3.370 \(\int \frac {\sinh ^{-1}(\sqrt {-1+b x^2})^n}{\sqrt {-1+b x^2}} \, dx\)

Optimal. Leaf size=37 \[ \frac {\sqrt {b x^2} \sinh ^{-1}\left (\sqrt {b x^2-1}\right )^{n+1}}{b (n+1) x} \]

[Out]

arcsinh((b*x^2-1)^(1/2))^(1+n)*(b*x^2)^(1/2)/b/(1+n)/x

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Rubi [A]  time = 0.07, antiderivative size = 37, 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 = {5894, 5675} \[ \frac {\sqrt {b x^2} \sinh ^{-1}\left (\sqrt {b x^2-1}\right )^{n+1}}{b (n+1) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rule 5894

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

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Mathematica [A]  time = 0.04, size = 37, normalized size = 1.00 \[ \frac {\sqrt {b x^2} \sinh ^{-1}\left (\sqrt {b x^2-1}\right )^{n+1}}{b (n+1) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.57, size = 108, normalized size = 2.92 \[ \frac {\sqrt {b x^{2}} \cosh \left (n \log \left (\log \left (\sqrt {b x^{2} - 1} + \sqrt {b x^{2}}\right )\right )\right ) \log \left (\sqrt {b x^{2} - 1} + \sqrt {b x^{2}}\right ) + \sqrt {b x^{2}} \log \left (\sqrt {b x^{2} - 1} + \sqrt {b x^{2}}\right ) \sinh \left (n \log \left (\log \left (\sqrt {b x^{2} - 1} + \sqrt {b x^{2}}\right )\right )\right )}{{\left (b n + b\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 0.25, size = 0, normalized size = 0.00 \[ \int \frac {\arcsinh \left (\sqrt {b \,x^{2}-1}\right )^{n}}{\sqrt {b \,x^{2}-1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (\sqrt {b x^{2} - 1}\right )^{n}}{\sqrt {b x^{2} - 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} - \frac {2 x}{\pi } & \text {for}\: b = 0 \wedge n = -1 \\- i x \left (\frac {i \pi }{2}\right )^{n} & \text {for}\: b = 0 \\\int \frac {1}{\sqrt {b x^{2} - 1} \operatorname {asinh}{\left (\sqrt {b x^{2} - 1} \right )}}\, dx & \text {for}\: n = -1 \\\frac {\sqrt {b} \sqrt {x^{2}} \operatorname {asinh}{\left (\sqrt {b x^{2} - 1} \right )} \operatorname {asinh}^{n}{\left (\sqrt {b x^{2} - 1} \right )}}{b n x + b x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*x/pi, Eq(b, 0) & Eq(n, -1)), (-I*x*(I*pi/2)**n, Eq(b, 0)), (Integral(1/(sqrt(b*x**2 - 1)*asinh(s
qrt(b*x**2 - 1))), x), Eq(n, -1)), (sqrt(b)*sqrt(x**2)*asinh(sqrt(b*x**2 - 1))*asinh(sqrt(b*x**2 - 1))**n/(b*n
*x + b*x), True))

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