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

Optimal. Leaf size=49 \[ \frac{x^5 \sqrt{x^4+\frac{1}{x^4}+2}}{3 \left (x^4+1\right )}-\frac{x \sqrt{x^4+\frac{1}{x^4}+2}}{x^4+1} \]

[Out]

-((x*Sqrt[2 + x^(-4) + x^4])/(1 + x^4)) + (x^5*Sqrt[2 + x^(-4) + x^4])/(3*(1 + x^4))

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Rubi [A]  time = 0.0148724, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1351, 1355, 14} \[ \frac{x^5 \sqrt{x^4+\frac{1}{x^4}+2}}{3 \left (x^4+1\right )}-\frac{x \sqrt{x^4+\frac{1}{x^4}+2}}{x^4+1} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + x^(-4) + x^4],x]

[Out]

-((x*Sqrt[2 + x^(-4) + x^4])/(1 + x^4)) + (x^5*Sqrt[2 + x^(-4) + x^4])/(3*(1 + x^4))

Rule 1351

Int[((a_) + (c_.)*(x_)^(n_.) + (b_.)*(x_)^(mn_))^(p_), x_Symbol] :> Dist[(x^(n*FracPart[p])*(a + b/x^n + c*x^n
)^FracPart[p])/(b + a*x^n + c*x^(2*n))^FracPart[p], Int[(b + a*x^n + c*x^(2*n))^p/x^(n*p), x], x] /; FreeQ[{a,
b, c, n, p}, x] && EqQ[mn, -n] &&  !IntegerQ[p] && PosQ[n]

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \sqrt{2+\frac{1}{x^4}+x^4} \, dx &=\frac{\left (x^2 \sqrt{2+\frac{1}{x^4}+x^4}\right ) \int \frac{\sqrt{1+2 x^4+x^8}}{x^2} \, dx}{\sqrt{1+2 x^4+x^8}}\\ &=\frac{\left (x^2 \sqrt{2+\frac{1}{x^4}+x^4}\right ) \int \frac{1+x^4}{x^2} \, dx}{1+x^4}\\ &=\frac{\left (x^2 \sqrt{2+\frac{1}{x^4}+x^4}\right ) \int \left (\frac{1}{x^2}+x^2\right ) \, dx}{1+x^4}\\ &=-\frac{x \sqrt{2+\frac{1}{x^4}+x^4}}{1+x^4}+\frac{x^5 \sqrt{2+\frac{1}{x^4}+x^4}}{3 \left (1+x^4\right )}\\ \end{align*}

Mathematica [A]  time = 0.0088593, size = 29, normalized size = 0.59 \[ \frac{x \left (x^4-3\right ) \sqrt{x^4+\frac{1}{x^4}+2}}{3 \left (x^4+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + x^(-4) + x^4],x]

[Out]

(x*(-3 + x^4)*Sqrt[2 + x^(-4) + x^4])/(3*(1 + x^4))

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Maple [A]  time = 0.004, size = 32, normalized size = 0.7 \begin{align*}{\frac{ \left ({x}^{4}-3 \right ) x}{3\,{x}^{4}+3}\sqrt{{\frac{{x}^{8}+2\,{x}^{4}+1}{{x}^{4}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x*(x^4-3)*((x^8+2*x^4+1)/x^4)^(1/2)/(x^4+1)

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Maxima [A]  time = 1.4332, size = 14, normalized size = 0.29 \begin{align*} \frac{x^{4} - 3}{3 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^4 - 3)/x

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Fricas [A]  time = 1.4382, size = 23, normalized size = 0.47 \begin{align*} \frac{x^{4} - 3}{3 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^4 - 3)/x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{4} + 2 + \frac{1}{x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x**4 + 2 + x**(-4)), x)

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Giac [A]  time = 1.09628, size = 15, normalized size = 0.31 \begin{align*} \frac{1}{3} \, x^{3} - \frac{1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3 - 1/x