3.130 \(\int x^2 (a^2+2 a b x^3+b^2 x^6)^p \, dx\)

Optimal. Leaf size=41 \[ \frac{\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b (2 p+1)} \]

[Out]

((a + b*x^3)*(a^2 + 2*a*b*x^3 + b^2*x^6)^p)/(3*b*(1 + 2*p))

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Rubi [A]  time = 0.0276851, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1352, 609} \[ \frac{\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b (2 p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]

[Out]

((a + b*x^3)*(a^2 + 2*a*b*x^3 + b^2*x^6)^p)/(3*b*(1 + 2*p))

Rule 1352

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rubi steps

\begin{align*} \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx,x,x^3\right )\\ &=\frac{\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b (1+2 p)}\\ \end{align*}

Mathematica [A]  time = 0.0062623, size = 32, normalized size = 0.78 \[ \frac{\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p}{3 b (2 p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]

[Out]

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

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Maple [A]  time = 0.004, size = 40, normalized size = 1. \begin{align*}{\frac{ \left ( b{x}^{3}+a \right ) \left ({b}^{2}{x}^{6}+2\,ab{x}^{3}+{a}^{2} \right ) ^{p}}{3\,b \left ( 1+2\,p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b^2*x^6+2*a*b*x^3+a^2)^p,x)

[Out]

1/3*(b*x^3+a)*(b^2*x^6+2*a*b*x^3+a^2)^p/b/(1+2*p)

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Maxima [A]  time = 1.04373, size = 41, normalized size = 1. \begin{align*} \frac{{\left (b x^{3} + a\right )}{\left (b x^{3} + a\right )}^{2 \, p}}{3 \, b{\left (2 \, p + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.55186, size = 80, normalized size = 1.95 \begin{align*} \frac{{\left (b x^{3} + a\right )}{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{3 \,{\left (2 \, b p + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="fricas")

[Out]

1/3*(b*x^3 + a)*(b^2*x^6 + 2*a*b*x^3 + a^2)^p/(2*b*p + b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b**2*x**6+2*a*b*x**3+a**2)**p,x)

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.11458, size = 78, normalized size = 1.9 \begin{align*} \frac{{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b x^{3} +{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a}{3 \,{\left (2 \, b p + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="giac")

[Out]

1/3*((b^2*x^6 + 2*a*b*x^3 + a^2)^p*b*x^3 + (b^2*x^6 + 2*a*b*x^3 + a^2)^p*a)/(2*b*p + b)