\left (\log \left (\sqrt {a^{2} x^{2} + 1} + 1\right ) - \log \left (\sqrt {a^{2} x^{2} + 1} - 1\right )\right )} - \frac {\log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 30, normalized size = 1.11 \[ a \left (-\frac {\arcsinh \left (a x \right )}{a x}-\arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 22, normalized size = 0.81 \[ -a \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {\operatorname {arsinh}\left (a x\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*arcsinh(1/(a*abs(x))) - arcsinh(a*x)/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {\mathrm {asinh}\left (a\,x\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(a*x)/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (a x \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x)/x**2, x)

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