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

Optimal. Leaf size=32 \[ \frac{a}{b^2 \sqrt{a+b x^2}}+\frac{\sqrt{a+b x^2}}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0223081, antiderivative size = 32, 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} \[ \frac{a}{b^2 \sqrt{a+b x^2}}+\frac{\sqrt{a+b x^2}}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*x^2)^(3/2),x]

[Out]

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

Mathematica [A]  time = 0.0108361, size = 24, normalized size = 0.75 \[ \frac{2 a+b x^2}{b^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b*x^2)^(3/2),x]

[Out]

(2*a + b*x^2)/(b^2*Sqrt[a + b*x^2])

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Maple [A]  time = 0.003, size = 23, normalized size = 0.7 \begin{align*}{\frac{b{x}^{2}+2\,a}{{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.29934, size = 66, normalized size = 2.06 \begin{align*} \frac{{\left (b x^{2} + 2 \, a\right )} \sqrt{b x^{2} + a}}{b^{3} x^{2} + a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.558642, size = 41, normalized size = 1.28 \begin{align*} \begin{cases} \frac{2 a}{b^{2} \sqrt{a + b x^{2}}} + \frac{x^{2}}{b \sqrt{a + b x^{2}}} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 a^{\frac{3}{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)**(3/2),x)

[Out]

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

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Giac [A]  time = 2.11577, size = 34, normalized size = 1.06 \begin{align*} \frac{\sqrt{b x^{2} + a} + \frac{a}{\sqrt{b x^{2} + a}}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(b*x^2 + a) + a/sqrt(b*x^2 + a))/b^2