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

Optimal. Leaf size=30 \[ \frac {\sqrt {-b x^2} \log \left (\sin ^{-1}\left (\sqrt {b x^2+1}\right )\right )}{b x} \]

[Out]

ln(arcsin((b*x^2+1)^(1/2)))*(-b*x^2)^(1/2)/b/x

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Rubi [A]  time = 0.06, antiderivative size = 30, 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 = {4834, 4639} \[ \frac {\sqrt {-b x^2} \log \left (\sin ^{-1}\left (\sqrt {b x^2+1}\right )\right )}{b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4639

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

Rule 4834

Int[ArcSin[Sqrt[1 + (b_.)*(x_)^2]]^(n_.)/Sqrt[1 + (b_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[-(b*x^2)]/(b*x), Subst
[Int[ArcSin[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} \sin ^{-1}\left (\sqrt {1+b x^2}\right )} \, dx &=\frac {\sqrt {-b x^2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sin ^{-1}(x)} \, dx,x,\sqrt {1+b x^2}\right )}{b x}\\ &=\frac {\sqrt {-b x^2} \log \left (\sin ^{-1}\left (\sqrt {1+b x^2}\right )\right )}{b x}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 26, normalized size = 0.87 \[ -\frac {x \log \left (\sin ^{-1}\left (\sqrt {b x^2+1}\right )\right )}{\sqrt {-b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 28, normalized size = 0.93 \[ \frac {\sqrt {-b x^{2}} \log \left (-\arcsin \left (\sqrt {b x^{2} + 1}\right )\right )}{b x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b x^{2} + 1} \arcsin \left (\sqrt {b x^{2} + 1}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {1}{\arcsin \left (\sqrt {b \,x^{2}+1}\right ) \sqrt {b \,x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found sqrt(-_SAGE_VAR_b)

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mupad [B]  time = 0.34, size = 26, normalized size = 0.87 \[ -\frac {\ln \left (\mathrm {asin}\left (\sqrt {b\,x^2+1}\right )\right )\,\sqrt {x^2}}{\sqrt {-b}\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b x^{2} + 1} \operatorname {asin}{\left (\sqrt {b x^{2} + 1} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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