\(\int x^5 \sqrt {1+x^3} \, dx\) [72]

Optimal result

Integrand size = 13, antiderivative size = 27 \[ \int x^5 \sqrt {1+x^3} \, dx=-\frac {2}{9} \left (1+x^3\right )^{3/2}+\frac {2}{15} \left (1+x^3\right )^{5/2} \]

[Out]

-2/9*(x^3+1)^(3/2)+2/15*(x^3+1)^(5/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, 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} \[ \int x^5 \sqrt {1+x^3} \, dx=\frac {2}{15} \left (x^3+1\right )^{5/2}-\frac {2}{9} \left (x^3+1\right )^{3/2} \]

[In]

Int[x^5*Sqrt[1 + x^3],x]

[Out]

(-2*(1 + x^3)^(3/2))/9 + (2*(1 + x^3)^(5/2))/15

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 x \sqrt {1+x} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (-\sqrt {1+x}+(1+x)^{3/2}\right ) \, dx,x,x^3\right ) \\ & = -\frac {2}{9} \left (1+x^3\right )^{3/2}+\frac {2}{15} \left (1+x^3\right )^{5/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int x^5 \sqrt {1+x^3} \, dx=\frac {2}{45} \left (1+x^3\right )^{3/2} \left (-2+3 x^3\right ) \]

[In]

Integrate[x^5*Sqrt[1 + x^3],x]

[Out]

(2*(1 + x^3)^(3/2)*(-2 + 3*x^3))/45

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63

method	result	size
pseudoelliptic	\(\frac {2 \left (x^{3}+1\right )^{\frac {3}{2}} \left (3 x^{3}-2\right )}{45}\)	\(17\)
risch	\(\frac {2 \left (3 x^{6}+x^{3}-2\right ) \sqrt {x^{3}+1}}{45}\)	\(20\)
trager	\(\left (\frac {2}{15} x^{6}+\frac {2}{45} x^{3}-\frac {4}{45}\right ) \sqrt {x^{3}+1}\)	\(21\)
gosper	\(\frac {2 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (3 x^{3}-2\right ) \sqrt {x^{3}+1}}{45}\)	\(28\)
meijerg	\(-\frac {-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (x^{3}+1\right )^{\frac {3}{2}} \left (-3 x^{3}+2\right )}{15}}{6 \sqrt {\pi }}\)	\(31\)
default	\(\frac {2 x^{6} \sqrt {x^{3}+1}}{15}+\frac {2 x^{3} \sqrt {x^{3}+1}}{45}-\frac {4 \sqrt {x^{3}+1}}{45}\)	\(35\)
elliptic	\(\frac {2 x^{6} \sqrt {x^{3}+1}}{15}+\frac {2 x^{3} \sqrt {x^{3}+1}}{45}-\frac {4 \sqrt {x^{3}+1}}{45}\)	\(35\)

[In]

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

[Out]

2/45*(x^3+1)^(3/2)*(3*x^3-2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int x^5 \sqrt {1+x^3} \, dx=\frac {2}{45} \, {\left (3 \, x^{6} + x^{3} - 2\right )} \sqrt {x^{3} + 1} \]

[In]

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

[Out]

2/45*(3*x^6 + x^3 - 2)*sqrt(x^3 + 1)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int x^5 \sqrt {1+x^3} \, dx=\frac {2 x^{6} \sqrt {x^{3} + 1}}{15} + \frac {2 x^{3} \sqrt {x^{3} + 1}}{45} - \frac {4 \sqrt {x^{3} + 1}}{45} \]

[In]

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

[Out]

2*x**6*sqrt(x**3 + 1)/15 + 2*x**3*sqrt(x**3 + 1)/45 - 4*sqrt(x**3 + 1)/45

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int x^5 \sqrt {1+x^3} \, dx=\frac {2}{15} \, {\left (x^{3} + 1\right )}^{\frac {5}{2}} - \frac {2}{9} \, {\left (x^{3} + 1\right )}^{\frac {3}{2}} \]

[In]

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

[Out]

2/15*(x^3 + 1)^(5/2) - 2/9*(x^3 + 1)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int x^5 \sqrt {1+x^3} \, dx=\frac {2}{15} \, {\left (x^{3} + 1\right )}^{\frac {5}{2}} - \frac {2}{9} \, {\left (x^{3} + 1\right )}^{\frac {3}{2}} \]

[In]

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

[Out]

2/15*(x^3 + 1)^(5/2) - 2/9*(x^3 + 1)^(3/2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int x^5 \sqrt {1+x^3} \, dx=\frac {2\,{\left (x^3+1\right )}^{5/2}}{15}-\frac {2\,{\left (x^3+1\right )}^{3/2}}{9} \]

[In]

int(x^5*(x^3 + 1)^(1/2),x)

[Out]

(2*(x^3 + 1)^(5/2))/15 - (2*(x^3 + 1)^(3/2))/9