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

Optimal. Leaf size=61 \[ \frac{x \left (x^2+1\right )}{2 \sqrt{1-x^4}}-\frac{\sqrt{1-x^2} \sqrt{x^2+1} E\left (\left .\sin ^{-1}(x)\right |-1\right )}{2 \sqrt{1-x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0406191, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1256, 471, 424} \[ \frac{x \left (x^2+1\right )}{2 \sqrt{1-x^4}}-\frac{\sqrt{1-x^2} \sqrt{x^2+1} E\left (\left .\sin ^{-1}(x)\right |-1\right )}{2 \sqrt{1-x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((1 - x^2)*Sqrt[1 - x^4]),x]

[Out]

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

Rule 1256

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + c*x^4)^Fr
acPart[p]/((d + e*x^2)^FracPart[p]*(a/d + (c*x^2)/e)^FracPart[p]), Int[(f*x)^m*(d + e*x^2)^(q + p)*(a/d + (c*x
^2)/e)^p, x], x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p]

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 424

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

Rubi steps

\begin{align*} \int \frac{x^2}{\left (1-x^2\right ) \sqrt{1-x^4}} \, dx &=\frac{\left (\sqrt{1-x^2} \sqrt{1+x^2}\right ) \int \frac{x^2}{\left (1-x^2\right )^{3/2} \sqrt{1+x^2}} \, dx}{\sqrt{1-x^4}}\\ &=\frac{x \left (1+x^2\right )}{2 \sqrt{1-x^4}}-\frac{\left (\sqrt{1-x^2} \sqrt{1+x^2}\right ) \int \frac{\sqrt{1+x^2}}{\sqrt{1-x^2}} \, dx}{2 \sqrt{1-x^4}}\\ &=\frac{x \left (1+x^2\right )}{2 \sqrt{1-x^4}}-\frac{\sqrt{1-x^2} \sqrt{1+x^2} E\left (\left .\sin ^{-1}(x)\right |-1\right )}{2 \sqrt{1-x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0812855, size = 37, normalized size = 0.61 \[ \frac{x^3-\sqrt{1-x^4} E\left (\left .\sin ^{-1}(x)\right |-1\right )+x}{2 \sqrt{1-x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x + x^3 - Sqrt[1 - x^4]*EllipticE[ArcSin[x], -1])/(2*Sqrt[1 - x^4])

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Maple [B]  time = 0.024, size = 143, normalized size = 2.3 \begin{align*} -{\frac{{\it EllipticF} \left ( x,i \right ) }{2}\sqrt{-{x}^{2}+1}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{-{x}^{4}+1}}}}-{\frac{-{x}^{3}-{x}^{2}-x-1}{4}{\frac{1}{\sqrt{ \left ( x-1 \right ) \left ( -{x}^{3}-{x}^{2}-x-1 \right ) }}}}+{\frac{{\it EllipticF} \left ( x,i \right ) -{\it EllipticE} \left ( x,i \right ) }{2}\sqrt{-{x}^{2}+1}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{-{x}^{4}+1}}}}-{\frac{-{x}^{3}+{x}^{2}-x+1}{4}{\frac{1}{\sqrt{ \left ( x+1 \right ) \left ( -{x}^{3}+{x}^{2}-x+1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-x^2+1)/(-x^4+1)^(1/2),x)

[Out]

-1/2*EllipticF(x,I)*(-x^2+1)^(1/2)*(x^2+1)^(1/2)/(-x^4+1)^(1/2)-1/4*(-x^3-x^2-x-1)/((x-1)*(-x^3-x^2-x-1))^(1/2
)+1/2*(-x^2+1)^(1/2)*(x^2+1)^(1/2)/(-x^4+1)^(1/2)*(EllipticF(x,I)-EllipticE(x,I))-1/4*(-x^3+x^2-x+1)/((x+1)*(-
x^3+x^2-x+1))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{x^{2}}{\sqrt{-x^{4} + 1}{\left (x^{2} - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-x^2+1)/(-x^4+1)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-x^{4} + 1} x^{2}}{x^{6} - x^{4} - x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-x^4 + 1)*x^2/(x^6 - x^4 - x^2 + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{2}}{x^{2} \sqrt{1 - x^{4}} - \sqrt{1 - x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x**2/(x**2*sqrt(1 - x**4) - sqrt(1 - x**4)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x^{2}}{\sqrt{-x^{4} + 1}{\left (x^{2} - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-x^2+1)/(-x^4+1)^(1/2),x, algorithm="giac")

[Out]

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