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

Optimal. Leaf size=70 \[ \frac{\left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{2}\right )}{4 \sqrt{x^4+1}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^4+1}}\right )}{2 \sqrt{2}} \]

[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 = 0.0594416, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1318, 220, 1699, 203} \[ \frac{\left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{4 \sqrt{x^4+1}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^4+1}}\right )}{2 \sqrt{2}} \]

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 &=\frac{1}{2} \int \frac{1}{\sqrt{1+x^4}} \, dx-\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 = 0.0904195, size = 40, normalized size = 0.57 \[ \sqrt [4]{-1} \left (\Pi \left (-i;\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )-\text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt [4]{-1} x\right ),-1\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.058, size = 110, normalized size = 1.6 \begin{align*}{\frac{{\it EllipticF} \left ( x \left ({\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) ,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}}}}+{ \left ( -1 \right ) ^{{\frac{3}{4}}}{\it EllipticPi} \left ( \sqrt [4]{-1}x,i,\sqrt{-i}- \left ( -1 \right ) ^{{\frac{3}{4}}} \right ) \sqrt{1-i{x}^{2}}\sqrt{1+i{x}^{2}}{\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(x*(1/2*2^(1/2)+1/2*I*2^(
1/2)),I)+(-1)^(3/4)*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticPi((-1)^(1/4)*x,I,(-I)^(1/2)/(-1)^(1
/4))

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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}}{\left (x^{2} + 1\right ) \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 + 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)