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

Optimal. Leaf size=38 \[ \frac {\tan ^{-1}\left (\frac {(k-1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{1-k} \]

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Rubi [C]  time = 0.78, antiderivative size = 111, normalized size of antiderivative = 2.92, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6719, 6725, 419, 537} \begin {gather*} \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 + k*x^2)/((-1 + k*x^2)*Sqrt[(1 - x^2)*(1 - k^2*x^2)]),x]

[Out]

(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]*S
qrt[1 - k^2*x^2]*EllipticPi[k, ArcSin[x], k^2])/Sqrt[(1 - x^2)*(1 - k^2*x^2)]

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 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 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1+k x^2}{\left (-1+k x^2\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+k x^2}{\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 (\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}{\sqrt {1-x^2} \left (-1+k x^2\right ) \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}{\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 )}}\\ \end {align*}

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Mathematica [C]  time = 0.22, size = 61, normalized size = 1.61 \begin {gather*} \frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (F\left (\sin ^{-1}(x)|k^2\right )-2 \Pi \left (k;\sin ^{-1}(x)|k^2\right )\right )}{\sqrt {\left (x^2-1\right ) \left (k^2 x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.59, size = 38, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {(-1+k) x}{\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 + k*x^2)/((-1 + k*x^2)*Sqrt[(1 - x^2)*(1 - k^2*x^2)]),x]

[Out]

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

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fricas [A]  time = 0.52, size = 37, normalized size = 0.97 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k - 1\right )} x}\right )}{k - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((k*x^2+1)/(k*x^2-1)/((-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 - 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^{2} + 1}{{\left (k x^{2} - 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 37, normalized size = 0.97

method	result	size
elliptic	\(\frac {\arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right )}{-1+k}\)	\(37\)
default	\(\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticF \left (x , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {2 \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , k , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {k\,x^2+1}{\left (k\,x^2-1\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*x^2 + 1)/((k*x^2 - 1)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)),x)

[Out]

int((k*x^2 + 1)/((k*x^2 - 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 {k x^{2} + 1}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k x^{2} - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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