3.289 \(\int \frac {\sinh ^{-1}(a x^2)}{x^3} \, dx\)

Optimal. Leaf size=33 \[ -\frac {1}{2} a \tanh ^{-1}\left (\sqrt {a^2 x^4+1}\right )-\frac {\sinh ^{-1}\left (a x^2\right )}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5902, 12, 266, 63, 208} \[ -\frac {1}{2} a \tanh ^{-1}\left (\sqrt {a^2 x^4+1}\right )-\frac {\sinh ^{-1}\left (a x^2\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x^2]/x^3,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

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 5902

Int[((a_.) + ArcSinh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin
h[u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 + u^2],
x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)
^(m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.00 \[ -\frac {1}{2} a \tanh ^{-1}\left (\sqrt {a^2 x^4+1}\right )-\frac {\sinh ^{-1}\left (a x^2\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x^2]/x^3,x]

[Out]

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

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fricas [B]  time = 1.05, size = 106, normalized size = 3.21 \[ -\frac {a x^{2} \log \left (-a x^{2} + \sqrt {a^{2} x^{4} + 1} + 1\right ) - a x^{2} \log \left (-a x^{2} + \sqrt {a^{2} x^{4} + 1} - 1\right ) - x^{2} \log \left (-a x^{2} + \sqrt {a^{2} x^{4} + 1}\right ) - {\left (x^{2} - 1\right )} \log \left (a x^{2} + \sqrt {a^{2} x^{4} + 1}\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x^2)/x^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.30, size = 58, normalized size = 1.76 \[ -\frac {1}{4} \, a {\left (\log \left (\sqrt {a^{2} x^{4} + 1} + 1\right ) - \log \left (\sqrt {a^{2} x^{4} + 1} - 1\right )\right )} - \frac {\log \left (a x^{2} + \sqrt {a^{2} x^{4} + 1}\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x^2)/x^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 28, normalized size = 0.85 \[ -\frac {\arcsinh \left (a \,x^{2}\right )}{2 x^{2}}-\frac {a \arctanh \left (\frac {1}{\sqrt {a^{2} x^{4}+1}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x^2)/x^3,x)

[Out]

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

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maxima [A]  time = 0.37, size = 46, normalized size = 1.39 \[ -\frac {1}{4} \, a {\left (\log \left (\sqrt {a^{2} x^{4} + 1} + 1\right ) - \log \left (\sqrt {a^{2} x^{4} + 1} - 1\right )\right )} - \frac {\operatorname {arsinh}\left (a x^{2}\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x^2)/x^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {asinh}\left (a\,x^2\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a*x^2)/x^3,x)

[Out]

int(asinh(a*x^2)/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (a x^{2} \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x**2)/x**3,x)

[Out]

Integral(asinh(a*x**2)/x**3, x)

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