Integrand size = 40, antiderivative size = 60 \[ \int \frac {-1+k^2 x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1+k^2 x^2\right )} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {1+k^2} \sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}{(-1+x) \left (-1+k^2 x\right )}\right )}{\sqrt {1+k^2}} \]
Time = 10.92 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.65 \[ \int \frac {-1+k^2 x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1+k^2 x^2\right )} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {1+k^2} x}{\sqrt {(-1+x) x \left (-1+k^2 x\right )}}\right )}{\sqrt {1+k^2}} \]
Time = 0.61 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.55, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2467, 25, 2035, 2537, 216}
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.
\(\displaystyle \int \frac {k^2 x^2-1}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (k^2 x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int -\frac {1-k^2 x^2}{\sqrt {x} \left (k^2 x^2+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \frac {1-k^2 x^2}{\sqrt {x} \left (k^2 x^2+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \frac {1-k^2 x^2}{\left (k^2 x^2+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}d\sqrt {x}}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\) |
\(\Big \downarrow \) 2537 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \frac {1}{\left (k^2+1\right ) x+1}d\frac {\sqrt {x}}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \arctan \left (\frac {\sqrt {k^2+1} \sqrt {x}}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )}{\sqrt {k^2+1} \sqrt {(1-x) x \left (1-k^2 x\right )}}\) |
(-2*Sqrt[x]*Sqrt[1 - (1 + k^2)*x + k^2*x^2]*ArcTan[(Sqrt[1 + k^2]*Sqrt[x]) /Sqrt[1 - (1 + k^2)*x + k^2*x^2]])/(Sqrt[1 + k^2]*Sqrt[(1 - x)*x*(1 - k^2* x)])
3.8.77.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[((u_)*((A_) + (B_.)*(x_)^4))/Sqrt[v_], x_Symbol] :> With[{a = Coeff[v, x, 0], b = Coeff[v, x, 2], c = Coeff[v, x, 4], d = Coeff[1/u, x, 0], e = Co eff[1/u, x, 2], f = Coeff[1/u, x, 4]}, Simp[A Subst[Int[1/(d - (b*d - a*e )*x^2), x], x, x/Sqrt[v]], x] /; EqQ[a*B + A*c, 0] && EqQ[c*d - a*f, 0]] /; FreeQ[{A, B}, x] && PolyQ[v, x^2, 2] && PolyQ[1/u, x^2, 2]
Time = 2.68 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.60
method | result | size |
default | \(\frac {2 \arctan \left (\frac {\sqrt {\left (x -1\right ) x \left (k^{2} x -1\right )}}{x \sqrt {k^{2}+1}}\right )}{\sqrt {k^{2}+1}}\) | \(36\) |
pseudoelliptic | \(\frac {2 \arctan \left (\frac {\sqrt {\left (x -1\right ) x \left (k^{2} x -1\right )}}{x \sqrt {k^{2}+1}}\right )}{\sqrt {k^{2}+1}}\) | \(36\) |
elliptic | \(-\frac {2 \sqrt {-k^{2} x +1}\, \sqrt {\frac {x}{\frac {1}{k^{2}}-1}-\frac {1}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticF}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{2} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}}-\frac {2 i \sqrt {-k^{2} x +1}\, \sqrt {\frac {x}{\frac {1}{k^{2}}-1}-\frac {1}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticPi}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}-\frac {i}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{3} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}-\frac {i}{k}\right )}+\frac {2 i \sqrt {-k^{2} x +1}\, \sqrt {\frac {x}{\frac {1}{k^{2}}-1}-\frac {1}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticPi}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}+\frac {i}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{3} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}+\frac {i}{k}\right )}\) | \(345\) |
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.53 \[ \int \frac {-1+k^2 x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1+k^2 x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, {\left (k^{2} + 1\right )} x + 1\right )} \sqrt {k^{2} + 1}}{2 \, {\left ({\left (k^{4} + k^{2}\right )} x^{3} - {\left (k^{4} + 2 \, k^{2} + 1\right )} x^{2} + {\left (k^{2} + 1\right )} x\right )}}\right )}{\sqrt {k^{2} + 1}} \]
arctan(1/2*sqrt(k^2*x^3 - (k^2 + 1)*x^2 + x)*(k^2*x^2 - 2*(k^2 + 1)*x + 1) *sqrt(k^2 + 1)/((k^4 + k^2)*x^3 - (k^4 + 2*k^2 + 1)*x^2 + (k^2 + 1)*x))/sq rt(k^2 + 1)
\[ \int \frac {-1+k^2 x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1+k^2 x^2\right )} \, dx=\int \frac {\left (k x - 1\right ) \left (k x + 1\right )}{\sqrt {x \left (x - 1\right ) \left (k^{2} x - 1\right )} \left (k^{2} x^{2} + 1\right )}\, dx \]
\[ \int \frac {-1+k^2 x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1+k^2 x^2\right )} \, dx=\int { \frac {k^{2} x^{2} - 1}{{\left (k^{2} x^{2} + 1\right )} \sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x}} \,d x } \]
\[ \int \frac {-1+k^2 x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1+k^2 x^2\right )} \, dx=\int { \frac {k^{2} x^{2} - 1}{{\left (k^{2} x^{2} + 1\right )} \sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x}} \,d x } \]
Time = 7.81 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07 \[ \int \frac {-1+k^2 x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1+k^2 x^2\right )} \, dx=\frac {\ln \left (\frac {k^2\,x^2-2\,x\,\left (k^2+1\right )+1+\sqrt {k^2+1}\,\sqrt {x\,\left (k^2\,x-1\right )\,\left (x-1\right )}\,2{}\mathrm {i}}{k^2\,x^2+1}\right )\,1{}\mathrm {i}}{\sqrt {k^2+1}} \]