3.1.98 \(\int \frac {1}{x \sqrt {-2+x^2}} \, dx\) [98]

Optimal. Leaf size=22 \[ \frac {\arctan \left (\frac {\sqrt {-2+x^2}}{\sqrt {2}}\right )}{\sqrt {2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {272, 65, 209} \begin {gather*} \frac {\arctan \left (\frac {\sqrt {x^2-2}}{\sqrt {2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[-2 + x^2]),x]

[Out]

ArcTan[Sqrt[-2 + x^2]/Sqrt[2]]/Sqrt[2]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {gather*} \begin {aligned} \text {Integral} &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {-2+x} x} \, dx,x,x^2\right )\\ &=\text {Subst}\left (\int \frac {1}{2+x^2} \, dx,x,\sqrt {-2+x^2}\right )\\ &=\frac {\arctan \left (\frac {\sqrt {-2+x^2}}{\sqrt {2}}\right )}{\sqrt {2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {-2+x^2}}{\sqrt {2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[-2 + x^2]),x]

[Out]

ArcTan[Sqrt[-2 + x^2]/Sqrt[2]]/Sqrt[2]

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Maple [A]
time = 0.14, size = 18, normalized size = 0.82

method	result	size
default	\(-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}}{\sqrt {x^{2}-2}}\right )}{2}\)	\(18\)
pseudoelliptic	\(\frac {\arctan \left (\frac {\sqrt {x^{2}-2}\, \sqrt {2}}{2}\right ) \sqrt {2}}{2}\)	\(19\)
trager	\(\frac {\mathit {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\mathit {RootOf}\left (\textit {\_Z}^{2}+2\right )+\sqrt {x^{2}-2}}{x}\right )}{2}\)	\(28\)
meijerg	\(\frac {\sqrt {2}\, \sqrt {-\mathrm {signum}\left (-1+\frac {x^{2}}{2}\right )}\, \left (\left (-3 \ln \left (2\right )+2 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1-\frac {x^{2}}{2}}}{2}\right )\right )}{4 \sqrt {\pi }\, \sqrt {\mathrm {signum}\left (-1+\frac {x^{2}}{2}\right )}}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(x^2-2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.42, size = 14, normalized size = 0.64 \begin {gather*} -\frac {1}{2} \, \sqrt {2} \arcsin \left (\frac {\sqrt {2}}{{\left | x \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*arcsin(sqrt(2)/abs(x))

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Fricas [A]
time = 0.57, size = 24, normalized size = 1.09 \begin {gather*} \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} x + \frac {1}{2} \, \sqrt {2} \sqrt {x^{2} - 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*arctan(-1/2*sqrt(2)*x + 1/2*sqrt(2)*sqrt(x^2 - 2))

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Sympy [C] Result contains complex when optimal does not.
time = 0.62, size = 39, normalized size = 1.77 \begin {gather*} \begin {cases} \frac {\sqrt {2} i \operatorname {acosh}{\left (\frac {\sqrt {2}}{x} \right )}}{2} & \text {for}\: \frac {1}{\left |{x^{2}}\right |} > \frac {1}{2} \\- \frac {\sqrt {2} \operatorname {asin}{\left (\frac {\sqrt {2}}{x} \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.46, size = 18, normalized size = 0.82 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {x^{2} - 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(x^2 - 2))

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Mupad [B]
time = 0.12, size = 18, normalized size = 0.82 \begin {gather*} \frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {x^2-2}}{2}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(x^2 - 2)^(1/2)),x)

[Out]

(2^(1/2)*atan((2^(1/2)*(x^2 - 2)^(1/2))/2))/2

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Chatgpt [F] Failed to verify
time = 1.00, size = 14, normalized size = 0.64 \begin {gather*} -\frac {\sqrt {2}\, \sqrt {-x^{2}+3}}{2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*2^(1/2)*(-x^2+3)^(1/2)

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