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

Optimal. Leaf size=36 \[ \frac{a}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac{1}{b^2 \sqrt{a+b x^2}} \]

[Out]

a/(3*b^2*(a + b*x^2)^(3/2)) - 1/(b^2*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.0215647, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac{a}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac{1}{b^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*x^2)^(5/2),x]

[Out]

a/(3*b^2*(a + b*x^2)^(3/2)) - 1/(b^2*Sqrt[a + b*x^2])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0109335, size = 28, normalized size = 0.78 \[ \frac{-2 a-3 b x^2}{3 b^2 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b*x^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.005, size = 25, normalized size = 0.7 \begin{align*} -{\frac{3\,b{x}^{2}+2\,a}{3\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.36117, size = 45, normalized size = 1.25 \begin{align*} -\frac{x^{2}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}} b} - \frac{2 \, a}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.3013, size = 97, normalized size = 2.69 \begin{align*} -\frac{{\left (3 \, b x^{2} + 2 \, a\right )} \sqrt{b x^{2} + a}}{3 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.01676, size = 92, normalized size = 2.56 \begin{align*} \begin{cases} - \frac{2 a}{3 a b^{2} \sqrt{a + b x^{2}} + 3 b^{3} x^{2} \sqrt{a + b x^{2}}} - \frac{3 b x^{2}}{3 a b^{2} \sqrt{a + b x^{2}} + 3 b^{3} x^{2} \sqrt{a + b x^{2}}} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 a^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*a/(3*a*b**2*sqrt(a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2)) - 3*b*x**2/(3*a*b**2*sqrt(a + b*x**
2) + 3*b**3*x**2*sqrt(a + b*x**2)), Ne(b, 0)), (x**4/(4*a**(5/2)), True))

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Giac [A]  time = 2.55065, size = 32, normalized size = 0.89 \begin{align*} -\frac{3 \, b x^{2} + 2 \, a}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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