3.1.13 \(\int x \sqrt [3]{1+x^2} \, dx\) [13]

Optimal. Leaf size=13 \[ \frac {3}{8} \left (1+x^2\right )^{4/3} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {267} \begin {gather*} \frac {3}{8} \left (x^2+1\right )^{4/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(1 + x^2)^(1/3),x]

[Out]

(3*(1 + x^2)^(4/3))/8

Rule 267

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

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 13, normalized size = 1.00 \begin {gather*} \frac {3}{8} \left (1+x^2\right )^{4/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(1 + x^2)^(1/3),x]

[Out]

(3*(1 + x^2)^(4/3))/8

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Maple [A]
time = 0.24, size = 10, normalized size = 0.77

method	result	size
gosper	\(\frac {3 \left (x^{2}+1\right )^{\frac {4}{3}}}{8}\)	\(10\)
derivativedivides	\(\frac {3 \left (x^{2}+1\right )^{\frac {4}{3}}}{8}\)	\(10\)
default	\(\frac {3 \left (x^{2}+1\right )^{\frac {4}{3}}}{8}\)	\(10\)
risch	\(\frac {3 \left (x^{2}+1\right )^{\frac {4}{3}}}{8}\)	\(10\)
trager	\(\left (\frac {3}{8}+\frac {3 x^{2}}{8}\right ) \left (x^{2}+1\right )^{\frac {1}{3}}\)	\(16\)
meijerg	\(\frac {x^{2} \hypergeom \left (\left [-\frac {1}{3}, 1\right ], \left [2\right ], -x^{2}\right )}{2}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^2+1)^(1/3),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.25, size = 9, normalized size = 0.69 \begin {gather*} \frac {3}{8} \, {\left (x^{2} + 1\right )}^{\frac {4}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(x^2 + 1)^(4/3)

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Fricas [A]
time = 0.35, size = 9, normalized size = 0.69 \begin {gather*} \frac {3}{8} \, {\left (x^{2} + 1\right )}^{\frac {4}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(x^2 + 1)^(4/3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (10) = 20\).
time = 0.06, size = 26, normalized size = 2.00 \begin {gather*} \frac {3 x^{2} \sqrt [3]{x^{2} + 1}}{8} + \frac {3 \sqrt [3]{x^{2} + 1}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x**2*(x**2 + 1)**(1/3)/8 + 3*(x**2 + 1)**(1/3)/8

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Giac [A]
time = 0.40, size = 9, normalized size = 0.69 \begin {gather*} \frac {3}{8} \, {\left (x^{2} + 1\right )}^{\frac {4}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(x^2 + 1)^(4/3)

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Mupad [B]
time = 0.05, size = 9, normalized size = 0.69 \begin {gather*} \frac {3\,{\left (x^2+1\right )}^{4/3}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^2 + 1)^(1/3),x)

[Out]

(3*(x^2 + 1)^(4/3))/8

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