3.355 \(\int x^3 \sqrt{a+b x^2} \, dx\)

Optimal. Leaf size=38 \[ \frac{\left (a+b x^2\right )^{5/2}}{5 b^2}-\frac{a \left (a+b x^2\right )^{3/2}}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0216558, antiderivative size = 38, 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} \[ \frac{\left (a+b x^2\right )^{5/2}}{5 b^2}-\frac{a \left (a+b x^2\right )^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[a + b*x^2],x]

[Out]

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

Mathematica [A]  time = 0.0137205, size = 28, normalized size = 0.74 \[ \frac{\left (a+b x^2\right )^{3/2} \left (3 b x^2-2 a\right )}{15 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[a + b*x^2],x]

[Out]

((a + b*x^2)^(3/2)*(-2*a + 3*b*x^2))/(15*b^2)

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Maple [A]  time = 0.003, size = 25, normalized size = 0.7 \begin{align*} -{\frac{-3\,b{x}^{2}+2\,a}{15\,{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)^(1/2),x)

[Out]

-1/15*(b*x^2+a)^(3/2)*(-3*b*x^2+2*a)/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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58821, size = 76, normalized size = 2. \begin{align*} \frac{{\left (3 \, b^{2} x^{4} + a b x^{2} - 2 \, a^{2}\right )} \sqrt{b x^{2} + a}}{15 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

1/15*(3*b^2*x^4 + a*b*x^2 - 2*a^2)*sqrt(b*x^2 + a)/b^2

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Sympy [A]  time = 0.303815, size = 63, normalized size = 1.66 \begin{align*} \begin{cases} - \frac{2 a^{2} \sqrt{a + b x^{2}}}{15 b^{2}} + \frac{a x^{2} \sqrt{a + b x^{2}}}{15 b} + \frac{x^{4} \sqrt{a + b x^{2}}}{5} & \text{for}\: b \neq 0 \\\frac{\sqrt{a} x^{4}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x**2+a)**(1/2),x)

[Out]

Piecewise((-2*a**2*sqrt(a + b*x**2)/(15*b**2) + a*x**2*sqrt(a + b*x**2)/(15*b) + x**4*sqrt(a + b*x**2)/5, Ne(b
, 0)), (sqrt(a)*x**4/4, True))

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Giac [A]  time = 1.96966, size = 39, normalized size = 1.03 \begin{align*} \frac{3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a}{15 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/15*(3*(b*x^2 + a)^(5/2) - 5*(b*x^2 + a)^(3/2)*a)/b^2