Integrand size = 24, antiderivative size = 147 \[ \int \frac {r}{\sqrt {-\alpha ^2+2 \left (\frac {5 \text {g0} \text {g2L} \text {L2}^5}{(\text {L1}+\text {L2})^6 r^2}+\frac {\text {g10} \text {g1L} \text {L2}^5}{(\text {L1}+\text {L2})^5 r^2}+\frac {\text {g0} \text {g2L} \text {L2}^6}{\text {L1} (\text {L1}+\text {L2})^5 r^2}\right ) r^2-2 k r^4}} \, dr=-\frac {\arctan \left (\frac {\frac {5 \text {g0} \text {g2L} \text {L2}^5}{(\text {L1}+\text {L2})^6 r^2}+\frac {\text {g10} \text {g1L} \text {L2}^5}{(\text {L1}+\text {L2})^5 r^2}+\frac {\text {g0} \text {g2L} \text {L2}^6}{\text {L1} (\text {L1}+\text {L2})^5 r^2}-2 k r^2}{\sqrt {2} \sqrt {k} \sqrt {-\alpha ^2+2 \left (\frac {5 \text {g0} \text {g2L} \text {L2}^5}{(\text {L1}+\text {L2})^6 r^2}+\frac {\text {g10} \text {g1L} \text {L2}^5}{(\text {L1}+\text {L2})^5 r^2}+\frac {\text {g0} \text {g2L} \text {L2}^6}{\text {L1} (\text {L1}+\text {L2})^5 r^2}\right ) r^2-2 k r^4}}\right )}{2 \sqrt {2} \sqrt {k}} \] Output:
-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)
Time = 10.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.63 \[ \int \frac {r}{\sqrt {-\alpha ^2+2 \left (\frac {5 \text {g0} \text {g2L} \text {L2}^5}{(\text {L1}+\text {L2})^6 r^2}+\frac {\text {g10} \text {g1L} \text {L2}^5}{(\text {L1}+\text {L2})^5 r^2}+\frac {\text {g0} \text {g2L} \text {L2}^6}{\text {L1} (\text {L1}+\text {L2})^5 r^2}\right ) r^2-2 k r^4}} \, dr=\frac {\arctan \left (\frac {\sqrt {2} \sqrt {k} r^2}{\sqrt {-\alpha ^2+\frac {2 \left (5 \text {g0} \text {g2L} \text {L1} \text {L2}^5+\text {g0} \text {g2L} \text {L1} \text {L2}^6+\text {g0} \text {g2L} \text {L2}^7+\text {g10} \text {g1L} \text {L1} \text {L2}^5 (\text {L1}+\text {L2})-k \text {L1} (\text {L1}+\text {L2})^6 r^4\right )}{\text {L1} (\text {L1}+\text {L2})^6}}}\right )}{2 \sqrt {2} \sqrt {k}} \] Input:
Integrate[r/Sqrt[-alpha^2 + 2*((5*g0*g2L*L2^5)/((L1 + L2)^6*r^2) + (g10*g1 L*L2^5)/((L1 + L2)^5*r^2) + (g0*g2L*L2^6)/(L1*(L1 + L2)^5*r^2))*r^2 - 2*k* r^4],r]
Output:
ArcTan[(Sqrt[2]*Sqrt[k]*r^2)/Sqrt[-alpha^2 + (2*(5*g0*g2L*L1*L2^5 + g0*g2L *L1*L2^6 + g0*g2L*L2^7 + g10*g1L*L1*L2^5*(L1 + L2) - k*L1*(L1 + L2)^6*r^4) )/(L1*(L1 + L2)^6)]]/(2*Sqrt[2]*Sqrt[k])
Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.38, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1432, 1092, 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.
\(\displaystyle \int \frac {r}{\sqrt {-\alpha ^2+2 e r^2-2 k r^4}} \, dr\) |
\(\Big \downarrow \) 1432 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt {-2 k r^4+2 e r^2-\alpha ^2}}dr^2\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \int \frac {1}{-8 k-r^4}d\frac {2 \left (e-2 k r^2\right )}{\sqrt {-\alpha ^2+2 e r^2-2 k r^4}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {\arctan \left (\frac {e-2 k r^2}{\sqrt {2} \sqrt {k} \sqrt {-\alpha ^2+2 e r^2-2 k r^4}}\right )}{2 \sqrt {2} \sqrt {k}}\) |
Input:
Int[r/Sqrt[-alpha^2 + 2*e*r^2 - 2*k*r^4],r]
Output:
-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])
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])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
Time = 0.47 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.31
method | result | size |
pseudoelliptic | \(-\frac {\arctan \left (\frac {\left (-2 k \,r^{2}+e \right ) \sqrt {2}}{2 \sqrt {k}\, \sqrt {-2 k \,r^{4}+2 e \,r^{2}-\alpha ^{2}}}\right ) \sqrt {2}}{4 \sqrt {k}}\) | \(46\) |
default | \(\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {k}\, \left (r^{2}-\frac {e}{2 k}\right )}{\sqrt {-2 k \,r^{4}+2 e \,r^{2}-\alpha ^{2}}}\right )}{4 \sqrt {k}}\) | \(47\) |
elliptic | \(\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {k}\, \left (r^{2}-\frac {e}{2 k}\right )}{\sqrt {-2 k \,r^{4}+2 e \,r^{2}-\alpha ^{2}}}\right )}{4 \sqrt {k}}\) | \(47\) |
Input:
int(r/(-2*k*r^4+2*e*r^2-alpha^2)^(1/2),r,method=_RETURNVERBOSE)
Output:
-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)
Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.03 \[ \int \frac {r}{\sqrt {-\alpha ^2+2 \left (\frac {5 \text {g0} \text {g2L} \text {L2}^5}{(\text {L1}+\text {L2})^6 r^2}+\frac {\text {g10} \text {g1L} \text {L2}^5}{(\text {L1}+\text {L2})^5 r^2}+\frac {\text {g0} \text {g2L} \text {L2}^6}{\text {L1} (\text {L1}+\text {L2})^5 r^2}\right ) r^2-2 k r^4}} \, dr=\left [-\frac {\sqrt {2} \sqrt {-k} \log \left (-8 \, k^{2} r^{4} + 8 \, e k r^{2} - 2 \, \alpha ^{2} k + 2 \, \sqrt {2} \sqrt {-2 \, k r^{4} + 2 \, e r^{2} - \alpha ^{2}} {\left (2 \, k r^{2} - e\right )} \sqrt {-k} - e^{2}\right )}{8 \, k}, -\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-2 \, k r^{4} + 2 \, e r^{2} - \alpha ^{2}} {\left (2 \, k r^{2} - e\right )} \sqrt {k}}{2 \, {\left (2 \, k^{2} r^{4} - 2 \, e k r^{2} + \alpha ^{2} k\right )}}\right )}{4 \, \sqrt {k}}\right ] \] Input:
integrate(r/(-2*k*r^4+2*e*r^2-alpha^2)^(1/2),r, algorithm="fricas")
Output:
[-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)]
\[ \int \frac {r}{\sqrt {-\alpha ^2+2 \left (\frac {5 \text {g0} \text {g2L} \text {L2}^5}{(\text {L1}+\text {L2})^6 r^2}+\frac {\text {g10} \text {g1L} \text {L2}^5}{(\text {L1}+\text {L2})^5 r^2}+\frac {\text {g0} \text {g2L} \text {L2}^6}{\text {L1} (\text {L1}+\text {L2})^5 r^2}\right ) r^2-2 k r^4}} \, dr=\int \frac {r}{\sqrt {- \alpha ^{2} + 2 e r^{2} - 2 k r^{4}}}\, dr \] Input:
integrate(r/(-2*k*r**4+2*e*r**2-alpha**2)**(1/2),r)
Output:
Integral(r/sqrt(-alpha**2 + 2*e*r**2 - 2*k*r**4), r)
Exception generated. \[ \int \frac {r}{\sqrt {-\alpha ^2+2 \left (\frac {5 \text {g0} \text {g2L} \text {L2}^5}{(\text {L1}+\text {L2})^6 r^2}+\frac {\text {g10} \text {g1L} \text {L2}^5}{(\text {L1}+\text {L2})^5 r^2}+\frac {\text {g0} \text {g2L} \text {L2}^6}{\text {L1} (\text {L1}+\text {L2})^5 r^2}\right ) r^2-2 k r^4}} \, dr=\text {Exception raised: ValueError} \] Input:
integrate(r/(-2*k*r^4+2*e*r^2-alpha^2)^(1/2),r, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(2*alpha^2*k-e^2>0)', see `assume ?` for mor
Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.39 \[ \int \frac {r}{\sqrt {-\alpha ^2+2 \left (\frac {5 \text {g0} \text {g2L} \text {L2}^5}{(\text {L1}+\text {L2})^6 r^2}+\frac {\text {g10} \text {g1L} \text {L2}^5}{(\text {L1}+\text {L2})^5 r^2}+\frac {\text {g0} \text {g2L} \text {L2}^6}{\text {L1} (\text {L1}+\text {L2})^5 r^2}\right ) r^2-2 k r^4}} \, dr=-\frac {\sqrt {2} \log \left ({\left | \sqrt {2} {\left (\sqrt {2} \sqrt {-k} r^{2} - \sqrt {-2 \, k r^{4} + 2 \, e r^{2} - \alpha ^{2}}\right )} \sqrt {-k} + e \right |}\right )}{4 \, \sqrt {-k}} \] Input:
integrate(r/(-2*k*r^4+2*e*r^2-alpha^2)^(1/2),r, algorithm="giac")
Output:
-1/4*sqrt(2)*log(abs(sqrt(2)*(sqrt(2)*sqrt(-k)*r^2 - sqrt(-2*k*r^4 + 2*e*r ^2 - alpha^2))*sqrt(-k) + e))/sqrt(-k)
Time = 0.50 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.34 \[ \int \frac {r}{\sqrt {-\alpha ^2+2 \left (\frac {5 \text {g0} \text {g2L} \text {L2}^5}{(\text {L1}+\text {L2})^6 r^2}+\frac {\text {g10} \text {g1L} \text {L2}^5}{(\text {L1}+\text {L2})^5 r^2}+\frac {\text {g0} \text {g2L} \text {L2}^6}{\text {L1} (\text {L1}+\text {L2})^5 r^2}\right ) r^2-2 k r^4}} \, dr=\frac {\sqrt {2}\,\ln \left (\sqrt {-\alpha ^2-2\,k\,r^4+2\,e\,r^2}+\frac {\sqrt {2}\,\left (e-2\,k\,r^2\right )}{2\,\sqrt {-k}}\right )}{4\,\sqrt {-k}} \] Input:
int(r/(2*e*r^2 - 2*k*r^4 - alpha^2)^(1/2),r)
Output:
(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))
\[ \int \frac {r}{\sqrt {-\alpha ^2+2 \left (\frac {5 \text {g0} \text {g2L} \text {L2}^5}{(\text {L1}+\text {L2})^6 r^2}+\frac {\text {g10} \text {g1L} \text {L2}^5}{(\text {L1}+\text {L2})^5 r^2}+\frac {\text {g0} \text {g2L} \text {L2}^6}{\text {L1} (\text {L1}+\text {L2})^5 r^2}\right ) r^2-2 k r^4}} \, dr=\int \frac {r}{\sqrt {-2 k \,r^{4}+2 e \,r^{2}-\alpha ^{2}}}d r \] Input:
int(r/(-2*k*r^4+2*e*r^2-alpha^2)^(1/2),r)
Output:
int(r/sqrt( - alpha**2 + 2*e*r**2 - 2*k*r**4),r)