3.111 \(\int \frac{x^2}{(a^2+2 a b x^3+b^2 x^6)^{5/2}} \, dx\)

Optimal. Leaf size=38 \[ -\frac{1}{12 b \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0295249, antiderivative size = 38, 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, 607} \[ -\frac{1}{12 b \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

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

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 607

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

Rubi steps

\begin{align*} \int \frac{x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx,x,x^3\right )\\ &=-\frac{1}{12 b \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0096304, size = 27, normalized size = 0.71 \[ -\frac{a+b x^3}{12 b \left (\left (a+b x^3\right )^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

-(a + b*x^3)/(12*b*((a + b*x^3)^2)^(5/2))

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Maple [A]  time = 0.006, size = 24, normalized size = 0.6 \begin{align*} -{\frac{b{x}^{3}+a}{12\,b} \left ( \left ( b{x}^{3}+a \right ) ^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

-1/12*(b*x^3+a)/b/((b*x^3+a)^2)^(5/2)

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Maxima [A]  time = 1.05502, size = 24, normalized size = 0.63 \begin{align*} -\frac{1}{12 \,{\left (x^{3} + \frac{a}{b}\right )}^{4}{\left (b^{2}\right )}^{\frac{5}{2}}} \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)^(5/2),x, algorithm="maxima")

[Out]

-1/12/((x^3 + a/b)^4*(b^2)^(5/2))

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Fricas [A]  time = 1.74862, size = 97, normalized size = 2.55 \begin{align*} -\frac{1}{12 \,{\left (b^{5} x^{12} + 4 \, a b^{4} x^{9} + 6 \, a^{2} b^{3} x^{6} + 4 \, a^{3} b^{2} x^{3} + a^{4} 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)^(5/2),x, algorithm="fricas")

[Out]

-1/12/(b^5*x^12 + 4*a*b^4*x^9 + 6*a^2*b^3*x^6 + 4*a^3*b^2*x^3 + a^4*b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac{5}{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)**(5/2),x)

[Out]

Integral(x**2/((a + b*x**3)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \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)^(5/2),x, algorithm="giac")

[Out]

sage0*x