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

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

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Rubi [A]  time = 0.03, antiderivative size = 38, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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 {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.71 \begin {gather*} -\frac {a+b x^2}{8 b \left (\left (a+b x^2\right )^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 0.75, size = 200, normalized size = 5.26 \begin {gather*} \frac {a^4 b+\sqrt {b^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a^3-a^2 b x^2+a b^2 x^4-b^3 x^6\right )+b^5 x^8}{b x^8 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-8 a^3 b^5-24 a^2 b^6 x^2-24 a b^7 x^4-8 b^8 x^6\right )+b \sqrt {b^2} x^8 \left (8 a^4 b^4+32 a^3 b^5 x^2+48 a^2 b^6 x^4+32 a b^7 x^6+8 b^8 x^8\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*b + b^5*x^8 + Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(a^3 - a^2*b*x^2 + a*b^2*x^4 - b^3*x^6))/(b*x^8*S
qrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-8*a^3*b^5 - 24*a^2*b^6*x^2 - 24*a*b^7*x^4 - 8*b^8*x^6) + b*Sqrt[b^2]*x^8*(8*a
^4*b^4 + 32*a^3*b^5*x^2 + 48*a^2*b^6*x^4 + 32*a*b^7*x^6 + 8*b^8*x^8))

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fricas [A]  time = 4.60, size = 48, normalized size = 1.26 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.21, size = 24, normalized size = 0.63 \begin {gather*} -\frac {1}{8 \, {\left (b x^{2} + a\right )}^{4} b \mathrm {sgn}\left (b x^{2} + a\right )} \end {gather*}

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]

-1/8/((b*x^2 + a)^4*b*sgn(b*x^2 + a))

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maple [A]  time = 0.01, size = 24, normalized size = 0.63 \begin {gather*} -\frac {b \,x^{2}+a}{8 \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} b} \end {gather*}

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 = 1.32, size = 48, normalized size = 1.26 \begin {gather*} -\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 {gather*}

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^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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mupad [B]  time = 4.27, size = 34, normalized size = 0.89 \begin {gather*} -\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,b\,{\left (b\,x^2+a\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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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