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

Optimal. Leaf size=25 \[ \frac{2 \left (x^2 \left (a+b x^3\right )\right )^{3/2}}{9 b x^3} \]

[Out]

(2*(x^2*(a + b*x^3))^(3/2))/(9*b*x^3)

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Rubi [A]  time = 0.010443, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {1588} \[ \frac{2 \left (x^2 \left (a+b x^3\right )\right )^{3/2}}{9 b x^3} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[x^2*(a + b*x^3)],x]

[Out]

(2*(x^2*(a + b*x^3))^(3/2))/(9*b*x^3)

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0119565, size = 25, normalized size = 1. \[ \frac{2 \left (x^2 \left (a+b x^3\right )\right )^{3/2}}{9 b x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[x^2*(a + b*x^3)],x]

[Out]

(2*(x^2*(a + b*x^3))^(3/2))/(9*b*x^3)

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Maple [A]  time = 0.005, size = 29, normalized size = 1.2 \begin{align*}{\frac{2\,b{x}^{3}+2\,a}{9\,bx}\sqrt{{x}^{2} \left ( b{x}^{3}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(b*x^3+a)*(x^2*(b*x^3+a))^(1/2)/b/x

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Maxima [A]  time = 1.24923, size = 19, normalized size = 0.76 \begin{align*} \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}}}{9 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(b*x^3 + a)^(3/2)/b

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Fricas [A]  time = 0.908743, size = 58, normalized size = 2.32 \begin{align*} \frac{2 \, \sqrt{b x^{5} + a x^{2}}{\left (b x^{3} + a\right )}}{9 \, b x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*sqrt(b*x^5 + a*x^2)*(b*x^3 + a)/(b*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.45822, size = 36, normalized size = 1.44 \begin{align*} \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} \mathrm{sgn}\left (x\right )}{9 \, b} - \frac{2 \, a^{\frac{3}{2}} \mathrm{sgn}\left (x\right )}{9 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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