3.139 \(\int x^2 (b+2 c x^3) (b x^3+c x^6)^p \, dx\)

Optimal. Leaf size=24 \[ \frac{\left (b x^3+c x^6\right )^{p+1}}{3 (p+1)} \]

[Out]

(b*x^3 + c*x^6)^(1 + p)/(3*(1 + p))

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Rubi [A]  time = 0.0196933, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {1588} \[ \frac{\left (b x^3+c x^6\right )^{p+1}}{3 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^3 + c*x^6)^(1 + p)/(3*(1 + p))

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

Mathematica [C]  time = 0.0757323, size = 97, normalized size = 4.04 \[ \frac{x^3 \left (x^3 \left (b+c x^3\right )\right )^p \left (\frac{c x^3}{b}+1\right )^{-p} \left (2 c (p+1) x^3 \, _2F_1\left (-p,p+2;p+3;-\frac{c x^3}{b}\right )+b (p+2) \, _2F_1\left (-p,p+1;p+2;-\frac{c x^3}{b}\right )\right )}{3 (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(x^3*(b + c*x^3))^p*(b*(2 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, -((c*x^3)/b)] + 2*c*(1 + p)*x^3*Hyperg
eometric2F1[-p, 2 + p, 3 + p, -((c*x^3)/b)]))/(3*(1 + p)*(2 + p)*(1 + (c*x^3)/b)^p)

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Maple [A]  time = 0.004, size = 31, normalized size = 1.3 \begin{align*}{\frac{{x}^{3} \left ( c{x}^{3}+b \right ) \left ( c{x}^{6}+b{x}^{3} \right ) ^{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)^p,x)

[Out]

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

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Maxima [A]  time = 1.1813, size = 47, normalized size = 1.96 \begin{align*} \frac{{\left (c x^{6} + b x^{3}\right )} e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right )\right )}}{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)^p,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.29856, size = 63, normalized size = 2.62 \begin{align*} \frac{{\left (c x^{6} + b x^{3}\right )}{\left (c x^{6} + b x^{3}\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)^p,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.12514, size = 59, normalized size = 2.46 \begin{align*} \frac{{\left (c x^{6} + b x^{3}\right )}^{p} c x^{6} +{\left (c x^{6} + b x^{3}\right )}^{p} b x^{3}}{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)^p,x, algorithm="giac")

[Out]

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