3.211 \(\int \frac {r}{\sqrt {-\alpha ^2+2 e r^2-2 k r^4}} \, dr\)

Optimal. Leaf size=56 \[ -\frac {\tan ^{-1}\left (\frac {e-2 k r^2}{\sqrt {2} \sqrt {k} \sqrt {-\alpha ^2+2 e r^2-2 k r^4}}\right )}{2 \sqrt {2} \sqrt {k}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1107, 621, 204} \[ -\frac {\tan ^{-1}\left (\frac {e-2 k r^2}{\sqrt {2} \sqrt {k} \sqrt {-\alpha ^2+2 e r^2-2 k r^4}}\right )}{2 \sqrt {2} \sqrt {k}} \]

Antiderivative was successfully verified.

[In]

Int[r/Sqrt[-alpha^2 + 2*e*r^2 - 2*k*r^4],r]

[Out]

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

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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1107

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

Rubi steps

\begin {align*} \int \frac {r}{\sqrt {-\alpha ^2+2 e r^2-2 k r^4}} \, dr &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-\alpha ^2+2 e r-2 k r^2}} \, dr,r,r^2\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{-8 k-r^2} \, dr,r,\frac {2 \left (e-2 k r^2\right )}{\sqrt {-\alpha ^2+2 e r^2-2 k r^4}}\right )\\ &=-\frac {\tan ^{-1}\left (\frac {e-2 k r^2}{\sqrt {2} \sqrt {k} \sqrt {-\alpha ^2+2 e r^2-2 k r^4}}\right )}{2 \sqrt {2} \sqrt {k}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 56, normalized size = 1.00 \[ -\frac {\tan ^{-1}\left (\frac {e-2 k r^2}{\sqrt {2} \sqrt {k} \sqrt {-\alpha ^2+2 e r^2-2 k r^4}}\right )}{2 \sqrt {2} \sqrt {k}} \]

Antiderivative was successfully verified.

[In]

Integrate[r/Sqrt[-alpha^2 + 2*e*r^2 - 2*k*r^4],r]

[Out]

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

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fricas [A]  time = 0.42, size = 152, normalized size = 2.71 \[ \left [-\frac {\sqrt {2} \sqrt {-k} \log \left (-8 \, k^{2} r^{4} + 8 \, e k r^{2} - 2 \, \alpha ^{2} k + 2 \, \sqrt {2} \sqrt {-2 \, k r^{4} + 2 \, e r^{2} - \alpha ^{2}} {\left (2 \, k r^{2} - e\right )} \sqrt {-k} - e^{2}\right )}{8 \, k}, -\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-2 \, k r^{4} + 2 \, e r^{2} - \alpha ^{2}} {\left (2 \, k r^{2} - e\right )} \sqrt {k}}{2 \, {\left (2 \, k^{2} r^{4} - 2 \, e k r^{2} + \alpha ^{2} k\right )}}\right )}{4 \, \sqrt {k}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*sqrt(2)*sqrt(-k)*log(-8*k^2*r^4 + 8*e*k*r^2 - 2*alpha^2*k + 2*sqrt(2)*sqrt(-2*k*r^4 + 2*e*r^2 - alpha^2)
*(2*k*r^2 - e)*sqrt(-k) - e^2)/k, -1/4*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-2*k*r^4 + 2*e*r^2 - alpha^2)*(2*k*r^2
- e)*sqrt(k)/(2*k^2*r^4 - 2*e*k*r^2 + alpha^2*k))/sqrt(k)]

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giac [A]  time = 1.22, size = 60, normalized size = 1.07 \[ -\frac {\sqrt {2} \log \left ({\left | \sqrt {2} {\left (\sqrt {2} \sqrt {-k} r^{2} - \sqrt {-2 \, k r^{4} + 2 \, r^{2} e - \alpha ^{2}}\right )} \sqrt {-k} + e \right |}\right )}{4 \, \sqrt {-k}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 47, normalized size = 0.84 \[ \frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (r^{2}-\frac {e}{2 k}\right ) \sqrt {k}}{\sqrt {-2 k \,r^{4}+2 e \,r^{2}-\alpha ^{2}}}\right )}{4 \sqrt {k}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(r/(-2*k*r^4+2*e*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-e^2>0)', see `assu
me?` for more details)Is 2*alpha^2*k-e^2 zero or nonzero?

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mupad [B]  time = 0.98, size = 50, normalized size = 0.89 \[ \frac {\sqrt {2}\,\ln \left (\sqrt {-\alpha ^2-2\,k\,r^4+2\,e\,r^2}+\frac {\sqrt {2}\,\left (e-2\,k\,r^2\right )}{2\,\sqrt {-k}}\right )}{4\,\sqrt {-k}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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