\(\int \frac {x^3}{(1+x^2)^{2/3}} \, dx\) [122]

Optimal result

Integrand size = 13, antiderivative size = 18 \[ \int \frac {x^3}{\left (1+x^2\right )^{2/3}} \, dx=\frac {3}{8} \left (-3+x^2\right ) \sqrt [3]{1+x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \frac {x^3}{\left (1+x^2\right )^{2/3}} \, dx=\frac {3}{8} \left (x^2+1\right )^{4/3}-\frac {3}{2} \sqrt [3]{x^2+1} \]

[In]

Int[x^3/(1 + x^2)^(2/3),x]

[Out]

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

Rule 45

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 272

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\left (1+x^2\right )^{2/3}} \, dx=\frac {3}{8} \left (-3+x^2\right ) \sqrt [3]{1+x^2} \]

[In]

Integrate[x^3/(1 + x^2)^(2/3),x]

[Out]

(3*(-3 + x^2)*(1 + x^2)^(1/3))/8

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {3 \left (x^{2}-3\right ) \left (x^{2}+1\right )^{\frac {1}{3}}}{8}\)	\(15\)
risch	\(\frac {3 \left (x^{2}-3\right ) \left (x^{2}+1\right )^{\frac {1}{3}}}{8}\)	\(15\)
pseudoelliptic	\(\frac {3 \left (x^{2}-3\right ) \left (x^{2}+1\right )^{\frac {1}{3}}}{8}\)	\(15\)
trager	\(\left (\frac {3 x^{2}}{8}-\frac {9}{8}\right ) \left (x^{2}+1\right )^{\frac {1}{3}}\)	\(16\)
meijerg	\(\frac {x^{4} \operatorname {hypergeom}\left (\left [\frac {2}{3}, 2\right ], \left [3\right ], -x^{2}\right )}{4}\)	\(17\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{\left (1+x^2\right )^{2/3}} \, dx=\frac {3}{8} \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 3\right )} \]

[In]

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

[Out]

3/8*(x^2 + 1)^(1/3)*(x^2 - 3)

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {x^3}{\left (1+x^2\right )^{2/3}} \, dx=\frac {3 x^{2} \sqrt [3]{x^{2} + 1}}{8} - \frac {9 \sqrt [3]{x^{2} + 1}}{8} \]

[In]

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

[Out]

3*x**2*(x**2 + 1)**(1/3)/8 - 9*(x**2 + 1)**(1/3)/8

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {x^3}{\left (1+x^2\right )^{2/3}} \, dx=\frac {3}{8} \, {\left (x^{2} + 1\right )}^{\frac {4}{3}} - \frac {3}{2} \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {x^3}{\left (1+x^2\right )^{2/3}} \, dx=\frac {3}{8} \, {\left (x^{2} + 1\right )}^{\frac {4}{3}} - \frac {3}{2} \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.98 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{\left (1+x^2\right )^{2/3}} \, dx=\frac {3\,{\left (x^2+1\right )}^{1/3}\,\left (x^2-3\right )}{8} \]

[In]

int(x^3/(x^2 + 1)^(2/3),x)

[Out]

(3*(x^2 + 1)^(1/3)*(x^2 - 3))/8