3.4.71 \(\int \frac {1}{\sqrt {-1+b x^2} \sinh ^{-1}(\sqrt {-1+b x^2})} \, dx\) [371]

Optimal. Leaf size=29 \[ \frac {\sqrt {b x^2} \log \left (\sinh ^{-1}\left (\sqrt {-1+b x^2}\right )\right )}{b x} \]

[Out]

ln(arcsinh((b*x^2-1)^(1/2)))*(b*x^2)^(1/2)/b/x

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Rubi [A]
time = 0.04, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {5871, 5782} \begin {gather*} \frac {\sqrt {b x^2} \log \left (\sinh ^{-1}\left (\sqrt {b x^2-1}\right )\right )}{b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[-1 + b*x^2]*ArcSinh[Sqrt[-1 + b*x^2]]),x]

[Out]

(Sqrt[b*x^2]*Log[ArcSinh[Sqrt[-1 + b*x^2]]])/(b*x)

Rule 5782

Int[1/(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Simp[(1/(b*c))*Simp[Sqrt[1
+ c^2*x^2]/Sqrt[d + e*x^2]]*Log[a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]

Rule 5871

Int[ArcSinh[Sqrt[-1 + (b_.)*(x_)^2]]^(n_.)/Sqrt[-1 + (b_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[b*x^2]/(b*x), Subst
[Int[ArcSinh[x]^n/Sqrt[1 + x^2], x], x, Sqrt[-1 + b*x^2]], x] /; FreeQ[{b, n}, x]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 24, normalized size = 0.83 \begin {gather*} \frac {x \log \left (\sinh ^{-1}\left (\sqrt {-1+b x^2}\right )\right )}{\sqrt {b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[-1 + b*x^2]*ArcSinh[Sqrt[-1 + b*x^2]]),x]

[Out]

(x*Log[ArcSinh[Sqrt[-1 + b*x^2]]])/Sqrt[b*x^2]

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {1}{\arcsinh \left (\sqrt {x^{2} b -1}\right ) \sqrt {x^{2} b -1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arcsinh((b*x^2-1)^(1/2))/(b*x^2-1)^(1/2),x)

[Out]

int(1/arcsinh((b*x^2-1)^(1/2))/(b*x^2-1)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsinh((b*x^2-1)^(1/2))/(b*x^2-1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x^2 - 1)*arcsinh(sqrt(b*x^2 - 1))), x)

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Fricas [A]
time = 0.35, size = 33, normalized size = 1.14 \begin {gather*} \frac {\sqrt {b x^{2}} \log \left (\log \left (\sqrt {b x^{2} - 1} + \sqrt {b x^{2}}\right )\right )}{b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsinh((b*x^2-1)^(1/2))/(b*x^2-1)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b x^{2} - 1} \operatorname {asinh}{\left (\sqrt {b x^{2} - 1} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/asinh((b*x**2-1)**(1/2))/(b*x**2-1)**(1/2),x)

[Out]

Integral(1/(sqrt(b*x**2 - 1)*asinh(sqrt(b*x**2 - 1))), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsinh((b*x^2-1)^(1/2))/(b*x^2-1)^(1/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 0.25, size = 23, normalized size = 0.79 \begin {gather*} \frac {\ln \left (\mathrm {asinh}\left (\sqrt {b\,x^2-1}\right )\right )\,\sqrt {x^2}}{\sqrt {b}\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(asinh((b*x^2 - 1)^(1/2))*(b*x^2 - 1)^(1/2)),x)

[Out]

(log(asinh((b*x^2 - 1)^(1/2)))*(x^2)^(1/2))/(b^(1/2)*x)

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