\(\int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x (1-k^2 x)}} \, dx\) [455]

Optimal result

Integrand size = 32, antiderivative size = 37 \[ \int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=-\frac {2 \arctan \left (\frac {(1+k) x}{\sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}\right )}{1+k} \]

[Out]

-2*arctan((1+k)*x/(x+(-k^2-1)*x^2+k^2*x^3)^(1/2))/(1+k)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.38 (sec) , antiderivative size = 154, normalized size of antiderivative = 4.16, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6850, 1621, 174, 552, 551, 116} \[ \int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x}\right ),k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {4 \sqrt {1-x} \sqrt {x} \sqrt {\frac {k^2 (1-x)}{1-k^2}+1} \operatorname {EllipticPi}\left (\frac {k}{k+1},\arcsin \left (\sqrt {1-x}\right ),-\frac {k^2}{1-k^2}\right )}{(k+1) \sqrt {(1-x) x \left (1-k^2 x\right )}} \]

[In]

Int[(-1 + k*x)/((1 + k*x)*Sqrt[(1 - x)*x*(1 - k^2*x)]),x]

[Out]

(2*Sqrt[1 - x]*Sqrt[x]*Sqrt[1 - k^2*x]*EllipticF[ArcSin[Sqrt[x]], k^2])/Sqrt[(1 - x)*x*(1 - k^2*x)] + (4*Sqrt[
1 + (k^2*(1 - x))/(1 - k^2)]*Sqrt[1 - x]*Sqrt[x]*EllipticPi[k/(1 + k), ArcSin[Sqrt[1 - x]], -(k^2/(1 - k^2))])
/((1 + k)*Sqrt[(1 - x)*x*(1 - k^2*x)])

Rule 116

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[(2/(b*Sqrt[e]))*Rt
[-b/d, 2]*EllipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-b/d, 2])], c*(f/(d*e))], x] /; FreeQ[{b, c, d, e, f}, x] &&
GtQ[c, 0] && GtQ[e, 0] && (GtQ[-b/d, 0] || LtQ[-b/f, 0])

Rule 174

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + f*(x^2/d), x]]*Sqrt[Simp[(d
*g - c*h)/d + h*(x^2/d), x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

Rule 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[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] && SimplerSqrtQ[-f/e, -d/c])

Rule 552

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[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]

Rule 1621

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.)*((g_.) + (h_.)*(x_)
)^(q_.), x_Symbol] :> Dist[PolynomialRemainder[Px, a + b*x, x], Int[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h
*x)^q, x], x] + Int[PolynomialQuotient[Px, a + b*x, x]*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q,
x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n, p, q}, x] && PolyQ[Px, x] && EqQ[m, -1]

Rule 6850

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.)*(z_)^(q_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m*w^n*z^q)^FracP
art[p]/(v^(m*FracPart[p])*w^(n*FracPart[p])*z^(q*FracPart[p]))), Int[u*v^(m*p)*w^(n*p)*z^(p*q), x], x] /; Free
Q[{a, m, n, p, q}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !FreeQ[w, x] &&  !FreeQ[z, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {-1+k x}{\sqrt {1-x} \sqrt {x} (1+k x) \sqrt {1-k^2 x}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}} \\ & = \frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} (1+k x) \sqrt {1-k^2 x}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}} \\ & = \frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x}\right ),k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {\left (4 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (1+k-k x^2\right ) \sqrt {1-k^2+k^2 x^2}} \, dx,x,\sqrt {1-x}\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}} \\ & = \frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x}\right ),k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {\left (4 \sqrt {1+\frac {k^2 (-1+x)}{-1+k^2}} \sqrt {1-x} \sqrt {x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (1+k-k x^2\right ) \sqrt {1+\frac {k^2 x^2}{1-k^2}}} \, dx,x,\sqrt {1-x}\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}} \\ & = \frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x}\right ),k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {4 \sqrt {1+\frac {k^2 (1-x)}{1-k^2}} \sqrt {1-x} \sqrt {x} \operatorname {EllipticPi}\left (\frac {k}{1+k},\arcsin \left (\sqrt {1-x}\right ),-\frac {k^2}{1-k^2}\right )}{(1+k) \sqrt {(1-x) x \left (1-k^2 x\right )}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=-\frac {2 \arctan \left (\frac {(1+k) x}{\sqrt {(-1+x) x \left (-1+k^2 x\right )}}\right )}{1+k} \]

[In]

Integrate[(-1 + k*x)/((1 + k*x)*Sqrt[(1 - x)*x*(1 - k^2*x)]),x]

[Out]

(-2*ArcTan[((1 + k)*x)/Sqrt[(-1 + x)*x*(-1 + k^2*x)]])/(1 + k)

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {2 \arctan \left (\frac {\sqrt {\left (x -1\right ) x \left (k^{2} x -1\right )}}{\left (1+k \right ) x}\right )}{1+k}\)	\(32\)
pseudoelliptic	\(\frac {2 \arctan \left (\frac {\sqrt {\left (x -1\right ) x \left (k^{2} x -1\right )}}{\left (1+k \right ) x}\right )}{1+k}\)	\(32\)
elliptic	\(-\frac {2 \sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}\, \sqrt {\frac {x -1}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticF}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{2} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}}+\frac {4 \sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}\, \sqrt {\frac {x -1}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticPi}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}+\frac {1}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{3} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}+\frac {1}{k}\right )}\)	\(206\)

[In]

int((k*x-1)/(k*x+1)/((1-x)*x*(-k^2*x+1))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*arctan(((x-1)*x*(k^2*x-1))^(1/2)/(1+k)/x)/(1+k)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (34) = 68\).

Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.19 \[ \int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\frac {\arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, {\left (k^{2} + k + 1\right )} x + 1\right )}}{2 \, {\left ({\left (k^{3} + k^{2}\right )} x^{3} - {\left (k^{3} + k^{2} + k + 1\right )} x^{2} + {\left (k + 1\right )} x\right )}}\right )}{k + 1} \]

[In]

integrate((k*x-1)/(k*x+1)/((1-x)*x*(-k^2*x+1))^(1/2),x, algorithm="fricas")

[Out]

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)

Sympy [F]

\[ \int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\int \frac {k x - 1}{\sqrt {x \left (x - 1\right ) \left (k^{2} x - 1\right )} \left (k x + 1\right )}\, dx \]

[In]

integrate((k*x-1)/(k*x+1)/((1-x)*x*(-k**2*x+1))**(1/2),x)

[Out]

Integral((k*x - 1)/(sqrt(x*(x - 1)*(k**2*x - 1))*(k*x + 1)), x)

Maxima [F]

\[ \int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\int { \frac {k x - 1}{\sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x} {\left (k x + 1\right )}} \,d x } \]

[In]

integrate((k*x-1)/(k*x+1)/((1-x)*x*(-k^2*x+1))^(1/2),x, algorithm="maxima")

[Out]

integrate((k*x - 1)/(sqrt((k^2*x - 1)*(x - 1)*x)*(k*x + 1)), x)

Giac [F]

\[ \int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\int { \frac {k x - 1}{\sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x} {\left (k x + 1\right )}} \,d x } \]

[In]

integrate((k*x-1)/(k*x+1)/((1-x)*x*(-k^2*x+1))^(1/2),x, algorithm="giac")

[Out]

integrate((k*x - 1)/(sqrt((k^2*x - 1)*(x - 1)*x)*(k*x + 1)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\text {Hanged} \]

[In]

int((k*x - 1)/((k*x + 1)*(x*(k^2*x - 1)*(x - 1))^(1/2)),x)

[Out]

\text{Hanged}