3.599 \(\int x^2 (a+b x^3)^p \, dx\)

Optimal. Leaf size=23 \[ \frac{\left (a+b x^3\right )^{p+1}}{3 b (p+1)} \]

[Out]

(a + b*x^3)^(1 + p)/(3*b*(1 + p))

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Rubi [A]  time = 0.0061137, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {261} \[ \frac{\left (a+b x^3\right )^{p+1}}{3 b (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^3)^p,x]

[Out]

(a + b*x^3)^(1 + p)/(3*b*(1 + p))

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 \left (a+b x^3\right )^p \, dx &=\frac{\left (a+b x^3\right )^{1+p}}{3 b (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0054607, size = 23, normalized size = 1. \[ \frac{\left (a+b x^3\right )^{p+1}}{3 b (p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^3)^p,x]

[Out]

(a + b*x^3)^(1 + p)/(3*b*(1 + p))

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Maple [A]  time = 0.004, size = 22, normalized size = 1. \begin{align*}{\frac{ \left ( b{x}^{3}+a \right ) ^{1+p}}{3\,b \left ( 1+p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^3+a)^(1+p)/b/(1+p)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54857, size = 55, normalized size = 2.39 \begin{align*} \frac{{\left (b x^{3} + a\right )}{\left (b x^{3} + a\right )}^{p}}{3 \,{\left (b p + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^3 + a)*(b*x^3 + a)^p/(b*p + b)

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Sympy [A]  time = 1.91722, size = 134, normalized size = 5.83 \begin{align*} \begin{cases} \frac{x^{3}}{3 a} & \text{for}\: b = 0 \wedge p = -1 \\\frac{a^{p} x^{3}}{3} & \text{for}\: b = 0 \\\frac{\log{\left (- \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{\frac{1}{b}} + x \right )}}{3 b} + \frac{\log{\left (4 \left (-1\right )^{\frac{2}{3}} a^{\frac{2}{3}} \left (\frac{1}{b}\right )^{\frac{2}{3}} + 4 \sqrt [3]{-1} \sqrt [3]{a} x \sqrt [3]{\frac{1}{b}} + 4 x^{2} \right )}}{3 b} & \text{for}\: p = -1 \\\frac{a \left (a + b x^{3}\right )^{p}}{3 b p + 3 b} + \frac{b x^{3} \left (a + b x^{3}\right )^{p}}{3 b p + 3 b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x**3+a)**p,x)

[Out]

Piecewise((x**3/(3*a), Eq(b, 0) & Eq(p, -1)), (a**p*x**3/3, Eq(b, 0)), (log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3)
+ x)/(3*b) + log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*(1/b)**(1/3) + 4*x**2)/(3*b),
Eq(p, -1)), (a*(a + b*x**3)**p/(3*b*p + 3*b) + b*x**3*(a + b*x**3)**p/(3*b*p + 3*b), True))

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Giac [A]  time = 1.09005, size = 28, normalized size = 1.22 \begin{align*} \frac{{\left (b x^{3} + a\right )}^{p + 1}}{3 \, b{\left (p + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^3 + a)^(p + 1)/(b*(p + 1))