3.649 \(\int \frac{x^3}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=69 \[ \frac{a}{8 b^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}-\frac{1}{6 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

[Out]

-1/(6*b^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)) + a/(8*b^2*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))

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Rubi [A]  time = 0.0537152, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1111, 640, 607} \[ \frac{a}{8 b^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}-\frac{1}{6 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

-1/(6*b^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)) + a/(8*b^2*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rule 640

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

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^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx,x,x^2\right )\\ &=-\frac{1}{6 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx,x,x^2\right )}{2 b}\\ &=-\frac{1}{6 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac{a}{8 b^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0136127, size = 39, normalized size = 0.57 \[ \frac{-a-4 b x^2}{24 b^2 \left (a+b x^2\right )^3 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(-a - 4*b*x^2)/(24*b^2*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])

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Maple [A]  time = 0.167, size = 32, normalized size = 0.5 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( 4\,b{x}^{2}+a \right ) }{24\,{b}^{2}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

-1/24*(b*x^2+a)*(4*b*x^2+a)/b^2/((b*x^2+a)^2)^(5/2)

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Maxima [A]  time = 0.963772, size = 65, normalized size = 0.94 \begin{align*} -\frac{1}{6 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{3}{2}} b^{2}} + \frac{a}{8 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x^{2} + \frac{a}{b}\right )}^{4} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

-1/6/((b^2*x^4 + 2*a*b*x^2 + a^2)^(3/2)*b^2) + 1/8*a/((b^2)^(5/2)*(x^2 + a/b)^4*b)

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Fricas [A]  time = 1.22864, size = 117, normalized size = 1.7 \begin{align*} -\frac{4 \, b x^{2} + a}{24 \,{\left (b^{6} x^{8} + 4 \, a b^{5} x^{6} + 6 \, a^{2} b^{4} x^{4} + 4 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/24*(4*b*x^2 + a)/(b^6*x^8 + 4*a*b^5*x^6 + 6*a^2*b^4*x^4 + 4*a^3*b^3*x^2 + a^4*b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**3/((a + b*x**2)**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^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x