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]

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

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Rubi [A]  time = 0.03, antiderivative size = 37, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.58, size = 22, normalized size = 0.59 \[ \frac {2 \, \sqrt {x^{3} + x^{2}} {\left (3 \, x^{2} + x - 2\right )}}{15 \, x} \]

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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giac [A]  time = 0.38, size = 48, normalized size = 1.30 \[ \frac {2}{15} \, {\left (3 \, {\left (x + 1\right )}^{\frac {5}{2}} - 10 \, {\left (x + 1\right )}^{\frac {3}{2}} + 15 \, \sqrt {x + 1}\right )} \mathrm {sgn}\relax (x) + \frac {2}{3} \, {\left ({\left (x + 1\right )}^{\frac {3}{2}} - 3 \, \sqrt {x + 1}\right )} \mathrm {sgn}\relax (x) + \frac {4}{15} \, \mathrm {sgn}\relax (x) \]

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

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maple [A]  time = 0.00, size = 23, normalized size = 0.62 \[ \frac {2 \left (x +1\right ) \left (3 x -2\right ) \sqrt {x^{3}+x^{2}}}{15 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 15, normalized size = 0.41 \[ \frac {2}{15} \, {\left (3 \, x^{2} + x - 2\right )} \sqrt {x + 1} \]

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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mupad [B]  time = 3.51, size = 22, normalized size = 0.59 \[ \frac {2\,\sqrt {x^3+x^2}\,\left (3\,x^2+x-2\right )}{15\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + x^3)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x^{3} + x^{2}}\, dx \]

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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