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

Optimal. Leaf size=54 \[ -\frac {2 \tan ^{-1}\left (\frac {(k-1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+2 \sqrt {k} x+1}\right )}{k-1} \]

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Rubi [C]  time = 1.39, antiderivative size = 199, normalized size of antiderivative = 3.69, number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {6719, 6742, 419, 2113, 537, 571, 93, 205} \begin {gather*} -\frac {2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \tan ^{-1}\left (\frac {\sqrt {k} \sqrt {1-x^2}}{\sqrt {1-k^2 x^2}}\right )}{(1-k) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (k;\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

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 571

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q, r}, x] && EqQ[m - n + 1, 0]

Rule 2113

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

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [C]  time = 0.34, size = 209, normalized size = 3.87 \begin {gather*} \frac {\sqrt {k-1} \sqrt {-((k-1) k)} \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )-2 \left (\sqrt {k} \sqrt {x^2-1} \sqrt {k^2 x^2-1} \tanh ^{-1}\left (\frac {\sqrt {-((k-1) k)} \sqrt {x^2-1}}{\sqrt {k-1} \sqrt {k^2 x^2-1}}\right )+\sqrt {k-1} \sqrt {-((k-1) k)} \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (k;\sin ^{-1}(x)|k^2\right )\right )}{\sqrt {k-1} \sqrt {-((k-1) k)} \sqrt {\left (x^2-1\right ) \left (k^2 x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 3.36, size = 54, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {(-1+k) x}{1+2 \sqrt {k} x+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{-1+k} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[((-1 + k)*x)/(1 + 2*Sqrt[k]*x + k*x^2 + Sqrt[1 + (-1 - k^2)*x^2 + k^2*x^4])])/(-1 + k)

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fricas [B]  time = 0.89, size = 100, normalized size = 1.85 \begin {gather*} \frac {\arctan \left (-\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} {\left ({\left (k^{3} + k^{2} - k - 1\right )} x - 2 \, {\left ({\left (k^{2} - k\right )} x^{2} + k - 1\right )} \sqrt {k}\right )}}{4 \, k^{3} x^{4} - {\left (k^{4} + 4 \, k^{3} - 2 \, k^{2} + 4 \, k + 1\right )} x^{2} + 4 \, k}\right )}{k - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(-sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1)*((k^3 + k^2 - k - 1)*x - 2*((k^2 - k)*x^2 + k - 1)*sqrt(k))/(4*k^3*x
^4 - (k^4 + 4*k^3 - 2*k^2 + 4*k + 1)*x^2 + 4*k))/(k - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.19, size = 157, normalized size = 2.91

method	result	size
elliptic	\(\frac {\arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right )}{-1+k}+\frac {\ln \left (\frac {-2 k^{2}+4 k -2+\left (-k^{3}+2 k^{2}-k \right ) \left (x^{2}-\frac {1}{k}\right )+2 \sqrt {-\left (-1+k \right )^{2}}\, \sqrt {k^{3} \left (x^{2}-\frac {1}{k}\right )^{2}+\left (-k^{3}+2 k^{2}-k \right ) \left (x^{2}-\frac {1}{k}\right )-k^{2}+2 k -1}}{x^{2}-\frac {1}{k}}\right )}{\sqrt {-\left (-1+k \right )^{2}}}\)	\(157\)
default	\(\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticF \left (x , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}+\frac {\ln \left (\frac {-2 k^{2}+4 k -2+\left (-k^{3}+2 k^{2}-k \right ) \left (x^{2}-\frac {1}{k}\right )+2 \sqrt {-\left (-1+k \right )^{2}}\, \sqrt {k^{3} \left (x^{2}-\frac {1}{k}\right )^{2}+\left (-k^{3}+2 k^{2}-k \right ) \left (x^{2}-\frac {1}{k}\right )-k^{2}+2 k -1}}{x^{2}-\frac {1}{k}}\right )}{\sqrt {-\left (-1+k \right )^{2}}}-\frac {2 \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , k , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}\)	\(221\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(-1+k)*arctan(((-x^2+1)*(-k^2*x^2+1))^(1/2)/x/(-1+k))+1/(-(-1+k)^2)^(1/2)*ln((-2*k^2+4*k-2+(-k^3+2*k^2-k)*(x
^2-1/k)+2*(-(-1+k)^2)^(1/2)*(k^3*(x^2-1/k)^2+(-k^3+2*k^2-k)*(x^2-1/k)-k^2+2*k-1)^(1/2))/(x^2-1/k))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {k}\,x-1}{\left (\sqrt {k}\,x+1\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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