3.30 \(\int x^2 (a^2+2 a b x^3+b^2 x^6)^{3/2} \, dx\)

Optimal. Leaf size=36 \[ \frac{\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{12 b} \]

[Out]

((a + b*x^3)*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2))/(12*b)

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Rubi [A]  time = 0.02958, antiderivative size = 36, 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 = {1352, 609} \[ \frac{\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{12 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2))/(12*b)

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 )^{3/2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, 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 )^{3/2}}{12 b}\\ \end{align*}

Mathematica [A]  time = 0.0182238, size = 60, normalized size = 1.67 \[ \frac{x^3 \sqrt{\left (a+b x^3\right )^2} \left (6 a^2 b x^3+4 a^3+4 a b^2 x^6+b^3 x^9\right )}{12 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*Sqrt[(a + b*x^3)^2]*(4*a^3 + 6*a^2*b*x^3 + 4*a*b^2*x^6 + b^3*x^9))/(12*(a + b*x^3))

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Maple [A]  time = 0.006, size = 57, normalized size = 1.6 \begin{align*}{\frac{{x}^{3} \left ({b}^{3}{x}^{9}+4\,a{b}^{2}{x}^{6}+6\,{a}^{2}b{x}^{3}+4\,{a}^{3} \right ) }{12\, \left ( b{x}^{3}+a \right ) ^{3}} \left ( \left ( b{x}^{3}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

1/12*x^3*(b^3*x^9+4*a*b^2*x^6+6*a^2*b*x^3+4*a^3)*((b*x^3+a)^2)^(3/2)/(b*x^3+a)^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.67701, size = 82, normalized size = 2.28 \begin{align*} \frac{1}{12} \, b^{3} x^{12} + \frac{1}{3} \, a b^{2} x^{9} + \frac{1}{2} \, a^{2} b x^{6} + \frac{1}{3} \, a^{3} x^{3} \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)^(3/2),x, algorithm="fricas")

[Out]

1/12*b^3*x^12 + 1/3*a*b^2*x^9 + 1/2*a^2*b*x^6 + 1/3*a^3*x^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac{3}{2}}\, dx \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)**(3/2),x)

[Out]

Integral(x**2*((a + b*x**3)**2)**(3/2), x)

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Giac [A]  time = 1.11209, size = 59, normalized size = 1.64 \begin{align*} \frac{1}{12} \,{\left (b^{3} x^{12} + 4 \, a b^{2} x^{9} + 6 \, a^{2} b x^{6} + 4 \, a^{3} x^{3}\right )} \mathrm{sgn}\left (b x^{3} + a\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)^(3/2),x, algorithm="giac")

[Out]

1/12*(b^3*x^12 + 4*a*b^2*x^9 + 6*a^2*b*x^6 + 4*a^3*x^3)*sgn(b*x^3 + a)