\(\int x^2 (a+b x^3)^p \, dx\) [599]

Optimal result

Integrand size = 13, antiderivative size = 23 \[ \int x^2 \left (a+b x^3\right )^p \, dx=\frac {\left (a+b x^3\right )^{1+p}}{3 b (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {267} \[ \int x^2 \left (a+b x^3\right )^p \, dx=\frac {\left (a+b x^3\right )^{p+1}}{3 b (p+1)} \]

[In]

Int[x^2*(a + b*x^3)^p,x]

[Out]

(a + b*x^3)^(1 + p)/(3*b*(1 + p))

Rule 267

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b x^3\right )^p \, dx=\frac {\left (a+b x^3\right )^{1+p}}{3 b (1+p)} \]

[In]

Integrate[x^2*(a + b*x^3)^p,x]

[Out]

(a + b*x^3)^(1 + p)/(3*b*(1 + p))

Maple [A] (verified)

Time = 3.85 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
gosper	\(\frac {\left (b \,x^{3}+a \right )^{1+p}}{3 b \left (1+p \right )}\)	\(22\)
derivativedivides	\(\frac {\left (b \,x^{3}+a \right )^{1+p}}{3 b \left (1+p \right )}\)	\(22\)
default	\(\frac {\left (b \,x^{3}+a \right )^{1+p}}{3 b \left (1+p \right )}\)	\(22\)
risch	\(\frac {\left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{p}}{3 b \left (1+p \right )}\)	\(27\)
parallelrisch	\(\frac {x^{3} \left (b \,x^{3}+a \right )^{p} a b +a^{2} \left (b \,x^{3}+a \right )^{p}}{3 a \left (1+p \right ) b}\)	\(43\)
norman	\(\frac {x^{3} {\mathrm e}^{p \ln \left (b \,x^{3}+a \right )}}{3+3 p}+\frac {a \,{\mathrm e}^{p \ln \left (b \,x^{3}+a \right )}}{3 b \left (1+p \right )}\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int x^2 \left (a+b x^3\right )^p \, dx=\frac {{\left (b x^{3} + a\right )} {\left (b x^{3} + a\right )}^{p}}{3 \, {\left (b p + b\right )}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (15) = 30\).

Time = 0.56 (sec) , antiderivative size = 104, normalized size of antiderivative = 4.52 \[ \int x^2 \left (a+b x^3\right )^p \, dx=\begin {cases} \frac {x^{3}}{3 a} & \text {for}\: b = 0 \wedge p = -1 \\\frac {a^{p} x^{3}}{3} & \text {for}\: b = 0 \\\frac {\log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 b} + \frac {\log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{3 b} & \text {for}\: p = -1 \\\frac {a \left (a + b x^{3}\right )^{p}}{3 b p + 3 b} + \frac {b x^{3} \left (a + b x^{3}\right )^{p}}{3 b p + 3 b} & \text {otherwise} \end {cases} \]

[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(x - (-a/b)**(1/3))/(3*b) + log(4*x
**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int x^2 \left (a+b x^3\right )^p \, dx=\frac {{\left (b x^{3} + a\right )}^{p + 1}}{3 \, b {\left (p + 1\right )}} \]

[In]

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

[Out]

1/3*(b*x^3 + a)^(p + 1)/(b*(p + 1))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int x^2 \left (a+b x^3\right )^p \, dx=\frac {{\left (b x^{3} + a\right )}^{p + 1}}{3 \, b {\left (p + 1\right )}} \]

[In]

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

[Out]

1/3*(b*x^3 + a)^(p + 1)/(b*(p + 1))

Mupad [B] (verification not implemented)

Time = 5.55 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int x^2 \left (a+b x^3\right )^p \, dx=\frac {{\left (b\,x^3+a\right )}^{p+1}}{3\,b\,\left (p+1\right )} \]

[In]

int(x^2*(a + b*x^3)^p,x)

[Out]

(a + b*x^3)^(p + 1)/(3*b*(p + 1))