3.213 \(\int \frac {1}{r \sqrt {-\alpha ^2+2 h r^2-2 k r^4}} \, dr\)

Optimal. Leaf size=44 \[ -\frac {\tan ^{-1}\left (\frac {\alpha ^2-h r^2}{\alpha \sqrt {-\alpha ^2+2 h r^2-2 k r^4}}\right )}{2 \alpha } \]

[Out]

-1/2*arctan((-h*r^2+alpha^2)/alpha/(-2*k*r^4+2*h*r^2-alpha^2)^(1/2))/alpha

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Rubi [A]  time = 0.04, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1114, 724, 204} \[ -\frac {\tan ^{-1}\left (\frac {\alpha ^2-h r^2}{\alpha \sqrt {-\alpha ^2+2 h r^2-2 k r^4}}\right )}{2 \alpha } \]

Antiderivative was successfully verified.

[In]

Int[1/(r*Sqrt[-alpha^2 + 2*h*r^2 - 2*k*r^4]),r]

[Out]

-ArcTan[(alpha^2 - h*r^2)/(alpha*Sqrt[-alpha^2 + 2*h*r^2 - 2*k*r^4])]/(2*alpha)

Rule 204

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

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 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 49, normalized size = 1.11 \[ \frac {\tan ^{-1}\left (\frac {2 h r^2-2 \alpha ^2}{2 \alpha \sqrt {-\alpha ^2+2 h r^2-2 k r^4}}\right )}{2 \alpha } \]

Antiderivative was successfully verified.

[In]

Integrate[1/(r*Sqrt[-alpha^2 + 2*h*r^2 - 2*k*r^4]),r]

[Out]

ArcTan[(-2*alpha^2 + 2*h*r^2)/(2*alpha*Sqrt[-alpha^2 + 2*h*r^2 - 2*k*r^4])]/(2*alpha)

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fricas [A]  time = 0.41, size = 58, normalized size = 1.32 \[ -\frac {\arctan \left (\frac {\sqrt {-2 \, k r^{4} + 2 \, h r^{2} - \alpha ^{2}} {\left (h r^{2} - \alpha ^{2}\right )}}{2 \, \alpha k r^{4} - 2 \, \alpha h r^{2} + \alpha ^{3}}\right )}{2 \, \alpha } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/r/(-2*k*r^4+2*h*r^2-alpha^2)^(1/2),r, algorithm="fricas")

[Out]

-1/2*arctan(sqrt(-2*k*r^4 + 2*h*r^2 - alpha^2)*(h*r^2 - alpha^2)/(2*alpha*k*r^4 - 2*alpha*h*r^2 + alpha^3))/al
pha

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giac [A]  time = 1.15, size = 45, normalized size = 1.02 \[ \frac {\arctan \left (-\frac {\sqrt {2} \sqrt {-k} r^{2} - \sqrt {-2 \, k r^{4} + 2 \, h r^{2} - \alpha ^{2}}}{\alpha }\right )}{\alpha } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/r/(-2*k*r^4+2*h*r^2-alpha^2)^(1/2),r, algorithm="giac")

[Out]

arctan(-(sqrt(2)*sqrt(-k)*r^2 - sqrt(-2*k*r^4 + 2*h*r^2 - alpha^2))/alpha)/alpha

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maple [A]  time = 0.02, size = 56, normalized size = 1.27 \[ -\frac {\ln \left (\frac {2 h \,r^{2}-2 \alpha ^{2}+2 \sqrt {-\alpha ^{2}}\, \sqrt {-2 k \,r^{4}+2 h \,r^{2}-\alpha ^{2}}}{r^{2}}\right )}{2 \sqrt {-\alpha ^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/r/(-2*k*r^4+2*h*r^2-alpha^2)^(1/2),r)

[Out]

-1/2/(-alpha^2)^(1/2)*ln((-2*alpha^2+2*h*r^2+2*(-alpha^2)^(1/2)*(-2*k*r^4+2*h*r^2-alpha^2)^(1/2))/r^2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/r/(-2*k*r^4+2*h*r^2-alpha^2)^(1/2),r, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(2*alpha^2*k-h^2>0)', see `assu
me?` for more details)Is 2*alpha^2*k-h^2 positive, negative or zero?

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mupad [B]  time = 0.40, size = 54, normalized size = 1.23 \[ -\frac {\ln \left (\frac {1}{r^2}\right )+\ln \left (h\,r^2-\alpha ^2+\sqrt {-\alpha ^2}\,\sqrt {-\alpha ^2-2\,k\,r^4+2\,h\,r^2}\right )}{2\,\sqrt {-\alpha ^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(r*(2*h*r^2 - 2*k*r^4 - alpha^2)^(1/2)),r)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{r \sqrt {- \alpha ^{2} + 2 h r^{2} - 2 k r^{4}}}\, dr \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/r/(-2*k*r**4+2*h*r**2-alpha**2)**(1/2),r)

[Out]

Integral(1/(r*sqrt(-alpha**2 + 2*h*r**2 - 2*k*r**4)), r)

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