### 3.234 $$\int \frac{x^3}{\sqrt {1+x^2}} \, dx$$

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

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Rubi [A]  time = 0.0103061, antiderivative size = 27, normalized size of antiderivative = 1., 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 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]]

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])

Rubi steps

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

Mathematica [A]  time = 0.0074507, 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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Maple [A]  time = 0.003, size = 17, normalized size = 0.6 \begin{align*}{\frac{6\,{x}^{2}-9}{20} \left ({x}^{2}+1 \right ) ^{{\frac{2}{3}}}} \end{align*}

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 = 0.945974, size = 26, normalized size = 0.96 \begin{align*} \frac{3}{10} \,{\left (x^{2} + 1\right )}^{\frac{5}{3}} - \frac{3}{4} \,{\left (x^{2} + 1\right )}^{\frac{2}{3}} \end{align*}

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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Fricas [A]  time = 2.09868, size = 46, normalized size = 1.7 \begin{align*} \frac{3}{20} \,{\left (2 \, x^{2} - 3\right )}{\left (x^{2} + 1\right )}^{\frac{2}{3}} \end{align*}

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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Sympy [A]  time = 0.798445, size = 26, normalized size = 0.96 \begin{align*} \frac{3 x^{2} \left (x^{2} + 1\right )^{\frac{2}{3}}}{10} - \frac{9 \left (x^{2} + 1\right )^{\frac{2}{3}}}{20} \end{align*}

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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Giac [A]  time = 1.06786, size = 26, normalized size = 0.96 \begin{align*} \frac{3}{10} \,{\left (x^{2} + 1\right )}^{\frac{5}{3}} - \frac{3}{4} \,{\left (x^{2} + 1\right )}^{\frac{2}{3}} \end{align*}

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)