3.3.7 \(\int \frac {1}{r \sqrt {-\alpha ^2-2 k r+2 h r^2}} \, dr\) [207]

Optimal. Leaf size=37 \[ -\frac {\tan ^{-1}\left (\frac {\alpha ^2+k r}{\alpha \sqrt {-\alpha ^2-2 k r+2 h r^2}}\right )}{\alpha } \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {738, 210} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\alpha ^2+k r}{\alpha \sqrt {-\alpha ^2+2 h r^2-2 k r}}\right )}{\alpha } \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(r*Sqrt[-alpha^2 - 2*k*r + 2*h*r^2]),r]

[Out]

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

Rule 210

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

Rule 738

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]

Rubi steps

\begin {align*} \int \frac {1}{r \sqrt {-\alpha ^2-2 k r+2 h r^2}} \, dr &=-\left (2 \text {Subst}\left (\int \frac {1}{-4 \alpha ^2-r^2} \, dr,r,\frac {-2 \alpha ^2-2 k r}{\sqrt {-\alpha ^2-2 k r+2 h r^2}}\right )\right )\\ &=-\frac {\tan ^{-1}\left (\frac {\alpha ^2+k r}{\alpha \sqrt {-\alpha ^2-2 k r+2 h r^2}}\right )}{\alpha }\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 45, normalized size = 1.22 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {h} r-\sqrt {-\alpha ^2-2 k r+2 h r^2}}{\alpha }\right )}{\alpha } \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[1/(r*Sqrt[2*h*r^2-alpha^2-2*k*r]),r]')

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

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Maple [A]
time = 0.12, size = 52, normalized size = 1.41

method result size
default \(-\frac {\ln \left (\frac {-2 \alpha ^{2}-2 k r +2 \sqrt {-\alpha ^{2}}\, \sqrt {2 h \,r^{2}-\alpha ^{2}-2 k r}}{r}\right )}{\sqrt {-\alpha ^{2}}}\) \(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.38, size = 40, normalized size = 1.08 \begin {gather*} -\frac {\arcsin \left (\frac {k}{\sqrt {2 \, \alpha ^{2} h + k^{2}}} + \frac {\alpha ^{2}}{\sqrt {2 \, \alpha ^{2} h + k^{2}} r}\right )}{\alpha } \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(k/sqrt(2*alpha^2*h + k^2) + alpha^2/(sqrt(2*alpha^2*h + k^2)*r))/alpha

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Fricas [A]
time = 0.32, size = 52, normalized size = 1.41 \begin {gather*} -\frac {\arctan \left (\frac {\sqrt {2 \, h r^{2} - \alpha ^{2} - 2 \, k r} {\left (\alpha ^{2} + k r\right )}}{2 \, \alpha h r^{2} - \alpha ^{3} - 2 \, \alpha k r}\right )}{\alpha } \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(sqrt(2*h*r^2 - alpha^2 - 2*k*r)*(alpha^2 + k*r)/(2*alpha*h*r^2 - alpha^3 - 2*alpha*k*r))/alpha

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{r \sqrt {- \alpha ^{2} + 2 h r^{2} - 2 k r}}\, dr \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(r*sqrt(-alpha**2 + 2*h*r**2 - 2*k*r)), r)

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Giac [A]
time = 0.01, size = 42, normalized size = 1.14 \begin {gather*} \frac {\frac {1}{2}\cdot 4 \arctan \left (\frac {\sqrt {-\alpha ^{2}+2 h r^{2}-2 k r}-\sqrt {2 h} r}{\alpha }\right )}{\alpha } \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(-(sqrt(2)*sqrt(h)*r - sqrt(2*h*r^2 - alpha^2 - 2*k*r))/alpha)/alpha

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Mupad [B]
time = 0.10, size = 51, normalized size = 1.38 \begin {gather*} -\frac {\ln \left (\frac {\sqrt {-\alpha ^2}\,\sqrt {-\alpha ^2+2\,h\,r^2-2\,k\,r}}{r}-\frac {\alpha ^2}{r}-k\right )}{\sqrt {-\alpha ^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(r*(2*h*r^2 - 2*k*r - alpha^2)^(1/2)),r)

[Out]

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

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