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

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

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 {gather*}

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 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)]

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]

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*}

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Mathematica [A]  time = 0.02, size = 60, normalized size = 1.67 \begin {gather*} \frac {x^3 \sqrt {\left (a+b x^3\right )^2} \left (4 a^3+6 a^2 b x^3+4 a b^2 x^6+b^3 x^9\right )}{12 \left (a+b x^3\right )} \end {gather*}

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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IntegrateAlgebraic [A]  time = 8.02, size = 60, normalized size = 1.67 \begin {gather*} \frac {x^3 \sqrt {\left (a+b x^3\right )^2} \left (4 a^3+6 a^2 b x^3+4 a b^2 x^6+b^3 x^9\right )}{12 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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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fricas [A]  time = 1.47, size = 35, normalized size = 0.97 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.41, size = 44, normalized size = 1.22 \begin {gather*} \frac {1}{12} \, {\left (2 \, {\left (b x^{6} + 2 \, a x^{3}\right )} a^{2} + {\left (b x^{6} + 2 \, a x^{3}\right )}^{2} b\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \end {gather*}

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*(2*(b*x^6 + 2*a*x^3)*a^2 + (b*x^6 + 2*a*x^3)^2*b)*sgn(b*x^3 + a)

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maple [A]  time = 0.01, size = 57, normalized size = 1.58 \begin {gather*} \frac {\left (b^{3} x^{9}+4 a \,b^{2} x^{6}+6 a^{2} b \,x^{3}+4 a^{3}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}} x^{3}}{12 \left (b \,x^{3}+a \right )^{3}} \end {gather*}

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 [A]  time = 0.45, size = 52, normalized size = 1.44 \begin {gather*} \frac {1}{12} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} x^{3} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a}{12 \, b} \end {gather*}

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]

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

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mupad [B]  time = 1.22, size = 36, normalized size = 1.00 \begin {gather*} \frac {\left (b^2\,x^3+a\,b\right )\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}}{12\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

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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