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

Optimal. Leaf size=54 \[ \frac{\sqrt{\sqrt{b x^2+1}-1} \sqrt{\sqrt{b x^2+1}+1} \log \left (\cosh ^{-1}\left (\sqrt{b x^2+1}\right )\right )}{b x} \]

[Out]

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

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Rubi [A]  time = 0.108626, antiderivative size = 54, normalized size of antiderivative = 1., 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, 5674} \[ \frac{\sqrt{\sqrt{b x^2+1}-1} \sqrt{\sqrt{b x^2+1}+1} \log \left (\cosh ^{-1}\left (\sqrt{b x^2+1}\right )\right )}{b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 5674

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

Rubi steps

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

Mathematica [A]  time = 0.0894676, size = 54, normalized size = 1. \[ \frac{\sqrt{\sqrt{b x^2+1}-1} \sqrt{\sqrt{b x^2+1}+1} \log \left (\cosh ^{-1}\left (\sqrt{b x^2+1}\right )\right )}{b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.217, size = 0, normalized size = 0. \begin{align*} \int{ \left ({\rm arccosh} \left (\sqrt{b{x}^{2}+1}\right ) \right ) ^{-1}{\frac{1}{\sqrt{b{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x^{2} + 1} \operatorname{arcosh}\left (\sqrt{b x^{2} + 1}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*x^2 + 1)*arccosh(sqrt(b*x^2 + 1))), x)

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Fricas [A]  time = 2.12005, size = 80, normalized size = 1.48 \begin{align*} \frac{\sqrt{b x^{2}} \log \left (\log \left (\sqrt{b x^{2} + 1} + \sqrt{b x^{2}}\right )\right )}{b x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccosh((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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x^{2} + 1} \operatorname{acosh}{\left (\sqrt{b x^{2} + 1} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(b*x**2 + 1)*acosh(sqrt(b*x**2 + 1))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x^{2} + 1} \operatorname{arcosh}\left (\sqrt{b x^{2} + 1}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*x^2 + 1)*arccosh(sqrt(b*x^2 + 1))), x)