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

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

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Rubi [A]  time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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 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 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])

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*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.57 \begin {gather*} \frac {-a-4 b x^2}{24 b^2 \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.69, size = 224, normalized size = 3.25 \begin {gather*} \frac {-3 a^5 b+\sqrt {b^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-3 a^4+3 a^3 b x^2-3 a^2 b^2 x^4+3 a b^3 x^6-4 b^4 x^8\right )+a b^5 x^8+4 b^6 x^{10}}{3 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-8 a^3 b^7-24 a^2 b^8 x^2-24 a b^9 x^4-8 b^{10} x^6\right )+3 \sqrt {b^2} x^8 \left (8 a^4 b^6+32 a^3 b^7 x^2+48 a^2 b^8 x^4+32 a b^9 x^6+8 b^{10} x^8\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^5*b + a*b^5*x^8 + 4*b^6*x^10 + Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-3*a^4 + 3*a^3*b*x^2 - 3*a^2*b
^2*x^4 + 3*a*b^3*x^6 - 4*b^4*x^8))/(3*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-8*a^3*b^7 - 24*a^2*b^8*x^2 - 24*a*
b^9*x^4 - 8*b^10*x^6) + 3*Sqrt[b^2]*x^8*(8*a^4*b^6 + 32*a^3*b^7*x^2 + 48*a^2*b^8*x^4 + 32*a*b^9*x^6 + 8*b^10*x
^8))

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fricas [A]  time = 1.21, size = 58, normalized size = 0.84 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.21, size = 32, normalized size = 0.46 \begin {gather*} -\frac {4 \, b x^{2} + a}{24 \, {\left (b x^{2} + a\right )}^{4} b^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \end {gather*}

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]

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

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maple [A]  time = 0.01, size = 32, normalized size = 0.46 \begin {gather*} -\frac {\left (b \,x^{2}+a \right ) \left (4 b \,x^{2}+a \right )}{24 \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} b^{2}} \end {gather*}

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 = 1.36, size = 58, normalized size = 0.84 \begin {gather*} -\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 {gather*}

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]

-1/24*(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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mupad [B]  time = 4.26, size = 42, normalized size = 0.61 \begin {gather*} -\frac {\left (4\,b\,x^2+a\right )\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{24\,b^2\,{\left (b\,x^2+a\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a + 4*b*x^2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(24*b^2*(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^{3}}{\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**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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