\(\int \frac {x^5}{\sqrt {a+b x^3}} \, dx\) [406]

Optimal result

Integrand size = 15, antiderivative size = 38 \[ \int \frac {x^5}{\sqrt {a+b x^3}} \, dx=-\frac {2 a \sqrt {a+b x^3}}{3 b^2}+\frac {2 \left (a+b x^3\right )^{3/2}}{9 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \frac {x^5}{\sqrt {a+b x^3}} \, dx=\frac {2 \left (a+b x^3\right )^{3/2}}{9 b^2}-\frac {2 a \sqrt {a+b x^3}}{3 b^2} \]

[In]

Int[x^5/Sqrt[a + b*x^3],x]

[Out]

(-2*a*Sqrt[a + b*x^3])/(3*b^2) + (2*(a + b*x^3)^(3/2))/(9*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*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {x}{\sqrt {a+b x}} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (-\frac {a}{b \sqrt {a+b x}}+\frac {\sqrt {a+b x}}{b}\right ) \, dx,x,x^3\right ) \\ & = -\frac {2 a \sqrt {a+b x^3}}{3 b^2}+\frac {2 \left (a+b x^3\right )^{3/2}}{9 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {x^5}{\sqrt {a+b x^3}} \, dx=\frac {2 \left (-2 a+b x^3\right ) \sqrt {a+b x^3}}{9 b^2} \]

[In]

Integrate[x^5/Sqrt[a + b*x^3],x]

[Out]

(2*(-2*a + b*x^3)*Sqrt[a + b*x^3])/(9*b^2)

Maple [A] (verified)

Time = 3.82 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66

method	result	size
gosper	\(-\frac {2 \sqrt {b \,x^{3}+a}\, \left (-b \,x^{3}+2 a \right )}{9 b^{2}}\)	\(25\)
trager	\(-\frac {2 \sqrt {b \,x^{3}+a}\, \left (-b \,x^{3}+2 a \right )}{9 b^{2}}\)	\(25\)
risch	\(-\frac {2 \sqrt {b \,x^{3}+a}\, \left (-b \,x^{3}+2 a \right )}{9 b^{2}}\)	\(25\)
pseudoelliptic	\(-\frac {2 \sqrt {b \,x^{3}+a}\, \left (-b \,x^{3}+2 a \right )}{9 b^{2}}\)	\(25\)
default	\(\frac {2 x^{3} \sqrt {b \,x^{3}+a}}{9 b}-\frac {4 a \sqrt {b \,x^{3}+a}}{9 b^{2}}\)	\(34\)
elliptic	\(\frac {2 x^{3} \sqrt {b \,x^{3}+a}}{9 b}-\frac {4 a \sqrt {b \,x^{3}+a}}{9 b^{2}}\)	\(34\)

[In]

int(x^5/(b*x^3+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.61 \[ \int \frac {x^5}{\sqrt {a+b x^3}} \, dx=\frac {2 \, \sqrt {b x^{3} + a} {\left (b x^{3} - 2 \, a\right )}}{9 \, b^{2}} \]

[In]

integrate(x^5/(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

2/9*sqrt(b*x^3 + a)*(b*x^3 - 2*a)/b^2

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int \frac {x^5}{\sqrt {a+b x^3}} \, dx=\begin {cases} - \frac {4 a \sqrt {a + b x^{3}}}{9 b^{2}} + \frac {2 x^{3} \sqrt {a + b x^{3}}}{9 b} & \text {for}\: b \neq 0 \\\frac {x^{6}}{6 \sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5/(b*x**3+a)**(1/2),x)

[Out]

Piecewise((-4*a*sqrt(a + b*x**3)/(9*b**2) + 2*x**3*sqrt(a + b*x**3)/(9*b), Ne(b, 0)), (x**6/(6*sqrt(a)), True)
)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79 \[ \int \frac {x^5}{\sqrt {a+b x^3}} \, dx=\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{9 \, b^{2}} - \frac {2 \, \sqrt {b x^{3} + a} a}{3 \, b^{2}} \]

[In]

integrate(x^5/(b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79 \[ \int \frac {x^5}{\sqrt {a+b x^3}} \, dx=\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{9 \, b^{2}} - \frac {2 \, \sqrt {b x^{3} + a} a}{3 \, b^{2}} \]

[In]

integrate(x^5/(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.79 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.63 \[ \int \frac {x^5}{\sqrt {a+b x^3}} \, dx=-\frac {2\,\sqrt {b\,x^3+a}\,\left (2\,a-b\,x^3\right )}{9\,b^2} \]

[In]

int(x^5/(a + b*x^3)^(1/2),x)

[Out]

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