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

Optimal. Leaf size=57 \[ \frac {x \left (x^2+1\right )}{2 \sqrt {x^4-1}}-\frac {\sqrt {1-x^2} \sqrt {x^2+1} E\left (\left .\sin ^{-1}(x)\right |-1\right )}{2 \sqrt {x^4-1}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1256, 471, 426, 424} \[ \frac {x \left (x^2+1\right )}{2 \sqrt {x^4-1}}-\frac {\sqrt {1-x^2} \sqrt {x^2+1} E\left (\left .\sin ^{-1}(x)\right |-1\right )}{2 \sqrt {x^4-1}} \]

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 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]

Rule 426

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

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 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]

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}{\sqrt {-1-x^2} \left (1-x^2\right )^{3/2}} \, 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 {\left (\left (-1-x^2\right ) \sqrt {1-x^2}\right ) \int \frac {\sqrt {1+x^2}}{\sqrt {1-x^2}} \, dx}{2 \sqrt {1+x^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*}

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Mathematica [A]  time = 0.07, size = 35, normalized size = 0.61 \[ \frac {-\sqrt {1-x^4} E\left (\left .\sin ^{-1}(x)\right |-1\right )+x^3+x}{2 \sqrt {x^4-1}} \]

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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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {x^{4} - 1} x^{2}}{x^{6} - x^{4} - x^{2} + 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x^{2}}{\sqrt {x^{4} - 1} {\left (x^{2} - 1\right )}}\,{d x} \]

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)

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maple [B]  time = 0.02, size = 134, normalized size = 2.35 \[ \frac {i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \EllipticF \left (i x , i\right )}{2 \sqrt {x^{4}-1}}+\frac {x^{3}-x^{2}+x -1}{4 \sqrt {\left (x +1\right ) \left (x^{3}-x^{2}+x -1\right )}}+\frac {i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (-\EllipticE \left (i x , i\right )+\EllipticF \left (i x , i\right )\right )}{2 \sqrt {x^{4}-1}}+\frac {x^{3}+x^{2}+x +1}{4 \sqrt {\left (x -1\right ) \left (x^{3}+x^{2}+x +1\right )}} \]

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*I*(x^2+1)^(1/2)*(-x^2+1)^(1/2)/(x^4-1)^(1/2)*EllipticF(I*x,I)+1/4*(x^3-x^2+x-1)/((x+1)*(x^3-x^2+x-1))^(1/2
)+1/2*I*(x^2+1)^(1/2)*(-x^2+1)^(1/2)/(x^4-1)^(1/2)*(EllipticF(I*x,I)-EllipticE(I*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 = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {x^{2}}{\sqrt {x^{4} - 1} {\left (x^{2} - 1\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ -\int \frac {x^2}{\left (x^2-1\right )\,\sqrt {x^4-1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{2}}{x^{2} \sqrt {x^{4} - 1} - \sqrt {x^{4} - 1}}\, dx \]

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(x**4 - 1) - sqrt(x**4 - 1)), x)

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