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

Optimal. Leaf size=32 \[ -\frac{\tanh ^{-1}\left (\frac{x+4}{2 \sqrt{2} \sqrt{-x^2+x+2}}\right )}{\sqrt{2}} \]

[Out]

-(ArcTanh[(4 + x)/(2*Sqrt[2]*Sqrt[2 + x - x^2])]/Sqrt[2])

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Rubi [A]  time = 0.0105747, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {724, 206} \[ -\frac{\tanh ^{-1}\left (\frac{x+4}{2 \sqrt{2} \sqrt{-x^2+x+2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[2 + x - x^2]),x]

[Out]

-(ArcTanh[(4 + x)/(2*Sqrt[2]*Sqrt[2 + x - x^2])]/Sqrt[2])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

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])

Rubi steps

\begin{align*} \int \frac{1}{x \sqrt{2+x-x^2}} \, dx &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{8-x^2} \, dx,x,\frac{4+x}{\sqrt{2+x-x^2}}\right )\right )\\ &=-\frac{\tanh ^{-1}\left (\frac{4+x}{2 \sqrt{2} \sqrt{2+x-x^2}}\right )}{\sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0055365, size = 32, normalized size = 1. \[ -\frac{\tanh ^{-1}\left (\frac{x+4}{2 \sqrt{2} \sqrt{-x^2+x+2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[2 + x - x^2]),x]

[Out]

-(ArcTanh[(4 + x)/(2*Sqrt[2]*Sqrt[2 + x - x^2])]/Sqrt[2])

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Maple [A]  time = 0.003, size = 25, normalized size = 0.8 \begin{align*} -{\frac{\sqrt{2}}{2}{\it Artanh} \left ({\frac{ \left ( 4+x \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{-{x}^{2}+x+2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(-x^2+x+2)^(1/2),x)

[Out]

-1/2*arctanh(1/4*(4+x)*2^(1/2)/(-x^2+x+2)^(1/2))*2^(1/2)

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Maxima [A]  time = 1.44071, size = 45, normalized size = 1.41 \begin{align*} -\frac{1}{2} \, \sqrt{2} \log \left (\frac{2 \, \sqrt{2} \sqrt{-x^{2} + x + 2}}{{\left | x \right |}} + \frac{4}{{\left | x \right |}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(-x^2+x+2)^(1/2),x, algorithm="maxima")

[Out]

-1/2*sqrt(2)*log(2*sqrt(2)*sqrt(-x^2 + x + 2)/abs(x) + 4/abs(x) + 1)

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Fricas [A]  time = 1.9215, size = 111, normalized size = 3.47 \begin{align*} \frac{1}{4} \, \sqrt{2} \log \left (-\frac{4 \, \sqrt{2} \sqrt{-x^{2} + x + 2}{\left (x + 4\right )} + 7 \, x^{2} - 16 \, x - 32}{x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(-x^2+x+2)^(1/2),x, algorithm="fricas")

[Out]

1/4*sqrt(2)*log(-(4*sqrt(2)*sqrt(-x^2 + x + 2)*(x + 4) + 7*x^2 - 16*x - 32)/x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(-x**2+x+2)**(1/2),x)

[Out]

Integral(1/(x*sqrt(-(x - 2)*(x + 1))), x)

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Giac [B]  time = 1.10022, size = 96, normalized size = 3. \begin{align*} -\frac{1}{2} \, \sqrt{2} \log \left (\frac{{\left | -4 \, \sqrt{2} + \frac{2 \,{\left (2 \, \sqrt{-x^{2} + x + 2} - 3\right )}}{2 \, x - 1} - 6 \right |}}{{\left | 4 \, \sqrt{2} + \frac{2 \,{\left (2 \, \sqrt{-x^{2} + x + 2} - 3\right )}}{2 \, x - 1} - 6 \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(-x^2+x+2)^(1/2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*log(abs(-4*sqrt(2) + 2*(2*sqrt(-x^2 + x + 2) - 3)/(2*x - 1) - 6)/abs(4*sqrt(2) + 2*(2*sqrt(-x^2 +
x + 2) - 3)/(2*x - 1) - 6))