### 3.131 $$\int x^2 (b+2 c x^3) (a+b x^3+c x^6)^p \, dx$$

Optimal. Leaf size=25 $\frac{\left (a+b x^3+c x^6\right )^{p+1}}{3 (p+1)}$

[Out]

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

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Rubi [A]  time = 0.0236525, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {1468, 629} $\frac{\left (a+b x^3+c x^6\right )^{p+1}}{3 (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 1468

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

Rule 629

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

Rubi steps

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

Mathematica [A]  time = 0.0123637, size = 25, normalized size = 1. $\frac{\left (a+b x^3+c x^6\right )^{p+1}}{3 (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 24, normalized size = 1. \begin{align*}{\frac{ \left ( c{x}^{6}+b{x}^{3}+a \right ) ^{1+p}}{3+3\,p}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.19067, size = 45, normalized size = 1.8 \begin{align*} \frac{{\left (c x^{6} + b x^{3} + a\right )}{\left (c x^{6} + b x^{3} + a\right )}^{p}}{3 \,{\left (p + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.08912, size = 74, normalized size = 2.96 \begin{align*} \frac{{\left (c x^{6} + b x^{3} + a\right )}{\left (c x^{6} + b x^{3} + a\right )}^{p}}{3 \,{\left (p + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.15765, size = 84, normalized size = 3.36 \begin{align*} \frac{{\left (c x^{6} + b x^{3} + a\right )}^{p} c x^{6} +{\left (c x^{6} + b x^{3} + a\right )}^{p} b x^{3} +{\left (c x^{6} + b x^{3} + a\right )}^{p} a}{3 \,{\left (p + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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