3.234 \(\int \sqrt{a x^2+b x^3} \, dx\)

Optimal. Leaf size=52 \[ \frac{2 \left (a x^2+b x^3\right )^{3/2}}{5 b x^2}-\frac{4 a \left (a x^2+b x^3\right )^{3/2}}{15 b^2 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0430886, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2002, 2014} \[ \frac{2 \left (a x^2+b x^3\right )^{3/2}}{5 b x^2}-\frac{4 a \left (a x^2+b x^3\right )^{3/2}}{15 b^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3],x]

[Out]

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

Mathematica [A]  time = 0.0139493, size = 31, normalized size = 0.6 \[ \frac{2 \left (x^2 (a+b x)\right )^{3/2} (3 b x-2 a)}{15 b^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2 + b*x^3],x]

[Out]

(2*(x^2*(a + b*x))^(3/2)*(-2*a + 3*b*x))/(15*b^2*x^3)

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Maple [A]  time = 0.003, size = 35, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,bx+2\,a \right ) \left ( -3\,bx+2\,a \right ) }{15\,{b}^{2}x}\sqrt{b{x}^{3}+a{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.976278, size = 41, normalized size = 0.79 \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{2} + a b x - 2 \, a^{2}\right )} \sqrt{b x + a}}{15 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.873243, size = 84, normalized size = 1.62 \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{2} + a b x - 2 \, a^{2}\right )} \sqrt{b x^{3} + a x^{2}}}{15 \, b^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*x**2 + b*x**3), x)

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Giac [A]  time = 1.1859, size = 51, normalized size = 0.98 \begin{align*} \frac{4 \, a^{\frac{5}{2}} \mathrm{sgn}\left (x\right )}{15 \, b^{2}} + \frac{2 \,{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} \mathrm{sgn}\left (x\right )}{15 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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