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

Optimal result

Integrand size = 24, antiderivative size = 37 \[ \int \frac {1}{r \sqrt {-\alpha ^2-2 k r+2 h r^2}} \, dr=-\frac {\arctan \left (\frac {\alpha ^2+k r}{\alpha \sqrt {-\alpha ^2-2 k r+2 h r^2}}\right )}{\alpha } \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.22 \[ \int \frac {1}{r \sqrt {-\alpha ^2-2 k r+2 h r^2}} \, dr=-\frac {2 \arctan \left (\frac {\sqrt {2} \sqrt {h} r-\sqrt {-\alpha ^2-2 k r+2 h r^2}}{\alpha }\right )}{\alpha } \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1154, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1154

\(\displaystyle -2 \int \frac {1}{-4 \alpha ^2-\frac {4 \left (\alpha ^2+k r\right )^2}{-\alpha ^2+2 h r^2-2 k r}}d\left (-\frac {2 \left (\alpha ^2+k r\right )}{\sqrt {-\alpha ^2+2 h r^2-2 k r}}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {\arctan \left (\frac {\alpha ^2+k r}{\alpha \sqrt {-\alpha ^2+2 h r^2-2 k r}}\right )}{\alpha }\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 217

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 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-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]
 

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.41

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

Input:

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

Output:

-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)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.41 \[ \int \frac {1}{r \sqrt {-\alpha ^2-2 k r+2 h r^2}} \, dr=-\frac {\arctan \left (\frac {\sqrt {2 \, h r^{2} - \alpha ^{2} - 2 \, k r} {\left (\alpha ^{2} + k r\right )}}{2 \, \alpha h r^{2} - \alpha ^{3} - 2 \, \alpha k r}\right )}{\alpha } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{r \sqrt {-\alpha ^2-2 k r+2 h r^2}} \, dr=\int \frac {1}{r \sqrt {- \alpha ^{2} + 2 h r^{2} - 2 k r}}\, dr \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.08 \[ \int \frac {1}{r \sqrt {-\alpha ^2-2 k r+2 h r^2}} \, dr=-\frac {\arcsin \left (\frac {k}{\sqrt {2 \, \alpha ^{2} h + k^{2}}} + \frac {\alpha ^{2}}{\sqrt {2 \, \alpha ^{2} h + k^{2}} r}\right )}{\alpha } \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.08 \[ \int \frac {1}{r \sqrt {-\alpha ^2-2 k r+2 h r^2}} \, dr=\frac {2 \, \arctan \left (-\frac {\sqrt {2} \sqrt {h} r - \sqrt {2 \, h r^{2} - \alpha ^{2} - 2 \, k r}}{\alpha }\right )}{\alpha } \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.38 \[ \int \frac {1}{r \sqrt {-\alpha ^2-2 k r+2 h r^2}} \, dr=-\frac {\ln \left (\frac {\sqrt {-\alpha ^2}\,\sqrt {-\alpha ^2+2\,h\,r^2-2\,k\,r}}{r}-\frac {\alpha ^2}{r}-k\right )}{\sqrt {-\alpha ^2}} \] Input:

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

Output:

-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)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.22 \[ \int \frac {1}{r \sqrt {-\alpha ^2-2 k r+2 h r^2}} \, dr=\frac {2 \mathit {atan} \left (\frac {\sqrt {h}\, \sqrt {2 h \,r^{2}-\alpha ^{2}-2 k r}\, \sqrt {2}+2 h r}{\sqrt {h}\, \sqrt {2}\, \alpha }\right )}{\alpha } \] Input:

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

Output:

(2*atan((sqrt(h)*sqrt( - alpha**2 + 2*h*r**2 - 2*k*r)*sqrt(2) + 2*h*r)/(sq 
rt(h)*sqrt(2)*alpha)))/alpha