3.49 \(\int x^3 \sqrt{1+x^2} \, dx\)

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

[Out]

-(1 + x^2)^(3/2)/3 + (1 + x^2)^(5/2)/5

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Rubi [A]  time = 0.0101029, 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} \[ \frac{1}{5} \left (x^2+1\right )^{5/2}-\frac{1}{3} \left (x^2+1\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[1 + x^2],x]

[Out]

-(1 + x^2)^(3/2)/3 + (1 + x^2)^(5/2)/5

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

Mathematica [A]  time = 0.0059802, size = 20, normalized size = 0.74 \[ \frac{1}{15} \left (x^2+1\right )^{3/2} \left (3 x^2-2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[1 + x^2],x]

[Out]

((1 + x^2)^(3/2)*(-2 + 3*x^2))/15

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Maple [A]  time = 0.004, size = 17, normalized size = 0.6 \begin{align*}{\frac{3\,{x}^{2}-2}{15} \left ({x}^{2}+1 \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(x^2+1)^(3/2)*(3*x^2-2)

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Maxima [A]  time = 1.475, size = 30, normalized size = 1.11 \begin{align*} \frac{1}{5} \,{\left (x^{2} + 1\right )}^{\frac{3}{2}} x^{2} - \frac{2}{15} \,{\left (x^{2} + 1\right )}^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^2 + 1)^(3/2)*x^2 - 2/15*(x^2 + 1)^(3/2)

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Fricas [A]  time = 1.98449, size = 51, normalized size = 1.89 \begin{align*} \frac{1}{15} \,{\left (3 \, x^{4} + x^{2} - 2\right )} \sqrt{x^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*x^4 + x^2 - 2)*sqrt(x^2 + 1)

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Sympy [A]  time = 0.663862, size = 37, normalized size = 1.37 \begin{align*} \frac{x^{4} \sqrt{x^{2} + 1}}{5} + \frac{x^{2} \sqrt{x^{2} + 1}}{15} - \frac{2 \sqrt{x^{2} + 1}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4*sqrt(x**2 + 1)/5 + x**2*sqrt(x**2 + 1)/15 - 2*sqrt(x**2 + 1)/15

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Giac [A]  time = 1.0514, size = 26, normalized size = 0.96 \begin{align*} \frac{1}{5} \,{\left (x^{2} + 1\right )}^{\frac{5}{2}} - \frac{1}{3} \,{\left (x^{2} + 1\right )}^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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