\(\int \frac {x^{11}}{\sqrt {-1-x^3}} \, dx\) [492]

Optimal result

Integrand size = 15, antiderivative size = 61 \[ \int \frac {x^{11}}{\sqrt {-1-x^3}} \, dx=\frac {2}{3} \sqrt {-1-x^3}+\frac {2}{3} \left (-1-x^3\right )^{3/2}+\frac {2}{5} \left (-1-x^3\right )^{5/2}+\frac {2}{21} \left (-1-x^3\right )^{7/2} \]

[Out]

2/3*(-x^3-1)^(3/2)+2/5*(-x^3-1)^(5/2)+2/21*(-x^3-1)^(7/2)+2/3*(-x^3-1)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, 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} \[ \int \frac {x^{11}}{\sqrt {-1-x^3}} \, dx=\frac {2}{21} \left (-x^3-1\right )^{7/2}+\frac {2}{5} \left (-x^3-1\right )^{5/2}+\frac {2}{3} \left (-x^3-1\right )^{3/2}+\frac {2}{3} \sqrt {-x^3-1} \]

[In]

Int[x^11/Sqrt[-1 - x^3],x]

[Out]

(2*Sqrt[-1 - x^3])/3 + (2*(-1 - x^3)^(3/2))/3 + (2*(-1 - x^3)^(5/2))/5 + (2*(-1 - x^3)^(7/2))/21

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^3}{\sqrt {-1-x}} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (-\frac {1}{\sqrt {-1-x}}-3 \sqrt {-1-x}-3 (-1-x)^{3/2}-(-1-x)^{5/2}\right ) \, dx,x,x^3\right ) \\ & = \frac {2}{3} \sqrt {-1-x^3}+\frac {2}{3} \left (-1-x^3\right )^{3/2}+\frac {2}{5} \left (-1-x^3\right )^{5/2}+\frac {2}{21} \left (-1-x^3\right )^{7/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.52 \[ \int \frac {x^{11}}{\sqrt {-1-x^3}} \, dx=-\frac {2}{105} \sqrt {-1-x^3} \left (-16+8 x^3-6 x^6+5 x^9\right ) \]

[In]

Integrate[x^11/Sqrt[-1 - x^3],x]

[Out]

(-2*Sqrt[-1 - x^3]*(-16 + 8*x^3 - 6*x^6 + 5*x^9))/105

Maple [A] (verified)

Time = 3.83 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.46

method	result	size
trager	\(\left (-\frac {2}{21} x^{9}+\frac {4}{35} x^{6}-\frac {16}{105} x^{3}+\frac {32}{105}\right ) \sqrt {-x^{3}-1}\)	\(28\)
pseudoelliptic	\(-\frac {2 \sqrt {-x^{3}-1}\, \left (5 x^{9}-6 x^{6}+8 x^{3}-16\right )}{105}\)	\(29\)
risch	\(\frac {2 \left (5 x^{9}-6 x^{6}+8 x^{3}-16\right ) \left (x^{3}+1\right )}{105 \sqrt {-x^{3}-1}}\)	\(34\)
gosper	\(\frac {2 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (5 x^{9}-6 x^{6}+8 x^{3}-16\right )}{105 \sqrt {-x^{3}-1}}\)	\(40\)
meijerg	\(-\frac {i \left (\frac {32 \sqrt {\pi }}{35}-\frac {\sqrt {\pi }\, \left (-40 x^{9}+48 x^{6}-64 x^{3}+128\right ) \sqrt {x^{3}+1}}{140}\right )}{3 \sqrt {\pi }}\)	\(42\)
default	\(-\frac {2 x^{9} \sqrt {-x^{3}-1}}{21}+\frac {4 x^{6} \sqrt {-x^{3}-1}}{35}-\frac {16 x^{3} \sqrt {-x^{3}-1}}{105}+\frac {32 \sqrt {-x^{3}-1}}{105}\)	\(55\)
elliptic	\(-\frac {2 x^{9} \sqrt {-x^{3}-1}}{21}+\frac {4 x^{6} \sqrt {-x^{3}-1}}{35}-\frac {16 x^{3} \sqrt {-x^{3}-1}}{105}+\frac {32 \sqrt {-x^{3}-1}}{105}\)	\(55\)

[In]

int(x^11/(-x^3-1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(-2/21*x^9+4/35*x^6-16/105*x^3+32/105)*(-x^3-1)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.46 \[ \int \frac {x^{11}}{\sqrt {-1-x^3}} \, dx=-\frac {2}{105} \, {\left (5 \, x^{9} - 6 \, x^{6} + 8 \, x^{3} - 16\right )} \sqrt {-x^{3} - 1} \]

[In]

integrate(x^11/(-x^3-1)^(1/2),x, algorithm="fricas")

[Out]

-2/105*(5*x^9 - 6*x^6 + 8*x^3 - 16)*sqrt(-x^3 - 1)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int \frac {x^{11}}{\sqrt {-1-x^3}} \, dx=- \frac {2 x^{9} \sqrt {- x^{3} - 1}}{21} + \frac {4 x^{6} \sqrt {- x^{3} - 1}}{35} - \frac {16 x^{3} \sqrt {- x^{3} - 1}}{105} + \frac {32 \sqrt {- x^{3} - 1}}{105} \]

[In]

integrate(x**11/(-x**3-1)**(1/2),x)

[Out]

-2*x**9*sqrt(-x**3 - 1)/21 + 4*x**6*sqrt(-x**3 - 1)/35 - 16*x**3*sqrt(-x**3 - 1)/105 + 32*sqrt(-x**3 - 1)/105

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \frac {x^{11}}{\sqrt {-1-x^3}} \, dx=\frac {2}{21} \, {\left (-x^{3} - 1\right )}^{\frac {7}{2}} + \frac {2}{5} \, {\left (-x^{3} - 1\right )}^{\frac {5}{2}} + \frac {2}{3} \, {\left (-x^{3} - 1\right )}^{\frac {3}{2}} + \frac {2}{3} \, \sqrt {-x^{3} - 1} \]

[In]

integrate(x^11/(-x^3-1)^(1/2),x, algorithm="maxima")

[Out]

2/21*(-x^3 - 1)^(7/2) + 2/5*(-x^3 - 1)^(5/2) + 2/3*(-x^3 - 1)^(3/2) + 2/3*sqrt(-x^3 - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97 \[ \int \frac {x^{11}}{\sqrt {-1-x^3}} \, dx=-\frac {2}{21} \, {\left (x^{3} + 1\right )}^{3} \sqrt {-x^{3} - 1} + \frac {2}{5} \, {\left (x^{3} + 1\right )}^{2} \sqrt {-x^{3} - 1} + \frac {2}{3} \, {\left (-x^{3} - 1\right )}^{\frac {3}{2}} + \frac {2}{3} \, \sqrt {-x^{3} - 1} \]

[In]

integrate(x^11/(-x^3-1)^(1/2),x, algorithm="giac")

[Out]

-2/21*(x^3 + 1)^3*sqrt(-x^3 - 1) + 2/5*(x^3 + 1)^2*sqrt(-x^3 - 1) + 2/3*(-x^3 - 1)^(3/2) + 2/3*sqrt(-x^3 - 1)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {x^{11}}{\sqrt {-1-x^3}} \, dx=\frac {4\,x^6\,\sqrt {-x^3-1}}{35}-\frac {16\,x^3\,\sqrt {-x^3-1}}{105}-\frac {2\,x^9\,\sqrt {-x^3-1}}{21}+\frac {32\,\sqrt {-x^3-1}}{105} \]

[In]

int(x^11/(- x^3 - 1)^(1/2),x)

[Out]

(4*x^6*(- x^3 - 1)^(1/2))/35 - (16*x^3*(- x^3 - 1)^(1/2))/105 - (2*x^9*(- x^3 - 1)^(1/2))/21 + (32*(- x^3 - 1)
^(1/2))/105