3.6.7 \(\int \frac {x^5}{(a+b x^2)^{5/2}} \, dx\) [507]

Optimal. Leaf size=54 \[ -\frac {a^2}{3 b^3 \left (a+b x^2\right )^{3/2}}+\frac {2 a}{b^3 \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2}}{b^3} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 54, 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^2}{3 b^3 \left (a+b x^2\right )^{3/2}}+\frac {2 a}{b^3 \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2}}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^2)^(5/2),x]

[Out]

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

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

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Mathematica [A]
time = 0.03, size = 39, normalized size = 0.72 \begin {gather*} \frac {8 a^2+12 a b x^2+3 b^2 x^4}{3 b^3 \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^2)^(5/2),x]

[Out]

(8*a^2 + 12*a*b*x^2 + 3*b^2*x^4)/(3*b^3*(a + b*x^2)^(3/2))

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Maple [A]
time = 0.05, size = 57, normalized size = 1.06

method	result	size
gosper	\(\frac {3 b^{2} x^{4}+12 a b \,x^{2}+8 a^{2}}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{3}}\)	\(36\)
trager	\(\frac {3 b^{2} x^{4}+12 a b \,x^{2}+8 a^{2}}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{3}}\)	\(36\)
default	\(\frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}} b}-\frac {4 a \left (-\frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}} b}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )}{b}\)	\(57\)
risch	\(\frac {\sqrt {b \,x^{2}+a}}{b^{3}}+\frac {\sqrt {b \,x^{2}+a}\, \left (6 b \,x^{2}+5 a \right ) a}{3 b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.32, size = 52, normalized size = 0.96 \begin {gather*} \frac {x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {4 \, a x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.64, size = 58, normalized size = 1.07 \begin {gather*} \frac {{\left (3 \, b^{2} x^{4} + 12 \, a b x^{2} + 8 \, a^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*b^2*x^4 + 12*a*b*x^2 + 8*a^2)*sqrt(b*x^2 + a)/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (48) = 96\).
time = 0.36, size = 138, normalized size = 2.56 \begin {gather*} \begin {cases} \frac {8 a^{2}}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} + \frac {12 a b x^{2}}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} + \frac {3 b^{2} x^{4}}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{6}}{6 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*a**2/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2)) + 12*a*b*x**2/(3*a*b**3*sqrt(a +
b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2)) + 3*b**2*x**4/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2
)), Ne(b, 0)), (x**6/(6*a**(5/2)), True))

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Giac [A]
time = 0.49, size = 44, normalized size = 0.81 \begin {gather*} \frac {\sqrt {b x^{2} + a}}{b^{3}} + \frac {6 \, {\left (b x^{2} + a\right )} a - a^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.20, size = 38, normalized size = 0.70 \begin {gather*} \frac {2\,a\,\left (b\,x^2+a\right )+{\left (b\,x^2+a\right )}^2-\frac {a^2}{3}}{b^3\,{\left (b\,x^2+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^2)^(5/2),x)

[Out]

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

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