Integrand size = 32, antiderivative size = 85 \[ \int \frac {x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^2 x^2\right )} \, dx=-\frac {\arctan \left (\frac {(-1+k) x}{\sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}\right )}{2 (-1+k) k}+\frac {\arctan \left (\frac {(1+k) x}{\sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}\right )}{2 k (1+k)} \]
-1/2*arctan((-1+k)*x/(x+(-k^2-1)*x^2+k^2*x^3)^(1/2))/(-1+k)/k+1/2*arctan(( 1+k)*x/(x+(-k^2-1)*x^2+k^2*x^3)^(1/2))/k/(1+k)
Time = 10.75 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \frac {x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^2 x^2\right )} \, dx=\frac {-\frac {\arctan \left (\frac {(-1+k) x}{\sqrt {(-1+x) x \left (-1+k^2 x\right )}}\right )}{-1+k}+\frac {\arctan \left (\frac {(1+k) x}{\sqrt {(-1+x) x \left (-1+k^2 x\right )}}\right )}{1+k}}{2 k} \]
(-(ArcTan[((-1 + k)*x)/Sqrt[(-1 + x)*x*(-1 + k^2*x)]]/(-1 + k)) + ArcTan[( (1 + k)*x)/Sqrt[(-1 + x)*x*(-1 + k^2*x)]]/(1 + k))/(2*k)
Time = 1.15 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.62, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2467, 25, 2035, 7276, 2009}
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.
\(\displaystyle \int \frac {x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (k^2 x^2-1\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int -\frac {\sqrt {x}}{\left (1-k^2 x^2\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \frac {\sqrt {x}}{\left (1-k^2 x^2\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \frac {x}{\left (1-k^2 x^2\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}d\sqrt {x}}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \left (-\frac {1}{2 k (k x-1) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}-\frac {1}{2 k (k x+1) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )d\sqrt {x}}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \left (\frac {\arctan \left (\frac {(1-k) \sqrt {x}}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )}{4 (1-k) k}-\frac {\arctan \left (\frac {(k+1) \sqrt {x}}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )}{4 k (k+1)}\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\) |
(-2*Sqrt[x]*Sqrt[1 - (1 + k^2)*x + k^2*x^2]*(ArcTan[((1 - k)*Sqrt[x])/Sqrt [1 - (1 + k^2)*x + k^2*x^2]]/(4*(1 - k)*k) - ArcTan[((1 + k)*Sqrt[x])/Sqrt [1 - (1 + k^2)*x + k^2*x^2]]/(4*k*(1 + k))))/Sqrt[(1 - x)*x*(1 - k^2*x)]
3.12.32.3.1 Defintions of rubi rules used
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_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 1.61 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84
method | result | size |
default | \(\frac {\left (1+k \right ) \arctan \left (\frac {\sqrt {\left (x -1\right ) x \left (k^{2} x -1\right )}}{\left (-1+k \right ) x}\right )-\arctan \left (\frac {\sqrt {\left (x -1\right ) x \left (k^{2} x -1\right )}}{\left (1+k \right ) x}\right ) \left (-1+k \right )}{2 k^{3}-2 k}\) | \(71\) |
pseudoelliptic | \(\frac {\left (1+k \right ) \arctan \left (\frac {\sqrt {\left (x -1\right ) x \left (k^{2} x -1\right )}}{\left (-1+k \right ) x}\right )-\arctan \left (\frac {\sqrt {\left (x -1\right ) x \left (k^{2} x -1\right )}}{\left (1+k \right ) x}\right ) \left (-1+k \right )}{2 k^{3}-2 k}\) | \(71\) |
elliptic | \(-\frac {\sqrt {-k^{2} x +1}\, \sqrt {\frac {x}{\frac {1}{k^{2}}-1}-\frac {1}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticPi}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}-\frac {1}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{4} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}-\frac {1}{k}\right )}-\frac {\sqrt {-k^{2} x +1}\, \sqrt {\frac {x}{\frac {1}{k^{2}}-1}-\frac {1}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticPi}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}+\frac {1}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{4} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}+\frac {1}{k}\right )}\) | \(240\) |
((1+k)*arctan(((x-1)*x*(k^2*x-1))^(1/2)/(-1+k)/x)-arctan(((x-1)*x*(k^2*x-1 ))^(1/2)/(1+k)/x)*(-1+k))/(2*k^3-2*k)
Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (75) = 150\).
Time = 0.30 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.08 \[ \int \frac {x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^2 x^2\right )} \, dx=-\frac {{\left (k - 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, {\left (k^{2} + k + 1\right )} x + 1\right )}}{2 \, {\left ({\left (k^{3} + k^{2}\right )} x^{3} - {\left (k^{3} + k^{2} + k + 1\right )} x^{2} + {\left (k + 1\right )} x\right )}}\right ) - {\left (k + 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, {\left (k^{2} - k + 1\right )} x + 1\right )}}{2 \, {\left ({\left (k^{3} - k^{2}\right )} x^{3} - {\left (k^{3} - k^{2} + k - 1\right )} x^{2} + {\left (k - 1\right )} x\right )}}\right )}{4 \, {\left (k^{3} - k\right )}} \]
-1/4*((k - 1)*arctan(1/2*sqrt(k^2*x^3 - (k^2 + 1)*x^2 + x)*(k^2*x^2 - 2*(k ^2 + k + 1)*x + 1)/((k^3 + k^2)*x^3 - (k^3 + k^2 + k + 1)*x^2 + (k + 1)*x) ) - (k + 1)*arctan(1/2*sqrt(k^2*x^3 - (k^2 + 1)*x^2 + x)*(k^2*x^2 - 2*(k^2 - k + 1)*x + 1)/((k^3 - k^2)*x^3 - (k^3 - k^2 + k - 1)*x^2 + (k - 1)*x))) /(k^3 - k)
\[ \int \frac {x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^2 x^2\right )} \, dx=\int \frac {x}{\sqrt {x \left (x - 1\right ) \left (k^{2} x - 1\right )} \left (k x - 1\right ) \left (k x + 1\right )}\, dx \]
\[ \int \frac {x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^2 x^2\right )} \, dx=\int { \frac {x}{{\left (k^{2} x^{2} - 1\right )} \sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x}} \,d x } \]
\[ \int \frac {x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^2 x^2\right )} \, dx=\int { \frac {x}{{\left (k^{2} x^{2} - 1\right )} \sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x}} \,d x } \]
Timed out. \[ \int \frac {x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^2 x^2\right )} \, dx=\text {Hanged} \]