3.5.21 \(\int \frac {x^{11}}{(a+b x^3)^{3/2}} \, dx\) [421]

Optimal. Leaf size=78 \[ \frac {2 a^3}{3 b^4 \sqrt {a+b x^3}}+\frac {2 a^2 \sqrt {a+b x^3}}{b^4}-\frac {2 a \left (a+b x^3\right )^{3/2}}{3 b^4}+\frac {2 \left (a+b x^3\right )^{5/2}}{15 b^4} \]

[Out]

-2/3*a*(b*x^3+a)^(3/2)/b^4+2/15*(b*x^3+a)^(5/2)/b^4+2/3*a^3/b^4/(b*x^3+a)^(1/2)+2*a^2*(b*x^3+a)^(1/2)/b^4

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Rubi [A]
time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \begin {gather*} \frac {2 a^3}{3 b^4 \sqrt {a+b x^3}}+\frac {2 a^2 \sqrt {a+b x^3}}{b^4}-\frac {2 a \left (a+b x^3\right )^{3/2}}{3 b^4}+\frac {2 \left (a+b x^3\right )^{5/2}}{15 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^11/(a + b*x^3)^(3/2),x]

[Out]

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

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}}{\left (a+b x^3\right )^{3/2}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x^3}{(a+b x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (-\frac {a^3}{b^3 (a+b x)^{3/2}}+\frac {3 a^2}{b^3 \sqrt {a+b x}}-\frac {3 a \sqrt {a+b x}}{b^3}+\frac {(a+b x)^{3/2}}{b^3}\right ) \, dx,x,x^3\right )\\ &=\frac {2 a^3}{3 b^4 \sqrt {a+b x^3}}+\frac {2 a^2 \sqrt {a+b x^3}}{b^4}-\frac {2 a \left (a+b x^3\right )^{3/2}}{3 b^4}+\frac {2 \left (a+b x^3\right )^{5/2}}{15 b^4}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 49, normalized size = 0.63 \begin {gather*} \frac {2 \left (16 a^3+8 a^2 b x^3-2 a b^2 x^6+b^3 x^9\right )}{15 b^4 \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^11/(a + b*x^3)^(3/2),x]

[Out]

(2*(16*a^3 + 8*a^2*b*x^3 - 2*a*b^2*x^6 + b^3*x^9))/(15*b^4*Sqrt[a + b*x^3])

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Maple [A]
time = 0.14, size = 75, normalized size = 0.96

method	result	size
gosper	\(\frac {\frac {2}{15} b^{3} x^{9}-\frac {4}{15} a \,b^{2} x^{6}+\frac {16}{15} a^{2} b \,x^{3}+\frac {32}{15} a^{3}}{\sqrt {b \,x^{3}+a}\, b^{4}}\)	\(46\)
trager	\(\frac {\frac {2}{15} b^{3} x^{9}-\frac {4}{15} a \,b^{2} x^{6}+\frac {16}{15} a^{2} b \,x^{3}+\frac {32}{15} a^{3}}{\sqrt {b \,x^{3}+a}\, b^{4}}\)	\(46\)
risch	\(\frac {2 \left (b^{2} x^{6}-3 a b \,x^{3}+11 a^{2}\right ) \sqrt {b \,x^{3}+a}}{15 b^{4}}+\frac {2 a^{3}}{3 b^{4} \sqrt {b \,x^{3}+a}}\)	\(53\)
default	\(\frac {2 a^{3}}{3 b^{4} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x^{6} \sqrt {b \,x^{3}+a}}{15 b^{2}}-\frac {2 a \,x^{3} \sqrt {b \,x^{3}+a}}{5 b^{3}}+\frac {22 a^{2} \sqrt {b \,x^{3}+a}}{15 b^{4}}\)	\(75\)
elliptic	\(\frac {2 a^{3}}{3 b^{4} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x^{6} \sqrt {b \,x^{3}+a}}{15 b^{2}}-\frac {2 a \,x^{3} \sqrt {b \,x^{3}+a}}{5 b^{3}}+\frac {22 a^{2} \sqrt {b \,x^{3}+a}}{15 b^{4}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(b*x^3+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3/b^4*a^3/((x^3+a/b)*b)^(1/2)+2/15/b^2*x^6*(b*x^3+a)^(1/2)-2/5*a/b^3*x^3*(b*x^3+a)^(1/2)+22/15*a^2*(b*x^3+a)
^(1/2)/b^4

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Maxima [A]
time = 0.30, size = 64, normalized size = 0.82 \begin {gather*} \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}}}{15 \, b^{4}} - \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a}{3 \, b^{4}} + \frac {2 \, \sqrt {b x^{3} + a} a^{2}}{b^{4}} + \frac {2 \, a^{3}}{3 \, \sqrt {b x^{3} + a} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^3+a)^(3/2),x, algorithm="maxima")

[Out]

2/15*(b*x^3 + a)^(5/2)/b^4 - 2/3*(b*x^3 + a)^(3/2)*a/b^4 + 2*sqrt(b*x^3 + a)*a^2/b^4 + 2/3*a^3/(sqrt(b*x^3 + a
)*b^4)

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Fricas [A]
time = 0.35, size = 57, normalized size = 0.73 \begin {gather*} \frac {2 \, {\left (b^{3} x^{9} - 2 \, a b^{2} x^{6} + 8 \, a^{2} b x^{3} + 16 \, a^{3}\right )} \sqrt {b x^{3} + a}}{15 \, {\left (b^{5} x^{3} + a b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

2/15*(b^3*x^9 - 2*a*b^2*x^6 + 8*a^2*b*x^3 + 16*a^3)*sqrt(b*x^3 + a)/(b^5*x^3 + a*b^4)

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Sympy [A]
time = 0.57, size = 94, normalized size = 1.21 \begin {gather*} \begin {cases} \frac {32 a^{3}}{15 b^{4} \sqrt {a + b x^{3}}} + \frac {16 a^{2} x^{3}}{15 b^{3} \sqrt {a + b x^{3}}} - \frac {4 a x^{6}}{15 b^{2} \sqrt {a + b x^{3}}} + \frac {2 x^{9}}{15 b \sqrt {a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {x^{12}}{12 a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11/(b*x**3+a)**(3/2),x)

[Out]

Piecewise((32*a**3/(15*b**4*sqrt(a + b*x**3)) + 16*a**2*x**3/(15*b**3*sqrt(a + b*x**3)) - 4*a*x**6/(15*b**2*sq
rt(a + b*x**3)) + 2*x**9/(15*b*sqrt(a + b*x**3)), Ne(b, 0)), (x**12/(12*a**(3/2)), True))

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Giac [A]
time = 1.47, size = 69, normalized size = 0.88 \begin {gather*} \frac {2 \, a^{3}}{3 \, \sqrt {b x^{3} + a} b^{4}} + \frac {2 \, {\left ({\left (b x^{3} + a\right )}^{\frac {5}{2}} b^{16} - 5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a b^{16} + 15 \, \sqrt {b x^{3} + a} a^{2} b^{16}\right )}}{15 \, b^{20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

2/3*a^3/(sqrt(b*x^3 + a)*b^4) + 2/15*((b*x^3 + a)^(5/2)*b^16 - 5*(b*x^3 + a)^(3/2)*a*b^16 + 15*sqrt(b*x^3 + a)
*a^2*b^16)/b^20

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Mupad [B]
time = 1.34, size = 70, normalized size = 0.90 \begin {gather*} \frac {22\,a^2\,\sqrt {b\,x^3+a}}{15\,b^4}+\frac {2\,a^3}{3\,b^4\,\sqrt {b\,x^3+a}}+\frac {2\,x^6\,\sqrt {b\,x^3+a}}{15\,b^2}-\frac {2\,a\,x^3\,\sqrt {b\,x^3+a}}{5\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(a + b*x^3)^(3/2),x)

[Out]

(22*a^2*(a + b*x^3)^(1/2))/(15*b^4) + (2*a^3)/(3*b^4*(a + b*x^3)^(1/2)) + (2*x^6*(a + b*x^3)^(1/2))/(15*b^2) -
(2*a*x^3*(a + b*x^3)^(1/2))/(5*b^3)

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