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

Optimal. Leaf size=22 \[ \frac {x \tan ^{-1}\left (2 \sqrt [4]{x^4}\right )}{2 \sqrt [4]{x^4}} \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {254, 203} \begin {gather*} \frac {x \tan ^{-1}\left (2 \sqrt [4]{x^4}\right )}{2 \sqrt [4]{x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 4*Sqrt[x^4])^(-1),x]

[Out]

(x*ArcTan[2*(x^4)^(1/4)])/(2*(x^4)^(1/4))

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 \sqrt {x^4}} \, dx &=\frac {x \operatorname {Subst}\left (\int \frac {1}{1+4 x^2} \, dx,x,\sqrt [4]{x^4}\right )}{\sqrt [4]{x^4}}\\ &=\frac {x \tan ^{-1}\left (2 \sqrt [4]{x^4}\right )}{2 \sqrt [4]{x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} \frac {x \tan ^{-1}\left (2 \sqrt [4]{x^4}\right )}{2 \sqrt [4]{x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 4*Sqrt[x^4])^(-1),x]

[Out]

(x*ArcTan[2*(x^4)^(1/4)])/(2*(x^4)^(1/4))

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IntegrateAlgebraic [B]  time = 0.17, size = 67, normalized size = 3.05 \begin {gather*} -\frac {1}{4} \tan ^{-1}\left (\frac {\sqrt {x^4}}{2 x^3}\right )-\frac {1}{4} \tanh ^{-1}\left (\frac {\sqrt {x^4}}{2 x^3}\right )-\frac {1}{8} \log (1-2 x)+\frac {1}{8} \log (2 x+1)+\frac {1}{4} \tan ^{-1}(2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + 4*Sqrt[x^4])^(-1),x]

[Out]

ArcTan[2*x]/4 - ArcTan[Sqrt[x^4]/(2*x^3)]/4 - ArcTanh[Sqrt[x^4]/(2*x^3)]/4 - Log[1 - 2*x]/8 + Log[1 + 2*x]/8

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fricas [A]  time = 1.01, size = 6, normalized size = 0.27 \begin {gather*} \frac {1}{2} \, \arctan \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(2*x)

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giac [A]  time = 0.15, size = 6, normalized size = 0.27 \begin {gather*} \frac {1}{2} \, \arctan \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(2*x)

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maple [A]  time = 0.05, size = 29, normalized size = 1.32 \begin {gather*} \frac {\arctan \left (2 \sqrt {\frac {\sqrt {x^{4}}}{x^{2}}}\, x \right )}{2 \sqrt {\frac {\sqrt {x^{4}}}{x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/((x^4)^(1/2)/x^2)^(1/2)*arctan(2*((x^4)^(1/2)/x^2)^(1/2)*x)

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maxima [A]  time = 1.36, size = 6, normalized size = 0.27 \begin {gather*} \frac {1}{2} \, \arctan \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(2*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {1}{4\,\sqrt {x^4}+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*(x^4)^(1/2) + 1),x)

[Out]

int(1/(4*(x^4)^(1/2) + 1), x)

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sympy [A]  time = 0.14, size = 5, normalized size = 0.23 \begin {gather*} \frac {\operatorname {atan}{\left (2 x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(2*x)/2

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