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

Optimal. Leaf size=62 \[ \frac {\sqrt {\sqrt {b x^2+1}-1} \sqrt {\sqrt {b x^2+1}+1} \cosh ^{-1}\left (\sqrt {b x^2+1}\right )^{n+1}}{b (n+1) x} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 62, 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 = {5895, 5676} \[ \frac {\sqrt {\sqrt {b x^2+1}-1} \sqrt {\sqrt {b x^2+1}+1} \cosh ^{-1}\left (\sqrt {b x^2+1}\right )^{n+1}}{b (n+1) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5676

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]), x_Symbol]
:> Simp[(a + b*ArcCosh[c*x])^(n + 1)/(b*c*Sqrt[-(d1*d2)]*(n + 1)), x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n},
x] && EqQ[e1, c*d1] && EqQ[e2, -(c*d2)] && GtQ[d1, 0] && LtQ[d2, 0] && NeQ[n, -1]

Rule 5895

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 62, normalized size = 1.00 \[ \frac {\sqrt {\sqrt {b x^2+1}-1} \sqrt {\sqrt {b x^2+1}+1} \cosh ^{-1}\left (\sqrt {b x^2+1}\right )^{n+1}}{b (n+1) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.65, size = 108, normalized size = 1.74 \[ \frac {\sqrt {b x^{2}} \cosh \left (n \log \left (\log \left (\sqrt {b x^{2} + 1} + \sqrt {b x^{2}}\right )\right )\right ) \log \left (\sqrt {b x^{2} + 1} + \sqrt {b x^{2}}\right ) + \sqrt {b x^{2}} \log \left (\sqrt {b x^{2} + 1} + \sqrt {b x^{2}}\right ) \sinh \left (n \log \left (\log \left (\sqrt {b x^{2} + 1} + \sqrt {b x^{2}}\right )\right )\right )}{{\left (b n + b\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh((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(arccosh((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 = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {arccosh}\left (\sqrt {b \,x^{2}+1}\right )^{n}}{\sqrt {b \,x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh((b*x^2+1)^(1/2))^n/(b*x^2+1)^(1/2),x)

[Out]

int(arccosh((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 \[ \int \frac {\operatorname {arcosh}\left (\sqrt {b x^{2} + 1}\right )^{n}}{\sqrt {b x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccosh(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 = -0.02 \[ \int \frac {{\mathrm {acosh}\left (\sqrt {b\,x^2+1}\right )}^n}{\sqrt {b\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh((b*x^2 + 1)^(1/2))^n/(b*x^2 + 1)^(1/2),x)

[Out]

int(acosh((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 \[ \begin {cases} \tilde {\infty } x & \text {for}\: b = 0 \wedge n = -1 \\0^{n} x & \text {for}\: b = 0 \\\int \frac {1}{\sqrt {b x^{2} + 1} \operatorname {acosh}{\left (\sqrt {b x^{2} + 1} \right )}}\, dx & \text {for}\: n = -1 \\\frac {\sqrt {b} \sqrt {x^{2}} \operatorname {acosh}{\left (\sqrt {b x^{2} + 1} \right )} \operatorname {acosh}^{n}{\left (\sqrt {b x^{2} + 1} \right )}}{b n x + b x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acosh((b*x**2+1)**(1/2))**n/(b*x**2+1)**(1/2),x)

[Out]

Piecewise((zoo*x, Eq(b, 0) & Eq(n, -1)), (0**n*x, Eq(b, 0)), (Integral(1/(sqrt(b*x**2 + 1)*acosh(sqrt(b*x**2 +
1))), x), Eq(n, -1)), (sqrt(b)*sqrt(x**2)*acosh(sqrt(b*x**2 + 1))*acosh(sqrt(b*x**2 + 1))**n/(b*n*x + b*x), T
rue))

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