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

Optimal. Leaf size=113 \[ \frac{\sqrt{x^2-1} \sqrt{x^2+1} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^2-1}}\right ),\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4-1}}-\frac{x \left (1-x^2\right )}{2 \sqrt{x^4-1}}-\frac{\sqrt{x^2+1} \sqrt{1-x^2} E\left (\left .\sin ^{-1}(x)\right |-1\right )}{2 \sqrt{x^4-1}} \]

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

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Rubi [A]  time = 0.0647771, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1256, 471, 423, 427, 424, 253, 222} \[ -\frac{x \left (1-x^2\right )}{2 \sqrt{x^4-1}}+\frac{\sqrt{x^2-1} \sqrt{x^2+1} F\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^2-1}}\right )|\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4-1}}-\frac{\sqrt{x^2+1} \sqrt{1-x^2} E\left (\left .\sin ^{-1}(x)\right |-1\right )}{2 \sqrt{x^4-1}} \]

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])
+ (Sqrt[-1 + x^2]*Sqrt[1 + x^2]*EllipticF[ArcSin[(Sqrt[2]*x)/Sqrt[-1 + x^2]], 1/2])/(Sqrt[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 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 427

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

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 253

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

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(a*b), 2]}, Simp[(Sqrt[-a + q*x^2]*Sqrt[(a + q*x^2
)/q]*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2])/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4]), x] /; IntegerQ[q]]
/; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 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}{\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{\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}}+\int \frac{1}{\sqrt{-1+x^4}} \, dx\\ &=-\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 (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{-1+x^2}}\right )|\frac{1}{2}\right )}{\sqrt{2} \sqrt{-1+x^4}}\\ \end{align*}

Mathematica [A]  time = 0.078912, size = 54, normalized size = 0.48 \[ \frac{2 \sqrt{1-x^4} \text{EllipticF}\left (\sin ^{-1}(x),-1\right )+x^3-\sqrt{1-x^4} E\left (\left .\sin ^{-1}(x)\right |-1\right )-x}{2 \sqrt{x^4-1}} \]

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

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Maple [A]  time = 0.015, size = 99, normalized size = 0.9 \begin{align*}{-{\frac{i}{2}}{\it EllipticF} \left ( ix,i \right ) \sqrt{{x}^{2}+1}\sqrt{-{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}-1}}}}+{\frac{x \left ({x}^{2}-1 \right ) }{2}{\frac{1}{\sqrt{ \left ({x}^{2}+1 \right ) \left ({x}^{2}-1 \right ) }}}}+{{\frac{i}{2}} \left ({\it EllipticF} \left ( ix,i \right ) -{\it EllipticE} \left ( ix,i \right ) \right ) \sqrt{{x}^{2}+1}\sqrt{-{x}^{2}+1}{\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]

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

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