3.2384 \(\int \frac {1}{1+x^{2/3}} \, dx\)

Optimal. Leaf size=16 \[ 3 \sqrt [3]{x}-3 \tan ^{-1}\left (\sqrt [3]{x}\right ) \]

[Out]

3*x^(1/3)-3*arctan(x^(1/3))

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {243, 321, 203} \[ 3 \sqrt [3]{x}-3 \tan ^{-1}\left (\sqrt [3]{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^(2/3))^(-1),x]

[Out]

3*x^(1/3) - 3*ArcTan[x^(1/3)]

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 243

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k - 1)*(a + b*
x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, p}, x] && FractionQ[n]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
 c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {1}{1+x^{2/3}} \, dx &=3 \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \sqrt [3]{x}-3 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \sqrt [3]{x}-3 \tan ^{-1}\left (\sqrt [3]{x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.00 \[ 3 \sqrt [3]{x}-3 \tan ^{-1}\left (\sqrt [3]{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^(2/3))^(-1),x]

[Out]

3*x^(1/3) - 3*ArcTan[x^(1/3)]

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fricas [A]  time = 0.62, size = 12, normalized size = 0.75 \[ 3 \, x^{\frac {1}{3}} - 3 \, \arctan \left (x^{\frac {1}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x^(2/3)),x, algorithm="fricas")

[Out]

3*x^(1/3) - 3*arctan(x^(1/3))

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giac [A]  time = 0.15, size = 12, normalized size = 0.75 \[ 3 \, x^{\frac {1}{3}} - 3 \, \arctan \left (x^{\frac {1}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x^(2/3)),x, algorithm="giac")

[Out]

3*x^(1/3) - 3*arctan(x^(1/3))

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maple [B]  time = 0.04, size = 41, normalized size = 2.56 \[ \arctan \relax (x )-2 \arctan \left (x^{\frac {1}{3}}\right )-\arctan \left (2 x^{\frac {1}{3}}-\sqrt {3}\right )-\arctan \left (2 x^{\frac {1}{3}}+\sqrt {3}\right )+3 x^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+x^(2/3)),x)

[Out]

arctan(x)+3*x^(1/3)-2*arctan(x^(1/3))-arctan(2*x^(1/3)-3^(1/2))-arctan(2*x^(1/3)+3^(1/2))

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maxima [A]  time = 1.20, size = 12, normalized size = 0.75 \[ 3 \, x^{\frac {1}{3}} - 3 \, \arctan \left (x^{\frac {1}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x^(2/3)),x, algorithm="maxima")

[Out]

3*x^(1/3) - 3*arctan(x^(1/3))

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mupad [B]  time = 1.12, size = 12, normalized size = 0.75 \[ 3\,x^{1/3}-3\,\mathrm {atan}\left (x^{1/3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(2/3) + 1),x)

[Out]

3*x^(1/3) - 3*atan(x^(1/3))

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sympy [A]  time = 0.17, size = 14, normalized size = 0.88 \[ 3 \sqrt [3]{x} - 3 \operatorname {atan}{\left (\sqrt [3]{x} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x**(2/3)),x)

[Out]

3*x**(1/3) - 3*atan(x**(1/3))

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