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]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 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])

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]]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.61, size = 34, normalized size = 0.89 \[ \frac {{\left (3 \, b^{2} x^{4} + a b x^{2} - 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{15 \, b^{2}} \]

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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giac [A]  time = 0.60, size = 29, normalized size = 0.76 \[ \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} - 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a}{15 \, b^{2}} \]

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

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maple [A]  time = 0.01, size = 25, normalized size = 0.66 \[ -\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (-3 b \,x^{2}+2 a \right )}{15 b^{2}} \]

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 [A]  time = 1.34, size = 33, normalized size = 0.87 \[ \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} x^{2}}{5 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a}{15 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.65, size = 33, normalized size = 0.87 \[ \sqrt {b\,x^2+a}\,\left (\frac {x^4}{5}-\frac {2\,a^2}{15\,b^2}+\frac {a\,x^2}{15\,b}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^2)^(1/2),x)

[Out]

(a + b*x^2)^(1/2)*(x^4/5 - (2*a^2)/(15*b^2) + (a*x^2)/(15*b))

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sympy [A]  time = 0.32, size = 63, normalized size = 1.66 \[ \begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases} \]

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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