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

Optimal. Leaf size=37 \[ -\frac {2 \tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^3+\left (-k^2-1\right ) x^2+x}}\right )}{k+1} \]

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Rubi [C]  time = 0.64, antiderivative size = 154, normalized size of antiderivative = 4.16, number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6718, 1607, 168, 538, 537, 115} \begin {gather*} \frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\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} \Pi \left (\frac {k}{k+1};\sin ^{-1}\left (\sqrt {1-x}\right )|-\frac {k^2}{1-k^2}\right )}{(k+1) \sqrt {(1-x) x \left (1-k^2 x\right )}} \end {gather*}

Antiderivative was successfully verified.

[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 115

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

Rule 168

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 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), 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 538

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 +
(d*x^2)/c]/Sqrt[c + d*x^2], Int[1/((a + b*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[e + f*x^2]), x], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[c, 0]

Rule 1607

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 6718

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*} \int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx &=\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} F\left (\sin ^{-1}\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 ) \operatorname {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} F\left (\sin ^{-1}\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 ) \operatorname {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} F\left (\sin ^{-1}\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} \Pi \left (\frac {k}{1+k};\sin ^{-1}\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*}

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Mathematica [C]  time = 0.85, size = 114, normalized size = 3.08 \begin {gather*} \frac {2 i \sqrt {\frac {1}{x-1}+1} (x-1)^{3/2} \sqrt {\frac {1-\frac {1}{k^2}}{x-1}+1} \left ((k-1) F\left (i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|1-\frac {1}{k^2}\right )+2 \Pi \left (1+\frac {1}{k};i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|1-\frac {1}{k^2}\right )\right )}{(k+1) \sqrt {(x-1) x \left (k^2 x-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*Sqrt[1 + (-1 + x)^(-1)]*Sqrt[1 + (1 - k^(-2))/(-1 + x)]*(-1 + x)^(3/2)*((-1 + k)*EllipticF[I*ArcSinh[1/
Sqrt[-1 + x]], 1 - k^(-2)] + 2*EllipticPi[1 + k^(-1), I*ArcSinh[1/Sqrt[-1 + x]], 1 - k^(-2)]))/((1 + k)*Sqrt[(
-1 + x)*x*(-1 + k^2*x)])

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IntegrateAlgebraic [A]  time = 0.14, size = 37, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {(1+k) x}{\sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}\right )}{1+k} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[((1 + k)*x)/Sqrt[x + (-1 - k^2)*x^2 + k^2*x^3]])/(1 + k)

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fricas [B]  time = 0.52, size = 81, normalized size = 2.19 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k x - 1}{\sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x} {\left (k x + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 0.11, size = 206, normalized size = 5.57

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/k^2*(-(x-1/k^2)*k^2)^(1/2)*((-1+x)/(1/k^2-1))^(1/2)*(k^2*x)^(1/2)/(k^2*x^3-k^2*x^2-x^2+x)^(1/2)*EllipticF((
-(x-1/k^2)*k^2)^(1/2),(1/k^2/(1/k^2-1))^(1/2))+4/k^3*(-(x-1/k^2)*k^2)^(1/2)*((-1+x)/(1/k^2-1))^(1/2)*(k^2*x)^(
1/2)/(k^2*x^3-k^2*x^2-x^2+x)^(1/2)/(1/k^2+1/k)*EllipticPi((-(x-1/k^2)*k^2)^(1/2),1/k^2/(1/k^2+1/k),(1/k^2/(1/k
^2-1))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k x - 1}{\sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x} {\left (k x + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k x - 1}{\sqrt {x \left (x - 1\right ) \left (k^{2} x - 1\right )} \left (k x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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