3.232 \(\int x^2 \sqrt{b x^2+c x^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0486033, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2016, 2000} \[ \frac{\left (b x^2+c x^4\right )^{3/2}}{5 c x}-\frac{2 b \left (b x^2+c x^4\right )^{3/2}}{15 c^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[b*x^2 + c*x^4],x]

[Out]

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

Rule 2016

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*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2000

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

Rubi steps

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

Mathematica [A]  time = 0.01787, size = 35, normalized size = 0.67 \[ \frac{\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (3 c x^2-2 b\right )}{15 c^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[b*x^2 + c*x^4],x]

[Out]

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

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Maple [A]  time = 0.047, size = 39, normalized size = 0.8 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -3\,c{x}^{2}+2\,b \right ) }{15\,{c}^{2}x}\sqrt{c{x}^{4}+b{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*x^4+b*x^2)^(1/2),x)

[Out]

-1/15*(c*x^2+b)*(-3*c*x^2+2*b)*(c*x^4+b*x^2)^(1/2)/c^2/x

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Maxima [A]  time = 0.997549, size = 46, normalized size = 0.88 \begin{align*} \frac{{\left (3 \, c^{2} x^{4} + b c x^{2} - 2 \, b^{2}\right )} \sqrt{c x^{2} + b}}{15 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/15*(3*c^2*x^4 + b*c*x^2 - 2*b^2)*sqrt(c*x^2 + b)/c^2

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Fricas [A]  time = 1.5388, size = 86, normalized size = 1.65 \begin{align*} \frac{{\left (3 \, c^{2} x^{4} + b c x^{2} - 2 \, b^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{15 \, c^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/15*(3*c^2*x^4 + b*c*x^2 - 2*b^2)*sqrt(c*x^4 + b*x^2)/(c^2*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(x**2*sqrt(x**2*(b + c*x**2)), x)

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Giac [A]  time = 1.26036, size = 57, normalized size = 1.1 \begin{align*} \frac{2 \, b^{\frac{5}{2}} \mathrm{sgn}\left (x\right )}{15 \, c^{2}} + \frac{{\left (3 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} - 5 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b\right )} \mathrm{sgn}\left (x\right )}{15 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

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