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

Optimal. Leaf size=74 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{-x^4-1}}\right )}{2 \sqrt{2}}-\frac{\left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{2}\right )}{4 \sqrt{-x^4-1}} \]

[Out]

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

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Rubi [A]  time = 0.0647003, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1318, 220, 1699, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{-x^4-1}}\right )}{2 \sqrt{2}}-\frac{\left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{4 \sqrt{-x^4-1}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((1 - x^2)*Sqrt[-1 - x^4]),x]

[Out]

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

Rule 1318

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

Rule 220

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

Rule 1699

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

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])

Rubi steps

\begin{align*} \int \frac{x^2}{\left (1-x^2\right ) \sqrt{-1-x^4}} \, dx &=-\left (\frac{1}{2} \int \frac{1}{\sqrt{-1-x^4}} \, dx\right )+\frac{1}{2} \int \frac{1+x^2}{\left (1-x^2\right ) \sqrt{-1-x^4}} \, dx\\ &=-\frac{\left (1+x^2\right ) \sqrt{\frac{1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{4 \sqrt{-1-x^4}}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+2 x^2} \, dx,x,\frac{x}{\sqrt{-1-x^4}}\right )\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{-1-x^4}}\right )}{2 \sqrt{2}}-\frac{\left (1+x^2\right ) \sqrt{\frac{1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{4 \sqrt{-1-x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0864783, size = 56, normalized size = 0.76 \[ \frac{\sqrt [4]{-1} \sqrt{x^4+1} \left (\text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt [4]{-1} x\right ),-1\right )-\Pi \left (i;\left .\sin ^{-1}\left ((-1)^{3/4} x\right )\right |-1\right )\right )}{\sqrt{-x^4-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(1/4)*Sqrt[1 + x^4]*(EllipticF[I*ArcSinh[(-1)^(1/4)*x], -1] - EllipticPi[I, ArcSin[(-1)^(3/4)*x], -1]))/
Sqrt[-1 - x^4]

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Maple [C]  time = 0.014, size = 115, normalized size = 1.6 \begin{align*} -{\frac{{\it EllipticF} \left ( \left ({\frac{\sqrt{2}}{2}}-{\frac{i}{2}}\sqrt{2} \right ) x,i \right ) }{{\frac{\sqrt{2}}{2}}-{\frac{i}{2}}\sqrt{2}}\sqrt{1+i{x}^{2}}\sqrt{1-i{x}^{2}}{\frac{1}{\sqrt{-{x}^{4}-1}}}}+{\frac{1}{\sqrt{-i}}\sqrt{1+i{x}^{2}}\sqrt{1-i{x}^{2}}{\it EllipticPi} \left ( \sqrt{-i}x,i,{\frac{\sqrt [4]{-1}}{\sqrt{-i}}} \right ){\frac{1}{\sqrt{-{x}^{4}-1}}}} \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/(1/2*2^(1/2)-1/2*I*2^(1/2))*(1+I*x^2)^(1/2)*(1-I*x^2)^(1/2)/(-x^4-1)^(1/2)*EllipticF((1/2*2^(1/2)-1/2*I*2^(
1/2))*x,I)+1/(-I)^(1/2)*(1+I*x^2)^(1/2)*(1-I*x^2)^(1/2)/(-x^4-1)^(1/2)*EllipticPi((-I)^(1/2)*x,I,(-1)^(1/4)/(-
I)^(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*} -\frac{1}{8} i \, \sqrt{2} \log \left (\frac{i \, \sqrt{2} x + \sqrt{-x^{4} - 1}}{x^{2} - 1}\right ) + \frac{1}{8} i \, \sqrt{2} \log \left (\frac{-i \, \sqrt{2} x + \sqrt{-x^{4} - 1}}{x^{2} - 1}\right ) +{\rm integral}\left (\frac{\sqrt{-x^{4} - 1}}{2 \,{\left (x^{4} + 1\right )}}, 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]

-1/8*I*sqrt(2)*log((I*sqrt(2)*x + sqrt(-x^4 - 1))/(x^2 - 1)) + 1/8*I*sqrt(2)*log((-I*sqrt(2)*x + sqrt(-x^4 - 1
))/(x^2 - 1)) + integral(1/2*sqrt(-x^4 - 1)/(x^4 + 1), x)

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