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*(2+1/x^4+x^4)^(1/2)/(x^4+1)+1/3*x^5*(2+1/x^4+x^4)^(1/2)/(x^4+1)

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Rubi [A]  time = 0.01, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]]

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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.40, size = 10, normalized size = 0.20 \[ \frac {x^{4} - 3}{3 \, x} \]

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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giac [A]  time = 1.12, size = 11, normalized size = 0.22 \[ \frac {1}{3} \, x^{3} - \frac {1}{x} \]

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

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maple [A]  time = 0.01, size = 32, normalized size = 0.65 \[ \frac {\left (x^{4}-3\right ) \sqrt {\frac {x^{8}+2 x^{4}+1}{x^{4}}}\, x}{3 x^{4}+3} \]

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 = 0.99, size = 10, normalized size = 0.20 \[ \frac {x^{4} - 3}{3 \, x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \sqrt {\frac {1}{x^4}+x^4+2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/x^4 + x^4 + 2)^(1/2),x)

[Out]

int((1/x^4 + x^4 + 2)^(1/2), x)

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

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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