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

Optimal. Leaf size=38 \[ -\frac{1}{8 b \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0251851, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1107, 607} \[ -\frac{1}{8 b \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1107

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

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}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx &=\frac{1}{2} \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 )\\ &=-\frac{1}{8 b \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 = 0.0106135, size = 27, normalized size = 0.71 \[ -\frac{a+b x^2}{8 b \left (\left (a+b x^2\right )^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + b*x^2)/(8*b*((a + b*x^2)^2)^(5/2))

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Maple [A]  time = 0.173, size = 24, normalized size = 0.6 \begin{align*} -{\frac{b{x}^{2}+a}{8\,b} \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/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

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

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Maxima [A]  time = 0.967018, size = 24, normalized size = 0.63 \begin{align*} -\frac{1}{8 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x^{2} + \frac{a}{b}\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/((b^2)^(5/2)*(x^2 + a/b)^4)

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Fricas [A]  time = 1.23912, size = 95, normalized size = 2.5 \begin{align*} -\frac{1}{8 \,{\left (b^{5} x^{8} + 4 \, a b^{4} x^{6} + 6 \, a^{2} b^{3} x^{4} + 4 \, a^{3} b^{2} x^{2} + a^{4} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\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/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

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

[Out]

sage0*x