3.3038 \(\int \frac {1}{1+4 (x^{2 n})^{\frac {1}{n}}} \, dx\)

Optimal. Leaf size=34 \[ \frac {1}{2} x \left (x^{2 n}\right )^{\left .-\frac {1}{2}\right /n} \tan ^{-1}\left (2 \left (x^{2 n}\right )^{\left .\frac {1}{2}\right /n}\right ) \]

[Out]

1/2*x*arctan(2*(x^(2*n))^(1/2/n))/((x^(2*n))^(1/2/n))

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Rubi [A]  time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {254, 203} \[ \frac {1}{2} x \left (x^{2 n}\right )^{\left .-\frac {1}{2}\right /n} \tan ^{-1}\left (2 \left (x^{2 n}\right )^{\left .\frac {1}{2}\right /n}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + 4*(x^(2*n))^n^(-1))^(-1),x]

[Out]

(x*ArcTan[2*(x^(2*n))^(1/(2*n))])/(2*(x^(2*n))^(1/(2*n)))

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])

Rule 254

Int[((a_) + (b_.)*((c_.)*(x_)^(q_.))^(n_))^(p_.), x_Symbol] :> Dist[x/(c*x^q)^(1/q), Subst[Int[(a + b*x^(n*q))
^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, n, p, q}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

Rubi steps

\begin {align*} \int \frac {1}{1+4 \left (x^{2 n}\right )^{\frac {1}{n}}} \, dx &=\left (x \left (x^{2 n}\right )^{\left .-\frac {1}{2}\right /n}\right ) \operatorname {Subst}\left (\int \frac {1}{1+4 x^2} \, dx,x,\left (x^{2 n}\right )^{\left .\frac {1}{2}\right /n}\right )\\ &=\frac {1}{2} x \left (x^{2 n}\right )^{\left .-\frac {1}{2}\right /n} \tan ^{-1}\left (2 \left (x^{2 n}\right )^{\left .\frac {1}{2}\right /n}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.00 \[ \frac {1}{2} x \left (x^{2 n}\right )^{\left .-\frac {1}{2}\right /n} \tan ^{-1}\left (2 \left (x^{2 n}\right )^{\left .\frac {1}{2}\right /n}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 4*(x^(2*n))^n^(-1))^(-1),x]

[Out]

(x*ArcTan[2*(x^(2*n))^(1/(2*n))])/(2*(x^(2*n))^(1/(2*n)))

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fricas [A]  time = 0.87, size = 6, normalized size = 0.18 \[ \frac {1}{2} \, \arctan \left (2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(2*x)

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giac [A]  time = 0.17, size = 6, normalized size = 0.18 \[ \frac {1}{2} \, \arctan \left (2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(2*x)

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maple [A]  time = 0.16, size = 42, normalized size = 1.24 \[ \frac {x \left (x^{2 n}\right )^{-\frac {1}{2 n}} \arctan \left (2 \left (x^{2 n}\right )^{\frac {1}{n}} \left (x^{2 n}\right )^{-\frac {1}{2 n}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x/((x^(2*n))^(1/2/n))*arctan(2*(x^(2*n))^(1/n)/((x^(2*n))^(1/2/n)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{4 \, {\left (x^{2 \, n}\right )}^{\left (\frac {1}{n}\right )} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{4\,{\left (x^{2\,n}\right )}^{1/n}+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*(x^(2*n))^(1/n) + 1),x)

[Out]

int(1/(4*(x^(2*n))^(1/n) + 1), x)

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sympy [A]  time = 0.13, size = 5, normalized size = 0.15 \[ \frac {\operatorname {atan}{\left (2 x \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(2*x)/2

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