3.3.13 \(\int \frac {1}{r \sqrt {-\alpha ^2+2 h r^2-2 k r^4}} \, dr\) [213]

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 = 0.03, 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 = {1128, 738, 210} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\alpha ^2-h r^2}{\alpha \sqrt {-\alpha ^2+2 h r^2-2 k r^4}}\right )}{2 \alpha } \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1128

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} \text {Subst}\left (\int \frac {1}{r \sqrt {-\alpha ^2+2 h r-2 k r^2}} \, dr,r,r^2\right )\\ &=-\text {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 = 0.08, size = 53, normalized size = 1.20 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {-k} r^2}{\alpha }-\frac {\sqrt {-\alpha ^2+2 h r^2-2 k r^4}}{\alpha }\right )}{\alpha } \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.03, size = 56, normalized size = 1.27

method	result	size
default	\(-\frac {\ln \left (\frac {-2 \alpha ^{2}+2 h \,r^{2}+2 \sqrt {-\alpha ^{2}}\, \sqrt {-2 k \,r^{4}+2 h \,r^{2}-\alpha ^{2}}}{r^{2}}\right )}{2 \sqrt {-\alpha ^{2}}}\)	\(56\)
elliptic	\(-\frac {\ln \left (\frac {-2 \alpha ^{2}+2 h \,r^{2}+2 \sqrt {-\alpha ^{2}}\, \sqrt {-2 k \,r^{4}+2 h \,r^{2}-\alpha ^{2}}}{r^{2}}\right )}{2 \sqrt {-\alpha ^{2}}}\)	\(56\)

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,method=_RETURNVERBOSE)

[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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 mor

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Fricas [A]
time = 0.58, size = 58, normalized size = 1.32 \begin {gather*} -\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 } \end {gather*}

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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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^{4}}}\, dr \end {gather*}

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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Giac [A]
time = 0.47, size = 45, normalized size = 1.02 \begin {gather*} \frac {\arctan \left (-\frac {\sqrt {2} \sqrt {-k} r^{2} - \sqrt {-2 \, k r^{4} + 2 \, h r^{2} - \alpha ^{2}}}{\alpha }\right )}{\alpha } \end {gather*}

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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Mupad [B]
time = 0.40, size = 54, normalized size = 1.23 \begin {gather*} -\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}} \end {gather*}

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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