3.5.38 \(\int \frac {x^{11}}{\sqrt {1+x^3}} \, dx\) [438]

Optimal. Leaf size=53 \[ -\frac {2}{3} \sqrt {1+x^3}+\frac {2}{3} \left (1+x^3\right )^{3/2}-\frac {2}{5} \left (1+x^3\right )^{5/2}+\frac {2}{21} \left (1+x^3\right )^{7/2} \]

[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)

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Rubi [A]
time = 0.01, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \begin {gather*} \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 \sqrt {x^3+1}}{3} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {x^{11}}{\sqrt {1+x^3}} \, dx &=\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*}

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Mathematica [A]
time = 0.02, size = 30, normalized size = 0.57 \begin {gather*} \frac {2}{105} \sqrt {1+x^3} \left (-16+8 x^3-6 x^6+5 x^9\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [A]
time = 0.23, size = 47, normalized size = 0.89

method	result	size
trager	\(\left (\frac {2}{21} x^{9}-\frac {4}{35} x^{6}+\frac {16}{105} x^{3}-\frac {32}{105}\right ) \sqrt {x^{3}+1}\)	\(26\)
risch	\(\frac {2 \left (5 x^{9}-6 x^{6}+8 x^{3}-16\right ) \sqrt {x^{3}+1}}{105}\)	\(27\)
gosper	\(\frac {2 \left (x +1\right ) \left (x^{2}-x +1\right ) \left (5 x^{9}-6 x^{6}+8 x^{3}-16\right )}{105 \sqrt {x^{3}+1}}\)	\(38\)
meijerg	\(\frac {\frac {32 \sqrt {\pi }}{35}-\frac {\sqrt {\pi }\, \left (-40 x^{9}+48 x^{6}-64 x^{3}+128\right ) \sqrt {x^{3}+1}}{140}}{3 \sqrt {\pi }}\)	\(41\)
default	\(\frac {2 x^{9} \sqrt {x^{3}+1}}{21}-\frac {4 x^{6} \sqrt {x^{3}+1}}{35}+\frac {16 x^{3} \sqrt {x^{3}+1}}{105}-\frac {32 \sqrt {x^{3}+1}}{105}\)	\(47\)
elliptic	\(\frac {2 x^{9} \sqrt {x^{3}+1}}{21}-\frac {4 x^{6} \sqrt {x^{3}+1}}{35}+\frac {16 x^{3} \sqrt {x^{3}+1}}{105}-\frac {32 \sqrt {x^{3}+1}}{105}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 37, normalized size = 0.70 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]
time = 0.34, size = 26, normalized size = 0.49 \begin {gather*} \frac {2}{105} \, {\left (5 \, x^{9} - 6 \, x^{6} + 8 \, x^{3} - 16\right )} \sqrt {x^{3} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]
time = 0.22, size = 56, normalized size = 1.06 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]
time = 1.08, size = 37, normalized size = 0.70 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Mupad [B]
time = 1.00, size = 46, normalized size = 0.87 \begin {gather*} \frac {16\,x^3\,\sqrt {x^3+1}}{105}-\frac {32\,\sqrt {x^3+1}}{105}-\frac {4\,x^6\,\sqrt {x^3+1}}{35}+\frac {2\,x^9\,\sqrt {x^3+1}}{21} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(x^3 + 1)^(1/2),x)

[Out]

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

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