3.14.6 \(\int \frac {x^2}{\sqrt {(1-x^2) (1-k^2 x^2)} (-1+k^2 x^4)} \, dx\)

Optimal. Leaf size=94 \[ \frac {\tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+1}\right )}{2 k (k+1)}-\frac {\tan ^{-1}\left (\frac {(k-1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{4 (k-1) k} \]

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Rubi [A]  time = 1.10, antiderivative size = 87, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 7, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {1992, 6725, 1210, 1103, 1698, 204, 203} \begin {gather*} \frac {\tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{4 k (k+1)}-\frac {\tan ^{-1}\left (\frac {(1-k) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{4 (1-k) k} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1210

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[
a + b*x^2 + c*x^4], x], x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x] /; Fr
eeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1698

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[
A, Subst[Int[1/(d - (b*d - 2*a*e)*x^2), x], x, x/Sqrt[a + b*x^2 + c*x^4]], x] /; FreeQ[{a, b, c, d, e, A, B},
x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 1992

Int[(u_)^(p_.)*((f_.)*(x_))^(m_.)*(z_)^(q_.), x_Symbol] :> Int[(f*x)^m*ExpandToSum[z, x]^q*ExpandToSum[u, x]^p
, x] /; FreeQ[{f, m, p, q}, x] && BinomialQ[z, x] && TrinomialQ[u, x] &&  !(BinomialMatchQ[z, x] && TrinomialM
atchQ[u, 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 {x^2}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx &=\int \frac {x^2}{\left (-1+k^2 x^4\right ) \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}} \, dx\\ &=\int \left (\frac {1}{2 k \left (-1+k x^2\right ) \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}+\frac {1}{2 k \left (1+k x^2\right ) \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}\right ) \, dx\\ &=\frac {\int \frac {1}{\left (-1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{2 k}+\frac {\int \frac {1}{\left (1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{2 k}\\ &=-\frac {\int \frac {-1-k x^2}{\left (-1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{4 k}+\frac {\int \frac {1-k x^2}{\left (1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{4 k}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{1-\left (-1-2 k-k^2\right ) x^2} \, dx,x,\frac {x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 k}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-\left (1-2 k+k^2\right ) x^2} \, dx,x,\frac {x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 k}\\ &=-\frac {\tan ^{-1}\left (\frac {(1-k) x}{\sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}\right )}{4 (1-k) k}+\frac {\tan ^{-1}\left (\frac {(1+k) x}{\sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}\right )}{4 k (1+k)}\\ \end {align*}

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Mathematica [C]  time = 0.39, size = 70, normalized size = 0.74 \begin {gather*} \frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (\Pi \left (-k;\sin ^{-1}(x)|k^2\right )-\Pi \left (k;\sin ^{-1}(x)|k^2\right )\right )}{2 k \sqrt {\left (x^2-1\right ) \left (k^2 x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.98, size = 94, 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 )}{4 (-1+k) k}+\frac {\tan ^{-1}\left (\frac {(1+k) x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 k (1+k)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 83, normalized size = 0.88 \begin {gather*} -\frac {{\left (k - 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k + 1\right )} x}\right ) - {\left (k + 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k - 1\right )} x}\right )}{4 \, {\left (k^{3} - k\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{{\left (k^{2} x^{4} - 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(x^2/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x, algorithm="giac")

[Out]

integrate(x^2/((k^2*x^4 - 1)*sqrt((k^2*x^2 - 1)*(x^2 - 1))), x)

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maple [A]  time = 0.12, size = 93, normalized size = 0.99

method	result	size
elliptic	\(\frac {\left (-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right )}{4 k \left (1+k \right )}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right )}{4 k \left (-1+k \right )}\right ) \sqrt {2}}{2}\)	\(93\)
default	\(\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , -k , k\right )}{2 k \sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , k , k\right )}{2 k \sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{{\left (k^{2} x^{4} - 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(x^2/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x, algorithm="maxima")

[Out]

integrate(x^2/((k^2*x^4 - 1)*sqrt((k^2*x^2 - 1)*(x^2 - 1))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{\left (k^2\,x^4-1\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(x^2/((k^2*x^4 - 1)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)),x)

[Out]

int(x^2/((k^2*x^4 - 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 {x^{2}}{\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 ) \left (k x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/((-x**2+1)*(-k**2*x**2+1))**(1/2)/(k**2*x**4-1),x)

[Out]

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

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