3.1.20 \(\int x^2 \sqrt [4]{1+x^3} \, dx\)

Optimal. Leaf size=13 \[ \frac {4}{15} \left (x^3+1\right )^{5/4} \]

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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 = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {261} \begin {gather*} \frac {4}{15} \left (x^3+1\right )^{5/4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(1 + x^3)^(1/4),x]

[Out]

(4*(1 + x^3)^(5/4))/15

Rule 261

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^2 \sqrt [4]{1+x^3} \, dx &=\frac {4}{15} \left (1+x^3\right )^{5/4}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \begin {gather*} \frac {4}{15} \left (x^3+1\right )^{5/4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(1 + x^3)^(1/4),x]

[Out]

(4*(1 + x^3)^(5/4))/15

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IntegrateAlgebraic [A]  time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} \frac {4}{15} \left (1+x^3\right )^{5/4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2*(1 + x^3)^(1/4),x]

[Out]

(4*(1 + x^3)^(5/4))/15

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fricas [A]  time = 0.45, size = 9, normalized size = 0.69 \begin {gather*} \frac {4}{15} \, {\left (x^{3} + 1\right )}^{\frac {5}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x^3+1)^(1/4),x, algorithm="fricas")

[Out]

4/15*(x^3 + 1)^(5/4)

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giac [A]  time = 0.36, size = 9, normalized size = 0.69 \begin {gather*} \frac {4}{15} \, {\left (x^{3} + 1\right )}^{\frac {5}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x^3+1)^(1/4),x, algorithm="giac")

[Out]

4/15*(x^3 + 1)^(5/4)

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maple [A]  time = 0.05, size = 10, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {4 \left (x^{3}+1\right )^{\frac {5}{4}}}{15}\)	\(10\)
default	\(\frac {4 \left (x^{3}+1\right )^{\frac {5}{4}}}{15}\)	\(10\)
risch	\(\frac {4 \left (x^{3}+1\right )^{\frac {5}{4}}}{15}\)	\(10\)
trager	\(\left (\frac {4}{15}+\frac {4 x^{3}}{15}\right ) \left (x^{3}+1\right )^{\frac {1}{4}}\)	\(16\)
meijerg	\(\frac {\hypergeom \left (\left [-\frac {1}{4}, 1\right ], \relax [2], -x^{3}\right ) x^{3}}{3}\)	\(17\)
gosper	\(\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (x^{3}+1\right )^{\frac {1}{4}}}{15}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^3+1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

4/15*(x^3+1)^(5/4)

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maxima [A]  time = 0.65, size = 9, normalized size = 0.69 \begin {gather*} \frac {4}{15} \, {\left (x^{3} + 1\right )}^{\frac {5}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x^3+1)^(1/4),x, algorithm="maxima")

[Out]

4/15*(x^3 + 1)^(5/4)

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mupad [B]  time = 0.12, size = 9, normalized size = 0.69 \begin {gather*} \frac {4\,{\left (x^3+1\right )}^{5/4}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^3 + 1)^(1/4),x)

[Out]

(4*(x^3 + 1)^(5/4))/15

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sympy [B]  time = 0.21, size = 26, normalized size = 2.00 \begin {gather*} \frac {4 x^{3} \sqrt [4]{x^{3} + 1}}{15} + \frac {4 \sqrt [4]{x^{3} + 1}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(x**3+1)**(1/4),x)

[Out]

4*x**3*(x**3 + 1)**(1/4)/15 + 4*(x**3 + 1)**(1/4)/15

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