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

Optimal. Leaf size=117 \[ -\frac {\sqrt {x^4+1}}{x}+\frac {\sqrt {x^4+1} x}{x^2+1}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {x^4+1}} \]

[Out]

-(x^4+1)^(1/2)/x+x*(x^4+1)^(1/2)/(x^2+1)-(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticE(sin(2*a
rctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/(x^4+1)^(1/2)+1/2*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*ar
ctan(x))*EllipticF(sin(2*arctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/(x^4+1)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {325, 305, 220, 1196} \[ \frac {\sqrt {x^4+1} x}{x^2+1}-\frac {\sqrt {x^4+1}}{x}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[1 + x^4]),x]

[Out]

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

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 305

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1196

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 20, normalized size = 0.17 \[ -\frac {\, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-x^4\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[1 + x^4]),x]

[Out]

-(Hypergeometric2F1[-1/4, 1/2, 3/4, -x^4]/x)

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 1}}{x^{6} + x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(x^4+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 1)/(x^6 + x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{4} + 1} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^4 + 1)*x^2), x)

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maple [C]  time = 0.01, size = 95, normalized size = 0.81 \[ -\frac {\sqrt {x^{4}+1}}{x}+\frac {i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (-\EllipticE \left (\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) x , i\right )+\EllipticF \left (\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) x , i\right )\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{4} + 1} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(x^4+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^4 + 1)*x^2), x)

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mupad [B]  time = 1.14, size = 15, normalized size = 0.13 \[ -\frac {{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ -x^4\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(x^4 + 1)^(1/2)),x)

[Out]

-hypergeom([-1/4, 1/2], 3/4, -x^4)/x

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sympy [C]  time = 1.28, size = 31, normalized size = 0.26 \[ \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(x**4+1)**(1/2),x)

[Out]

gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), x**4*exp_polar(I*pi))/(4*x*gamma(3/4))

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