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

Optimal. Leaf size=55 \[ -\frac {a^2}{b^3 \sqrt {a+b x^2}}-\frac {2 a \sqrt {a+b x^2}}{b^3}+\frac {\left (a+b x^2\right )^{3/2}}{3 b^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 55, 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} \[ -\frac {a^2}{b^3 \sqrt {a+b x^2}}-\frac {2 a \sqrt {a+b x^2}}{b^3}+\frac {\left (a+b x^2\right )^{3/2}}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^2)^(3/2),x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 38, normalized size = 0.69 \[ \frac {-8 a^2-4 a b x^2+b^2 x^4}{3 b^3 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^2)^(3/2),x]

[Out]

(-8*a^2 - 4*a*b*x^2 + b^2*x^4)/(3*b^3*Sqrt[a + b*x^2])

________________________________________________________________________________________

fricas [A]  time = 0.93, size = 46, normalized size = 0.84 \[ \frac {{\left (b^{2} x^{4} - 4 \, a b x^{2} - 8 \, a^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b^2*x^4 - 4*a*b*x^2 - 8*a^2)*sqrt(b*x^2 + a)/(b^4*x^2 + a*b^3)

________________________________________________________________________________________

giac [A]  time = 0.98, size = 52, normalized size = 0.95 \[ -\frac {a^{2}}{\sqrt {b x^{2} + a} b^{3}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{6} - 6 \, \sqrt {b x^{2} + a} a b^{6}}{3 \, b^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2/(sqrt(b*x^2 + a)*b^3) + 1/3*((b*x^2 + a)^(3/2)*b^6 - 6*sqrt(b*x^2 + a)*a*b^6)/b^9

________________________________________________________________________________________

maple [A]  time = 0.01, size = 36, normalized size = 0.65 \[ -\frac {-b^{2} x^{4}+4 a b \,x^{2}+8 a^{2}}{3 \sqrt {b \,x^{2}+a}\, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-b^2*x^4+4*a*b*x^2+8*a^2)/(b*x^2+a)^(1/2)/b^3

________________________________________________________________________________________

maxima [A]  time = 1.31, size = 53, normalized size = 0.96 \[ \frac {x^{4}}{3 \, \sqrt {b x^{2} + a} b} - \frac {4 \, a x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} - \frac {8 \, a^{2}}{3 \, \sqrt {b x^{2} + a} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^4/(sqrt(b*x^2 + a)*b) - 4/3*a*x^2/(sqrt(b*x^2 + a)*b^2) - 8/3*a^2/(sqrt(b*x^2 + a)*b^3)

________________________________________________________________________________________

mupad [B]  time = 4.72, size = 41, normalized size = 0.75 \[ -\frac {6\,a\,\left (b\,x^2+a\right )-{\left (b\,x^2+a\right )}^2+3\,a^2}{3\,b^3\,\sqrt {b\,x^2+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^2)^(3/2),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.94, size = 68, normalized size = 1.24 \[ \begin {cases} - \frac {8 a^{2}}{3 b^{3} \sqrt {a + b x^{2}}} - \frac {4 a x^{2}}{3 b^{2} \sqrt {a + b x^{2}}} + \frac {x^{4}}{3 b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{6}}{6 a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-8*a**2/(3*b**3*sqrt(a + b*x**2)) - 4*a*x**2/(3*b**2*sqrt(a + b*x**2)) + x**4/(3*b*sqrt(a + b*x**2)
), Ne(b, 0)), (x**6/(6*a**(3/2)), True))

________________________________________________________________________________________