3.3.5 \(\int \frac {1}{\sqrt {-\alpha ^2+2 h r^2}} \, dr\) [205]

Optimal. Leaf size=40 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {h} r}{\sqrt {-\alpha ^2+2 h r^2}}\right )}{\sqrt {2} \sqrt {h}} \]

[Out]

1/2*arctanh(r*2^(1/2)*h^(1/2)/(2*h*r^2-alpha^2)^(1/2))*2^(1/2)/h^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, 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 = {223, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {h} r}{\sqrt {2 h r^2-\alpha ^2}}\right )}{\sqrt {2} \sqrt {h}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-alpha^2 + 2*h*r^2],r]

[Out]

ArcTanh[(Sqrt[2]*Sqrt[h]*r)/Sqrt[-alpha^2 + 2*h*r^2]]/(Sqrt[2]*Sqrt[h])

Rule 212

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

Rule 223

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 {-\alpha ^2+2 h r^2}} \, dr &=\text {Subst}\left (\int \frac {1}{1-2 h r^2} \, dr,r,\frac {r}{\sqrt {-\alpha ^2+2 h r^2}}\right )\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {h} r}{\sqrt {-\alpha ^2+2 h r^2}}\right )}{\sqrt {2} \sqrt {h}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 40, normalized size = 1.00 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {h} r}{\sqrt {-\alpha ^2+2 h r^2}}\right )}{\sqrt {2} \sqrt {h}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-alpha^2 + 2*h*r^2],r]

[Out]

ArcTanh[(Sqrt[2]*Sqrt[h]*r)/Sqrt[-alpha^2 + 2*h*r^2]]/(Sqrt[2]*Sqrt[h])

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Maple [A]
time = 0.04, size = 33, normalized size = 0.82

method	result	size
default	\(\frac {\ln \left (\sqrt {h}\, r \sqrt {2}+\sqrt {2 h \,r^{2}-\alpha ^{2}}\right ) \sqrt {2}}{2 \sqrt {h}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*h*r^2-alpha^2)^(1/2),r,method=_RETURNVERBOSE)

[Out]

1/2*ln(h^(1/2)*r*2^(1/2)+(2*h*r^2-alpha^2)^(1/2))*2^(1/2)/h^(1/2)

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Maxima [A]
time = 1.31, size = 36, normalized size = 0.90 \begin {gather*} \frac {\sqrt {2} \log \left (4 \, h r + 2 \, \sqrt {2} \sqrt {2 \, h r^{2} - \alpha ^{2}} \sqrt {h}\right )}{2 \, \sqrt {h}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*h*r^2-alpha^2)^(1/2),r, algorithm="maxima")

[Out]

1/2*sqrt(2)*log(4*h*r + 2*sqrt(2)*sqrt(2*h*r^2 - alpha^2)*sqrt(h))/sqrt(h)

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Fricas [A]
time = 0.92, size = 85, normalized size = 2.12 \begin {gather*} \left [\frac {\sqrt {2} \log \left (4 \, h r^{2} + 2 \, \sqrt {2} \sqrt {2 \, h r^{2} - \alpha ^{2}} \sqrt {h} r - \alpha ^{2}\right )}{4 \, \sqrt {h}}, -\frac {1}{2} \, \sqrt {2} \sqrt {-\frac {1}{h}} \arctan \left (\frac {\sqrt {2} h r \sqrt {-\frac {1}{h}}}{\sqrt {2 \, h r^{2} - \alpha ^{2}}}\right )\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*h*r^2-alpha^2)^(1/2),r, algorithm="fricas")

[Out]

[1/4*sqrt(2)*log(4*h*r^2 + 2*sqrt(2)*sqrt(2*h*r^2 - alpha^2)*sqrt(h)*r - alpha^2)/sqrt(h), -1/2*sqrt(2)*sqrt(-
1/h)*arctan(sqrt(2)*h*r*sqrt(-1/h)/sqrt(2*h*r^2 - alpha^2))]

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Sympy [C] Result contains complex when optimal does not.
time = 0.49, size = 66, normalized size = 1.65 \begin {gather*} \begin {cases} \frac {\sqrt {2} \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {h} r}{\alpha } \right )}}{2 \sqrt {h}} & \text {for}\: \left |{\frac {h r^{2}}{\alpha ^{2}}}\right | > \frac {1}{2} \\- \frac {\sqrt {2} i \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {h} r}{\alpha } \right )}}{2 \sqrt {h}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*h*r**2-alpha**2)**(1/2),r)

[Out]

Piecewise((sqrt(2)*acosh(sqrt(2)*sqrt(h)*r/alpha)/(2*sqrt(h)), Abs(h*r**2/alpha**2) > 1/2), (-sqrt(2)*I*asin(s
qrt(2)*sqrt(h)*r/alpha)/(2*sqrt(h)), True))

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Giac [A]
time = 0.57, size = 55, normalized size = 1.38 \begin {gather*} \frac {\sqrt {2} \alpha ^{2} \log \left ({\left | -\sqrt {2} \sqrt {h} r + \sqrt {2 \, h r^{2} - \alpha ^{2}} \right |}\right )}{4 \, \sqrt {h}} + \frac {1}{2} \, \sqrt {2 \, h r^{2} - \alpha ^{2}} r \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*h*r^2-alpha^2)^(1/2),r, algorithm="giac")

[Out]

1/4*sqrt(2)*alpha^2*log(abs(-sqrt(2)*sqrt(h)*r + sqrt(2*h*r^2 - alpha^2)))/sqrt(h) + 1/2*sqrt(2*h*r^2 - alpha^
2)*r

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Mupad [B]
time = 0.48, size = 32, normalized size = 0.80 \begin {gather*} \frac {\sqrt {2}\,\ln \left (\sqrt {2\,h\,r^2-\alpha ^2}+\sqrt {2}\,\sqrt {h}\,r\right )}{2\,\sqrt {h}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*h*r^2 - alpha^2)^(1/2),r)

[Out]

(2^(1/2)*log((2*h*r^2 - alpha^2)^(1/2) + 2^(1/2)*h^(1/2)*r))/(2*h^(1/2))

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