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

Optimal. Leaf size=99 \[ -\frac {x \left (1-x^2\right )}{2 \sqrt {1-x^4}}+\frac {\sqrt {x^2+1} \sqrt {1-x^2} F\left (\left .\sin ^{-1}(x)\right |-1\right )}{\sqrt {1-x^4}}-\frac {\sqrt {x^2+1} \sqrt {1-x^2} E\left (\left .\sin ^{-1}(x)\right |-1\right )}{2 \sqrt {1-x^4}} \]

[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)+EllipticF(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 = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1256, 471, 423, 424, 248, 221} \[ -\frac {x \left (1-x^2\right )}{2 \sqrt {1-x^4}}+\frac {\sqrt {x^2+1} \sqrt {1-x^2} F\left (\left .\sin ^{-1}(x)\right |-1\right )}{\sqrt {1-x^4}}-\frac {\sqrt {x^2+1} \sqrt {1-x^2} E\left (\left .\sin ^{-1}(x)\right |-1\right )}{2 \sqrt {1-x^4}} \]

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

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 248

Int[((a1_.) + (b1_.)*(x_)^(n_))^(p_.)*((a2_.) + (b2_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[(a1*a2 + b1*b2*x^(2*
n))^p, x] /; FreeQ[{a1, b1, a2, b2, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a
2, 0]))

Rule 423

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

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 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 (\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 {\left (\sqrt {1-x^2} \sqrt {1+x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2}} \, dx}{\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}}+\frac {\left (\sqrt {1-x^2} \sqrt {1+x^2}\right ) \int \frac {1}{\sqrt {1-x^4}} \, dx}{\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}}+\frac {\sqrt {1-x^2} \sqrt {1+x^2} F\left (\left .\sin ^{-1}(x)\right |-1\right )}{\sqrt {1-x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 46, normalized size = 0.46 \[ \frac {1}{2} \left (-\frac {x}{\sqrt {1-x^4}}+\frac {x^3}{\sqrt {1-x^4}}+2 F\left (\left .\sin ^{-1}(x)\right |-1\right )-E\left (\left .\sin ^{-1}(x)\right |-1\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.19, 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 [A]  time = 0.02, size = 96, normalized size = 0.97 \[ -\frac {\left (-x^{2}+1\right ) x}{2 \sqrt {\left (-x^{2}+1\right ) \left (x^{2}+1\right )}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \EllipticF \left (x , i\right )}{2 \sqrt {-x^{4}+1}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (x , i\right )+\EllipticF \left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}} \]

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

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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 = -0.01 \[ \int \frac {x^2}{\left (x^2+1\right )\,\sqrt {1-x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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