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

Optimal. Leaf size=37 \[ \frac{2 \left (x^3+x^2\right )^{3/2}}{5 x^2}-\frac{4 \left (x^3+x^2\right )^{3/2}}{15 x^3} \]

[Out]

(-4*(x^2 + x^3)^(3/2))/(15*x^3) + (2*(x^2 + x^3)^(3/2))/(5*x^2)

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Rubi [A]  time = 0.026565, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2002, 2014} \[ \frac{2 \left (x^3+x^2\right )^{3/2}}{5 x^2}-\frac{4 \left (x^3+x^2\right )^{3/2}}{15 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x^2 + x^3],x]

[Out]

(-4*(x^2 + x^3)^(3/2))/(15*x^3) + (2*(x^2 + x^3)^(3/2))/(5*x^2)

Rule 2002

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[(b*(n*p + n - j + 1))/(a*(j*p + 1)), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

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

Mathematica [A]  time = 0.0084559, size = 23, normalized size = 0.62 \[ \frac{2 \left (x^2 (x+1)\right )^{3/2} (3 x-2)}{15 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x^2 + x^3],x]

[Out]

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

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Maple [A]  time = 0.002, size = 23, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2+2\,x \right ) \left ( 3\,x-2 \right ) }{15\,x}\sqrt{{x}^{3}+{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03624, size = 20, normalized size = 0.54 \begin{align*} \frac{2}{15} \,{\left (3 \, x^{2} + x - 2\right )} \sqrt{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*x^2 + x - 2)*sqrt(x + 1)

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Fricas [A]  time = 1.42531, size = 54, normalized size = 1.46 \begin{align*} \frac{2 \, \sqrt{x^{3} + x^{2}}{\left (3 \, x^{2} + x - 2\right )}}{15 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{3} + x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x**3 + x**2), x)

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Giac [A]  time = 1.15465, size = 32, normalized size = 0.86 \begin{align*} \frac{2}{15} \,{\left (3 \,{\left (x + 1\right )}^{\frac{5}{2}} - 5 \,{\left (x + 1\right )}^{\frac{3}{2}}\right )} \mathrm{sgn}\left (x\right ) + \frac{4}{15} \, \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*(x + 1)^(5/2) - 5*(x + 1)^(3/2))*sgn(x) + 4/15*sgn(x)