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

Optimal. Leaf size=27 \[ \frac {3}{10} \left (x^2+1\right )^{5/3}-\frac {3}{4} \left (x^2+1\right )^{2/3} \]

[Out]

-3/4*(x^2+1)^(2/3)+3/10*(x^2+1)^(5/3)

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {266, 43} \[ \frac {3}{10} \left (x^2+1\right )^{5/3}-\frac {3}{4} \left (x^2+1\right )^{2/3} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(1 + x^2)^(1/3),x]

[Out]

(-3*(1 + x^2)^(2/3))/4 + (3*(1 + x^2)^(5/3))/10

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt [3]{1+x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{\sqrt [3]{1+x}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {1}{\sqrt [3]{1+x}}+(1+x)^{2/3}\right ) \, dx,x,x^2\right )\\ &=-\frac {3}{4} \left (1+x^2\right )^{2/3}+\frac {3}{10} \left (1+x^2\right )^{5/3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.74 \[ \frac {3}{20} \left (x^2+1\right )^{2/3} \left (2 x^2-3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(1 + x^2)^(1/3),x]

[Out]

(3*(1 + x^2)^(2/3)*(-3 + 2*x^2))/20

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fricas [A]  time = 0.41, size = 16, normalized size = 0.59 \[ \frac {3}{20} \, {\left (2 \, x^{2} - 3\right )} {\left (x^{2} + 1\right )}^{\frac {2}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/20*(2*x^2 - 3)*(x^2 + 1)^(2/3)

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giac [A]  time = 1.10, size = 19, normalized size = 0.70 \[ \frac {3}{10} \, {\left (x^{2} + 1\right )}^{\frac {5}{3}} - \frac {3}{4} \, {\left (x^{2} + 1\right )}^{\frac {2}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(x^2 + 1)^(5/3) - 3/4*(x^2 + 1)^(2/3)

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maple [A]  time = 0.00, size = 17, normalized size = 0.63 \[ \frac {3 \left (x^{2}+1\right )^{\frac {2}{3}} \left (2 x^{2}-3\right )}{20} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/20*(x^2+1)^(2/3)*(2*x^2-3)

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maxima [A]  time = 0.48, size = 19, normalized size = 0.70 \[ \frac {3}{10} \, {\left (x^{2} + 1\right )}^{\frac {5}{3}} - \frac {3}{4} \, {\left (x^{2} + 1\right )}^{\frac {2}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(x^2 + 1)^(5/3) - 3/4*(x^2 + 1)^(2/3)

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mupad [B]  time = 0.26, size = 16, normalized size = 0.59 \[ \frac {3\,{\left (x^2+1\right )}^{2/3}\,\left (2\,x^2-3\right )}{20} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^2 + 1)^(1/3),x)

[Out]

(3*(x^2 + 1)^(2/3)*(2*x^2 - 3))/20

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sympy [A]  time = 0.94, size = 26, normalized size = 0.96 \[ \frac {3 x^{2} \left (x^{2} + 1\right )^{\frac {2}{3}}}{10} - \frac {9 \left (x^{2} + 1\right )^{\frac {2}{3}}}{20} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x**2*(x**2 + 1)**(2/3)/10 - 9*(x**2 + 1)**(2/3)/20

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