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

Optimal. Leaf size=31 \[ -\frac{\sqrt{-b x^2} \log \left (\cos ^{-1}\left (\sqrt{b x^2+1}\right )\right )}{b x} \]

[Out]

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

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Rubi [A]  time = 0.0608216, antiderivative size = 31, 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 = {4835, 4640} \[ -\frac{\sqrt{-b x^2} \log \left (\cos ^{-1}\left (\sqrt{b x^2+1}\right )\right )}{b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4835

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

Rule 4640

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

Rubi steps

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

Mathematica [A]  time = 0.0239597, size = 25, normalized size = 0.81 \[ \frac{x \log \left (\cos ^{-1}\left (\sqrt{b x^2+1}\right )\right )}{\sqrt{-b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.184, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \arccos \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/arccos((b*x^2+1)^(1/2))/(b*x^2+1)^(1/2),x)

[Out]

int(1/arccos((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} \arccos \left (\sqrt{b x^{2} + 1}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.57099, size = 68, normalized size = 2.19 \begin{align*} -\frac{\sqrt{-b x^{2}} \log \left (\arccos \left (\sqrt{b x^{2} + 1}\right )\right )}{b x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos((b*x^2+1)^(1/2))/(b*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-sqrt(-b*x^2)*log(arccos(sqrt(b*x^2 + 1)))/(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{acos}{\left (\sqrt{b x^{2} + 1} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(b*x**2 + 1)*acos(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} \arccos \left (\sqrt{b x^{2} + 1}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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