\(\int \frac {x^5}{\sqrt {5+x^2}} \, dx\) [40]

Optimal result

Integrand size = 13, antiderivative size = 38 \[ \int \frac {x^5}{\sqrt {5+x^2}} \, dx=25 \sqrt {5+x^2}-\frac {10}{3} \left (5+x^2\right )^{3/2}+\frac {1}{5} \left (5+x^2\right )^{5/2} \]

[Out]

-10/3*(x^2+5)^(3/2)+1/5*(x^2+5)^(5/2)+25*(x^2+5)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \frac {x^5}{\sqrt {5+x^2}} \, dx=\frac {1}{5} \left (x^2+5\right )^{5/2}-\frac {10}{3} \left (x^2+5\right )^{3/2}+25 \sqrt {x^2+5} \]

[In]

Int[x^5/Sqrt[5 + x^2],x]

[Out]

25*Sqrt[5 + x^2] - (10*(5 + x^2)^(3/2))/3 + (5 + x^2)^(5/2)/5

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\sqrt {5+x}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {25}{\sqrt {5+x}}-10 \sqrt {5+x}+(5+x)^{3/2}\right ) \, dx,x,x^2\right ) \\ & = 25 \sqrt {5+x^2}-\frac {10}{3} \left (5+x^2\right )^{3/2}+\frac {1}{5} \left (5+x^2\right )^{5/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {x^5}{\sqrt {5+x^2}} \, dx=\frac {1}{15} \sqrt {5+x^2} \left (200-20 x^2+3 x^4\right ) \]

[In]

Integrate[x^5/Sqrt[5 + x^2],x]

[Out]

(Sqrt[5 + x^2]*(200 - 20*x^2 + 3*x^4))/15

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.55

method	result	size
trager	\(\sqrt {x^{2}+5}\, \left (\frac {1}{5} x^{4}-\frac {4}{3} x^{2}+\frac {40}{3}\right )\)	\(21\)
gosper	\(\frac {\sqrt {x^{2}+5}\, \left (3 x^{4}-20 x^{2}+200\right )}{15}\)	\(22\)
risch	\(\frac {\sqrt {x^{2}+5}\, \left (3 x^{4}-20 x^{2}+200\right )}{15}\)	\(22\)
pseudoelliptic	\(\frac {\sqrt {x^{2}+5}\, \left (3 x^{4}-20 x^{2}+200\right )}{15}\)	\(22\)
default	\(\frac {x^{4} \sqrt {x^{2}+5}}{5}-\frac {4 x^{2} \sqrt {x^{2}+5}}{3}+\frac {40 \sqrt {x^{2}+5}}{3}\)	\(35\)
meijerg	\(\frac {25 \sqrt {5}\, \left (-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (\frac {6}{25} x^{4}-\frac {8}{5} x^{2}+16\right ) \sqrt {1+\frac {x^{2}}{5}}}{15}\right )}{2 \sqrt {\pi }}\)	\(41\)

[In]

int(x^5/(x^2+5)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(x^2+5)^(1/2)*(1/5*x^4-4/3*x^2+40/3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.55 \[ \int \frac {x^5}{\sqrt {5+x^2}} \, dx=\frac {1}{15} \, {\left (3 \, x^{4} - 20 \, x^{2} + 200\right )} \sqrt {x^{2} + 5} \]

[In]

integrate(x^5/(x^2+5)^(1/2),x, algorithm="fricas")

[Out]

1/15*(3*x^4 - 20*x^2 + 200)*sqrt(x^2 + 5)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int \frac {x^5}{\sqrt {5+x^2}} \, dx=\frac {x^{4} \sqrt {x^{2} + 5}}{5} - \frac {4 x^{2} \sqrt {x^{2} + 5}}{3} + \frac {40 \sqrt {x^{2} + 5}}{3} \]

[In]

integrate(x**5/(x**2+5)**(1/2),x)

[Out]

x**4*sqrt(x**2 + 5)/5 - 4*x**2*sqrt(x**2 + 5)/3 + 40*sqrt(x**2 + 5)/3

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {x^5}{\sqrt {5+x^2}} \, dx=\frac {1}{5} \, \sqrt {x^{2} + 5} x^{4} - \frac {4}{3} \, \sqrt {x^{2} + 5} x^{2} + \frac {40}{3} \, \sqrt {x^{2} + 5} \]

[In]

integrate(x^5/(x^2+5)^(1/2),x, algorithm="maxima")

[Out]

1/5*sqrt(x^2 + 5)*x^4 - 4/3*sqrt(x^2 + 5)*x^2 + 40/3*sqrt(x^2 + 5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {x^5}{\sqrt {5+x^2}} \, dx=\frac {1}{5} \, {\left (x^{2} + 5\right )}^{\frac {5}{2}} - \frac {10}{3} \, {\left (x^{2} + 5\right )}^{\frac {3}{2}} + 25 \, \sqrt {x^{2} + 5} \]

[In]

integrate(x^5/(x^2+5)^(1/2),x, algorithm="giac")

[Out]

1/5*(x^2 + 5)^(5/2) - 10/3*(x^2 + 5)^(3/2) + 25*sqrt(x^2 + 5)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.53 \[ \int \frac {x^5}{\sqrt {5+x^2}} \, dx=\sqrt {x^2+5}\,\left (\frac {x^4}{5}-\frac {4\,x^2}{3}+\frac {40}{3}\right ) \]

[In]

int(x^5/(x^2 + 5)^(1/2),x)

[Out]

(x^2 + 5)^(1/2)*(x^4/5 - (4*x^2)/3 + 40/3)