3.2944 \(\int x^2 \sqrt{a+b (c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=36 \[ \frac{2 x^3 \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2}}{9 b \left (c x^2\right )^{3/2}} \]

[Out]

(2*x^3*(a + b*(c*x^2)^(3/2))^(3/2))/(9*b*(c*x^2)^(3/2))

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Rubi [A]  time = 0.0173357, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {368, 261} \[ \frac{2 x^3 \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2}}{9 b \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a + b*(c*x^2)^(3/2)],x]

[Out]

(2*x^3*(a + b*(c*x^2)^(3/2))^(3/2))/(9*b*(c*x^2)^(3/2))

Rule 368

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*((c*x^q
)^(1/q))^(m + 1)), Subst[Int[x^m*(a + b*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q
}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int x^2 \sqrt{a+b \left (c x^2\right )^{3/2}} \, dx &=\frac{x^3 \operatorname{Subst}\left (\int x^2 \sqrt{a+b x^3} \, dx,x,\sqrt{c x^2}\right )}{\left (c x^2\right )^{3/2}}\\ &=\frac{2 x^3 \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2}}{9 b \left (c x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0105376, size = 36, normalized size = 1. \[ \frac{2 x^3 \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2}}{9 b \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[a + b*(c*x^2)^(3/2)],x]

[Out]

(2*x^3*(a + b*(c*x^2)^(3/2))^(3/2))/(9*b*(c*x^2)^(3/2))

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Maple [A]  time = 0.006, size = 29, normalized size = 0.8 \begin{align*}{\frac{2\,{x}^{3}}{9\,b} \left ( a+b \left ( c{x}^{2} \right ) ^{{\frac{3}{2}}} \right ) ^{{\frac{3}{2}}} \left ( c{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*x^3*(a+b*(c*x^2)^(3/2))^(3/2)/b/(c*x^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\frac{2 \,{\left (b c^{\frac{3}{2}} x^{3} + a\right )}^{\frac{3}{2}}{\left (c - \sqrt{c}\right )}}{9 \, b c^{\frac{3}{2}}}}{c + 1} + \frac{{\left (b c^{\frac{3}{2}} x^{3} + a\right )}^{\frac{3}{2}}}{3 \,{\left (c^{2} + c\right )} b \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c - sqrt(c))*integrate(sqrt(b*c^(3/2)*x^3 + a)*x^2, x)/(c + 1) + 1/3*(b*c^(3/2)*x^3 + a)^(3/2)/((c^2 + c)*b*s
qrt(c))

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Fricas [A]  time = 1.28523, size = 99, normalized size = 2.75 \begin{align*} \frac{2 \,{\left (b c^{2} x^{4} + \sqrt{c x^{2}} a\right )} \sqrt{\sqrt{c x^{2}} b c x^{2} + a}}{9 \, b c^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(b*c^2*x^4 + sqrt(c*x^2)*a)*sqrt(sqrt(c*x^2)*b*c*x^2 + a)/(b*c^2*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(a + b*(c*x**2)**(3/2)), x)

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Giac [A]  time = 1.17058, size = 27, normalized size = 0.75 \begin{align*} \frac{2 \,{\left (b c^{\frac{3}{2}} x^{3} + a\right )}^{\frac{3}{2}}}{9 \, b c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(b*c^(3/2)*x^3 + a)^(3/2)/(b*c^(3/2))