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

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

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {2 \left (x^2 \left (a+b x^3\right )\right )^{3/2}}{9 b x^3} \end {gather*}

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*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \begin {gather*} \frac {2 \left (x^2 \left (a+b x^3\right )\right )^{3/2}}{9 b x^3} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.02, size = 25, normalized size = 1.00 \begin {gather*} \frac {2 \left (a x^2+b x^5\right )^{3/2}}{9 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*Sqrt[x^2*(a + b*x^3)],x]

[Out]

(2*(a*x^2 + b*x^5)^(3/2))/(9*b*x^3)

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fricas [A]  time = 0.38, size = 28, normalized size = 1.12 \begin {gather*} \frac {2 \, \sqrt {b x^{5} + a x^{2}} {\left (b x^{3} + a\right )}}{9 \, b x} \end {gather*}

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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giac [A]  time = 0.17, size = 27, normalized size = 1.08 \begin {gather*} \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} \mathrm {sgn}\relax (x)}{9 \, b} - \frac {2 \, a^{\frac {3}{2}} \mathrm {sgn}\relax (x)}{9 \, b} \end {gather*}

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

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maple [A]  time = 0.05, size = 29, normalized size = 1.16 \begin {gather*} \frac {2 \left (b \,x^{3}+a \right ) \sqrt {\left (b \,x^{3}+a \right ) x^{2}}}{9 b x} \end {gather*}

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 = 1.46, size = 14, normalized size = 0.56 \begin {gather*} \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{9 \, b} \end {gather*}

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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mupad [B]  time = 5.23, size = 22, normalized size = 0.88 \begin {gather*} \frac {2\,{\left (b\,x^3+a\right )}^{3/2}\,\sqrt {x^2}}{9\,b\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^2*(a + b*x^3))^(1/2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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