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

Optimal. Leaf size=28 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {-b x^2-\pi }}\right )}{\sqrt {b}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {217, 203} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {-b x^2-\pi }}\right )}{\sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-Pi - b*x^2],x]

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 {\tan ^{-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 = 28, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {-b x^2-\pi }}\right )}{\sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-Pi - b*x^2],x]

[Out]

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

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fricas [A]  time = 0.99, size = 74, normalized size = 2.64 \[ \left [-\frac {\sqrt {-b} \log \left (-\pi - 2 \, b x^{2} + 2 \, \sqrt {-\pi - b x^{2}} \sqrt {-b} x\right )}{2 \, b}, -\frac {\arctan \left (\frac {\sqrt {-\pi - b x^{2}} \sqrt {b} x}{\pi + b x^{2}}\right )}{\sqrt {b}}\right ] \]

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

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giac [A]  time = 1.16, size = 30, normalized size = 1.07 \[ -\frac {\log \left ({\left | -\sqrt {-b} x + \sqrt {-\pi - b x^{2}} \right |}\right )}{\sqrt {-b}} \]

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 = 0.82 \[ \frac {\arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}-\pi }}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 1.35, size = 14, normalized size = 0.50 \[ -\frac {i \, \operatorname {arsinh}\left (\frac {b x}{\sqrt {\pi b}}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*arcsinh(b*x/sqrt(pi*b))/sqrt(b)

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mupad [B]  time = 5.09, size = 27, normalized size = 0.96 \[ \frac {\ln \left (\sqrt {-b\,x^2-\Pi }+\sqrt {-b}\,x\right )}{\sqrt {-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(- Pi - b*x^2)^(1/2),x)

[Out]

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

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sympy [C]  time = 1.00, size = 20, normalized size = 0.71 \[ - \frac {i \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {\pi }} \right )}}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*asinh(sqrt(b)*x/sqrt(pi))/sqrt(b)

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