3.5.98 \(\int \frac {x^3}{(a+b x^2)^{3/2}} \, dx\) [498]

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

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Rubi [A]
time = 0.02, antiderivative size = 32, 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 = {272, 45} \begin {gather*} \frac {a}{b^2 \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2}}{b^2} \end {gather*}

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 45

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 272

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^3}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x}{(a+b x)^{3/2}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {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*}

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Mathematica [A]
time = 0.02, size = 24, normalized size = 0.75 \begin {gather*} \frac {2 a+b x^2}{b^2 \sqrt {a+b x^2}} \end {gather*}

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 = 0.04, size = 33, normalized size = 1.03

method	result	size
gosper	\(\frac {b \,x^{2}+2 a}{\sqrt {b \,x^{2}+a}\, b^{2}}\)	\(23\)
trager	\(\frac {b \,x^{2}+2 a}{\sqrt {b \,x^{2}+a}\, b^{2}}\)	\(23\)
risch	\(\frac {a}{b^{2} \sqrt {b \,x^{2}+a}}+\frac {\sqrt {b \,x^{2}+a}}{b^{2}}\)	\(29\)
default	\(\frac {x^{2}}{\sqrt {b \,x^{2}+a}\, b}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 32, normalized size = 1.00 \begin {gather*} \frac {x^{2}}{\sqrt {b x^{2} + a} b} + \frac {2 \, a}{\sqrt {b x^{2} + a} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.14, size = 34, normalized size = 1.06 \begin {gather*} \frac {{\left (b x^{2} + 2 \, a\right )} \sqrt {b x^{2} + a}}{b^{3} x^{2} + a b^{2}} \end {gather*}

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 = 0.25, size = 41, normalized size = 1.28 \begin {gather*} \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 {gather*}

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 = 0.61, size = 32, normalized size = 1.00 \begin {gather*} \frac {\frac {\sqrt {b x^{2} + a}}{b} + \frac {a}{\sqrt {b x^{2} + a} b}}{b} \end {gather*}

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)/b + a/(sqrt(b*x^2 + a)*b))/b

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Mupad [B]
time = 4.70, size = 22, normalized size = 0.69 \begin {gather*} \frac {b\,x^2+2\,a}{b^2\,\sqrt {b\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b*x^2)^(3/2),x)

[Out]

(2*a + b*x^2)/(b^2*(a + b*x^2)^(1/2))

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