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

Optimal. Leaf size=27 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2-\pi }}\right )}{\sqrt {b}} \]

[Out]

arctanh(x*b^(1/2)/(b*x^2-Pi)^(1/2))/b^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {217, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2-\pi }}\right )}{\sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-Pi + b*x^2],x]

[Out]

ArcTanh[(Sqrt[b]*x)/Sqrt[-Pi + b*x^2]]/Sqrt[b]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2-\pi }}\right )}{\sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-Pi + b*x^2],x]

[Out]

ArcTanh[(Sqrt[b]*x)/Sqrt[-Pi + b*x^2]]/Sqrt[b]

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fricas [A]  time = 0.64, size = 74, normalized size = 2.74 \[ \left [\frac {\log \left (-\pi + 2 \, b x^{2} + 2 \, \sqrt {-\pi + b x^{2}} \sqrt {b} x\right )}{2 \, \sqrt {b}}, -\frac {\sqrt {-b} \arctan \left (-\frac {\sqrt {-\pi + b x^{2}} \sqrt {-b} x}{\pi - b x^{2}}\right )}{b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*log(-pi + 2*b*x^2 + 2*sqrt(-pi + b*x^2)*sqrt(b)*x)/sqrt(b), -sqrt(-b)*arctan(-sqrt(-pi + b*x^2)*sqrt(-b)*
x/(pi - b*x^2))/b]

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giac [A]  time = 1.14, size = 25, normalized size = 0.93 \[ -\frac {\log \left ({\left | -\sqrt {b} x + \sqrt {-\pi + b x^{2}} \right |}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(abs(-sqrt(b)*x + sqrt(-pi + b*x^2)))/sqrt(b)

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maple [A]  time = 0.00, size = 23, normalized size = 0.85 \[ \frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}-\pi }\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.39, size = 26, normalized size = 0.96 \[ \frac {\log \left (2 \, b x + 2 \, \sqrt {-\pi + b x^{2}} \sqrt {b}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.13, size = 22, normalized size = 0.81 \[ \frac {\ln \left (\sqrt {b\,x^2-\Pi }+\sqrt {b}\,x\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2 - Pi)^(1/2),x)

[Out]

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

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sympy [A]  time = 1.05, size = 46, normalized size = 1.70 \[ \begin {cases} \frac {\operatorname {acosh}{\left (\frac {\sqrt {b} x}{\sqrt {\pi }} \right )}}{\sqrt {b}} & \text {for}\: \frac {\left |{b x^{2}}\right |}{\pi } > 1 \\- \frac {i \operatorname {asin}{\left (\frac {\sqrt {b} x}{\sqrt {\pi }} \right )}}{\sqrt {b}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2-pi)**(1/2),x)

[Out]

Piecewise((acosh(sqrt(b)*x/sqrt(pi))/sqrt(b), Abs(b*x**2)/pi > 1), (-I*asin(sqrt(b)*x/sqrt(pi))/sqrt(b), True)
)

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