Integrand size = 32, antiderivative size = 33 \[ \int \frac {1}{\sqrt {3-5 x^2} \sqrt {-2+2 x^2} \left (1+3 x^2\right )} \, dx=\frac {\sqrt {1-x^2} \operatorname {EllipticPi}\left (-3,\arcsin (x),\frac {5}{3}\right )}{\sqrt {6} \sqrt {-1+x^2}} \] Output:
1/6*(-x^2+1)^(1/2)*EllipticPi(x,-3,1/3*15^(1/2))*6^(1/2)/(x^2-1)^(1/2)
Time = 0.50 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\sqrt {3-5 x^2} \sqrt {-2+2 x^2} \left (1+3 x^2\right )} \, dx=\frac {\sqrt {1-x^2} \operatorname {EllipticPi}\left (-\frac {9}{5},\arcsin \left (\sqrt {\frac {5}{3}} x\right ),\frac {3}{5}\right )}{\sqrt {10} \sqrt {-1+x^2}} \] Input:
Integrate[1/(Sqrt[3 - 5*x^2]*Sqrt[-2 + 2*x^2]*(1 + 3*x^2)),x]
Output:
(Sqrt[1 - x^2]*EllipticPi[-9/5, ArcSin[Sqrt[5/3]*x], 3/5])/(Sqrt[10]*Sqrt[ -1 + x^2])
Time = 0.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {413, 412}
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.
\(\displaystyle \int \frac {1}{\sqrt {3-5 x^2} \sqrt {2 x^2-2} \left (3 x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 413 |
\(\displaystyle \frac {\sqrt {1-x^2} \int \frac {1}{\sqrt {3-5 x^2} \sqrt {1-x^2} \left (3 x^2+1\right )}dx}{\sqrt {2} \sqrt {x^2-1}}\) |
\(\Big \downarrow \) 412 |
\(\displaystyle \frac {\sqrt {1-x^2} \operatorname {EllipticPi}\left (-3,\arcsin (x),\frac {5}{3}\right )}{\sqrt {6} \sqrt {x^2-1}}\) |
Input:
Int[1/(Sqrt[3 - 5*x^2]*Sqrt[-2 + 2*x^2]*(1 + 3*x^2)),x]
Output:
(Sqrt[1 - x^2]*EllipticPi[-3, ArcSin[x], 5/3])/(Sqrt[6]*Sqrt[-1 + x^2])
Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x _)^2]), x_Symbol] :> Simp[(1/(a*Sqrt[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b* (c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, f}, x] && !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] && !( !GtQ[f/e, 0] && S implerSqrtQ[-f/e, -d/c])
Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x _)^2]), x_Symbol] :> Simp[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*x^2] Int[1/((a + b*x^2)*Sqrt[1 + (d/c)*x^2]*Sqrt[e + f*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && !GtQ[c, 0]
Time = 4.57 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09
method | result | size |
default | \(\frac {\operatorname {EllipticPi}\left (x , -3, \frac {\sqrt {3}\, \sqrt {5}}{3}\right ) \sqrt {3}\, \sqrt {-x^{2}+1}\, \sqrt {2}}{6 \sqrt {x^{2}-1}}\) | \(36\) |
elliptic | \(\frac {\sqrt {-\left (5 x^{2}-3\right ) \left (x^{2}-1\right )}\, \sqrt {-x^{2}+1}\, \sqrt {1-\frac {5 x^{2}}{3}}\, \operatorname {EllipticPi}\left (x , -3, \frac {\sqrt {3}\, \sqrt {5}}{3}\right )}{\sqrt {-5 x^{2}+3}\, \sqrt {x^{2}-1}\, \sqrt {-10 x^{4}+16 x^{2}-6}}\) | \(77\) |
Input:
int(1/(-5*x^2+3)^(1/2)/(2*x^2-2)^(1/2)/(3*x^2+1),x,method=_RETURNVERBOSE)
Output:
1/6*EllipticPi(x,-3,1/3*3^(1/2)*5^(1/2))*3^(1/2)*(-x^2+1)^(1/2)/(x^2-1)^(1 /2)*2^(1/2)
\[ \int \frac {1}{\sqrt {3-5 x^2} \sqrt {-2+2 x^2} \left (1+3 x^2\right )} \, dx=\int { \frac {1}{{\left (3 \, x^{2} + 1\right )} \sqrt {2 \, x^{2} - 2} \sqrt {-5 \, x^{2} + 3}} \,d x } \] Input:
integrate(1/(-5*x^2+3)^(1/2)/(2*x^2-2)^(1/2)/(3*x^2+1),x, algorithm="frica s")
Output:
integral(-1/2*sqrt(2*x^2 - 2)*sqrt(-5*x^2 + 3)/(15*x^6 - 19*x^4 + x^2 + 3) , x)
\[ \int \frac {1}{\sqrt {3-5 x^2} \sqrt {-2+2 x^2} \left (1+3 x^2\right )} \, dx=\frac {\sqrt {2} \int \frac {1}{3 x^{2} \sqrt {3 - 5 x^{2}} \sqrt {x^{2} - 1} + \sqrt {3 - 5 x^{2}} \sqrt {x^{2} - 1}}\, dx}{2} \] Input:
integrate(1/(-5*x**2+3)**(1/2)/(2*x**2-2)**(1/2)/(3*x**2+1),x)
Output:
sqrt(2)*Integral(1/(3*x**2*sqrt(3 - 5*x**2)*sqrt(x**2 - 1) + sqrt(3 - 5*x* *2)*sqrt(x**2 - 1)), x)/2
\[ \int \frac {1}{\sqrt {3-5 x^2} \sqrt {-2+2 x^2} \left (1+3 x^2\right )} \, dx=\int { \frac {1}{{\left (3 \, x^{2} + 1\right )} \sqrt {2 \, x^{2} - 2} \sqrt {-5 \, x^{2} + 3}} \,d x } \] Input:
integrate(1/(-5*x^2+3)^(1/2)/(2*x^2-2)^(1/2)/(3*x^2+1),x, algorithm="maxim a")
Output:
integrate(1/((3*x^2 + 1)*sqrt(2*x^2 - 2)*sqrt(-5*x^2 + 3)), x)
\[ \int \frac {1}{\sqrt {3-5 x^2} \sqrt {-2+2 x^2} \left (1+3 x^2\right )} \, dx=\int { \frac {1}{{\left (3 \, x^{2} + 1\right )} \sqrt {2 \, x^{2} - 2} \sqrt {-5 \, x^{2} + 3}} \,d x } \] Input:
integrate(1/(-5*x^2+3)^(1/2)/(2*x^2-2)^(1/2)/(3*x^2+1),x, algorithm="giac" )
Output:
integrate(1/((3*x^2 + 1)*sqrt(2*x^2 - 2)*sqrt(-5*x^2 + 3)), x)
Timed out. \[ \int \frac {1}{\sqrt {3-5 x^2} \sqrt {-2+2 x^2} \left (1+3 x^2\right )} \, dx=\int \frac {1}{\sqrt {2\,x^2-2}\,\left (3\,x^2+1\right )\,\sqrt {3-5\,x^2}} \,d x \] Input:
int(1/((2*x^2 - 2)^(1/2)*(3*x^2 + 1)*(3 - 5*x^2)^(1/2)),x)
Output:
int(1/((2*x^2 - 2)^(1/2)*(3*x^2 + 1)*(3 - 5*x^2)^(1/2)), x)
\[ \int \frac {1}{\sqrt {3-5 x^2} \sqrt {-2+2 x^2} \left (1+3 x^2\right )} \, dx=-\frac {\sqrt {2}\, \left (\int \frac {\sqrt {-5 x^{2}+3}\, \sqrt {x^{2}-1}}{15 x^{6}-19 x^{4}+x^{2}+3}d x \right )}{2} \] Input:
int(1/(-5*x^2+3)^(1/2)/(2*x^2-2)^(1/2)/(3*x^2+1),x)
Output:
( - sqrt(2)*int((sqrt( - 5*x**2 + 3)*sqrt(x**2 - 1))/(15*x**6 - 19*x**4 + x**2 + 3),x))/2