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

Optimal. Leaf size=81 \[ \frac{\sqrt{-\alpha ^2+2 e r^2-2 k r}}{2 e}-\frac{k \tanh ^{-1}\left (\frac{k-2 e r}{\sqrt{2} \sqrt{e} \sqrt{-\alpha ^2+2 e r^2-2 k r}}\right )}{2 \sqrt{2} e^{3/2}} \]

[Out]

Sqrt[-alpha^2 - 2*k*r + 2*e*r^2]/(2*e) - (k*ArcTanh[(k - 2*e*r)/(Sqrt[2]*Sqrt[e]*Sqrt[-alpha^2 - 2*k*r + 2*e*r
^2])])/(2*Sqrt[2]*e^(3/2))

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Rubi [A]  time = 0.0275925, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {640, 621, 206} \[ \frac{\sqrt{-\alpha ^2+2 e r^2-2 k r}}{2 e}-\frac{k \tanh ^{-1}\left (\frac{k-2 e r}{\sqrt{2} \sqrt{e} \sqrt{-\alpha ^2+2 e r^2-2 k r}}\right )}{2 \sqrt{2} e^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[-alpha^2 - 2*k*r + 2*e*r^2]/(2*e) - (k*ArcTanh[(k - 2*e*r)/(Sqrt[2]*Sqrt[e]*Sqrt[-alpha^2 - 2*k*r + 2*e*r
^2])])/(2*Sqrt[2]*e^(3/2))

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

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 206

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

Rubi steps

\begin{align*} \int \frac{r}{\sqrt{-\alpha ^2-2 k r+2 e r^2}} \, dr &=\frac{\sqrt{-\alpha ^2-2 k r+2 e r^2}}{2 e}+\frac{k \int \frac{1}{\sqrt{-\alpha ^2-2 k r+2 e r^2}} \, dr}{2 e}\\ &=\frac{\sqrt{-\alpha ^2-2 k r+2 e r^2}}{2 e}+\frac{k \operatorname{Subst}\left (\int \frac{1}{8 e-r^2} \, dr,r,\frac{-2 k+4 e r}{\sqrt{-\alpha ^2-2 k r+2 e r^2}}\right )}{e}\\ &=\frac{\sqrt{-\alpha ^2-2 k r+2 e r^2}}{2 e}-\frac{k \tanh ^{-1}\left (\frac{k-2 e r}{\sqrt{2} \sqrt{e} \sqrt{-\alpha ^2-2 k r+2 e r^2}}\right )}{2 \sqrt{2} e^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.100761, size = 82, normalized size = 1.01 \[ \frac{1}{4} \left (\frac{\sqrt{2} k \tanh ^{-1}\left (\frac{2 e r-k}{\sqrt{2} \sqrt{e} \sqrt{2 r (e r-k)-\alpha ^2}}\right )}{e^{3/2}}+\frac{2 \sqrt{2 r (e r-k)-\alpha ^2}}{e}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[-alpha^2 + 2*r*(-k + e*r)])/e + (Sqrt[2]*k*ArcTanh[(-k + 2*e*r)/(Sqrt[2]*Sqrt[e]*Sqrt[-alpha^2 + 2*r*
(-k + e*r)])])/e^(3/2))/4

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Maple [A]  time = 0.005, size = 70, normalized size = 0.9 \begin{align*}{\frac{1}{2\,e}\sqrt{2\,e{r}^{2}-{\alpha }^{2}-2\,kr}}+{\frac{k\sqrt{2}}{4}\ln \left ({\frac{ \left ( 2\,er-k \right ) \sqrt{2}}{2}{\frac{1}{\sqrt{e}}}}+\sqrt{2\,e{r}^{2}-{\alpha }^{2}-2\,kr} \right ){e}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*e*r^2-alpha^2-2*k*r)^(1/2)/e+1/4*k/e^(3/2)*ln(1/2*(2*e*r-k)*2^(1/2)/e^(1/2)+(2*e*r^2-alpha^2-2*k*r)^(1/
2))*2^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65361, size = 481, normalized size = 5.94 \begin{align*} \left [\frac{\sqrt{2} \sqrt{e} k \log \left (8 \, e^{2} r^{2} - 2 \, \alpha ^{2} e - 8 \, e k r + 2 \, \sqrt{2} \sqrt{2 \, e r^{2} - \alpha ^{2} - 2 \, k r}{\left (2 \, e r - k\right )} \sqrt{e} + k^{2}\right ) + 4 \, \sqrt{2 \, e r^{2} - \alpha ^{2} - 2 \, k r} e}{8 \, e^{2}}, -\frac{\sqrt{2} \sqrt{-e} k \arctan \left (\frac{\sqrt{2} \sqrt{2 \, e r^{2} - \alpha ^{2} - 2 \, k r}{\left (2 \, e r - k\right )} \sqrt{-e}}{2 \,{\left (2 \, e^{2} r^{2} - \alpha ^{2} e - 2 \, e k r\right )}}\right ) - 2 \, \sqrt{2 \, e r^{2} - \alpha ^{2} - 2 \, k r} e}{4 \, e^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{r}{\sqrt{- \alpha ^{2} + 2 e r^{2} - 2 k r}}\, dr \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24285, size = 97, normalized size = 1.2 \begin{align*} -\frac{1}{4} \, \sqrt{2} k e^{\left (-\frac{3}{2}\right )} \log \left ({\left | -\sqrt{2}{\left (\sqrt{2} r e^{\frac{1}{2}} - \sqrt{2 \, r^{2} e - \alpha ^{2} - 2 \, k r}\right )} e^{\frac{1}{2}} + k \right |}\right ) + \frac{1}{2} \, \sqrt{2 \, r^{2} e - \alpha ^{2} - 2 \, k r} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*k*e^(-3/2)*log(abs(-sqrt(2)*(sqrt(2)*r*e^(1/2) - sqrt(2*r^2*e - alpha^2 - 2*k*r))*e^(1/2) + k)) +
1/2*sqrt(2*r^2*e - alpha^2 - 2*k*r)*e^(-1)