3.936 \(\int \frac{1}{x^6 \sqrt{1+x^4}} \, dx\)

Optimal. Leaf size=140 \[ -\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{2}\right )}{10 \sqrt{x^4+1}}-\frac{3 \sqrt{x^4+1} x}{5 \left (x^2+1\right )}+\frac{3 \sqrt{x^4+1}}{5 x}-\frac{\sqrt{x^4+1}}{5 x^5}+\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{5 \sqrt{x^4+1}} \]

[Out]

-Sqrt[1 + x^4]/(5*x^5) + (3*Sqrt[1 + x^4])/(5*x) - (3*x*Sqrt[1 + x^4])/(5*(1 + x^2)) + (3*(1 + x^2)*Sqrt[(1 +
x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/(5*Sqrt[1 + x^4]) - (3*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*El
lipticF[2*ArcTan[x], 1/2])/(10*Sqrt[1 + x^4])

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Rubi [A]  time = 0.0238988, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {325, 305, 220, 1196} \[ -\frac{3 \sqrt{x^4+1} x}{5 \left (x^2+1\right )}+\frac{3 \sqrt{x^4+1}}{5 x}-\frac{\sqrt{x^4+1}}{5 x^5}-\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{10 \sqrt{x^4+1}}+\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{5 \sqrt{x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*Sqrt[1 + x^4]),x]

[Out]

-Sqrt[1 + x^4]/(5*x^5) + (3*Sqrt[1 + x^4])/(5*x) - (3*x*Sqrt[1 + x^4])/(5*(1 + x^2)) + (3*(1 + x^2)*Sqrt[(1 +
x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/(5*Sqrt[1 + x^4]) - (3*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*El
lipticF[2*ArcTan[x], 1/2])/(10*Sqrt[1 + x^4])

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 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 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 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^6 \sqrt{1+x^4}} \, dx &=-\frac{\sqrt{1+x^4}}{5 x^5}-\frac{3}{5} \int \frac{1}{x^2 \sqrt{1+x^4}} \, dx\\ &=-\frac{\sqrt{1+x^4}}{5 x^5}+\frac{3 \sqrt{1+x^4}}{5 x}-\frac{3}{5} \int \frac{x^2}{\sqrt{1+x^4}} \, dx\\ &=-\frac{\sqrt{1+x^4}}{5 x^5}+\frac{3 \sqrt{1+x^4}}{5 x}-\frac{3}{5} \int \frac{1}{\sqrt{1+x^4}} \, dx+\frac{3}{5} \int \frac{1-x^2}{\sqrt{1+x^4}} \, dx\\ &=-\frac{\sqrt{1+x^4}}{5 x^5}+\frac{3 \sqrt{1+x^4}}{5 x}-\frac{3 x \sqrt{1+x^4}}{5 \left (1+x^2\right )}+\frac{3 \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 )}{5 \sqrt{1+x^4}}-\frac{3 \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 )}{10 \sqrt{1+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0026391, size = 22, normalized size = 0.16 \[ -\frac{\, _2F_1\left (-\frac{5}{4},\frac{1}{2};-\frac{1}{4};-x^4\right )}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*Sqrt[1 + x^4]),x]

[Out]

-Hypergeometric2F1[-5/4, 1/2, -1/4, -x^4]/(5*x^5)

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Maple [C]  time = 0.021, size = 107, normalized size = 0.8 \begin{align*} -{\frac{1}{5\,{x}^{5}}\sqrt{{x}^{4}+1}}+{\frac{3}{5\,x}\sqrt{{x}^{4}+1}}-{\frac{{\frac{3\,i}{5}} \left ({\it EllipticF} \left ( x \left ({\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) ,i \right ) -{\it EllipticE} \left ( x \left ({\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) ,i \right ) \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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(x^4+1)^(1/2),x)

[Out]

-1/5*(x^4+1)^(1/2)/x^5+3/5*(x^4+1)^(1/2)/x-3/5*I/(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)-EllipticE(x*(1/2*2^(1/2)+1/2*I*2^(1/2)),I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{4} + 1} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(x^4+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^4 + 1)*x^6), 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^{10} + x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(x^4+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 1)/(x^10 + x^6), x)

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Sympy [C]  time = 1.08858, size = 36, normalized size = 0.26 \begin{align*} \frac{\Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{x^{4} e^{i \pi }} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(x**4+1)**(1/2),x)

[Out]

gamma(-5/4)*hyper((-5/4, 1/2), (-1/4,), x**4*exp_polar(I*pi))/(4*x**5*gamma(-1/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{4} + 1} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^4 + 1)*x^6), x)