3.2.99 \(\int \frac {x^8 (A+B x^3)}{(a+b x^3)^{5/2}} \, dx\)

Optimal. Leaf size=103 \[ -\frac {2 a^2 (A b-a B)}{9 b^4 \left (a+b x^3\right )^{3/2}}+\frac {2 a (2 A b-3 a B)}{3 b^4 \sqrt {a+b x^3}}+\frac {2 \sqrt {a+b x^3} (A b-3 a B)}{3 b^4}+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b^4} \]

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Rubi [A]  time = 0.08, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {446, 77} \begin {gather*} -\frac {2 a^2 (A b-a B)}{9 b^4 \left (a+b x^3\right )^{3/2}}+\frac {2 a (2 A b-3 a B)}{3 b^4 \sqrt {a+b x^3}}+\frac {2 \sqrt {a+b x^3} (A b-3 a B)}{3 b^4}+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^8*(A + B*x^3))/(a + b*x^3)^(5/2),x]

[Out]

(-2*a^2*(A*b - a*B))/(9*b^4*(a + b*x^3)^(3/2)) + (2*a*(2*A*b - 3*a*B))/(3*b^4*Sqrt[a + b*x^3]) + (2*(A*b - 3*a
*B)*Sqrt[a + b*x^3])/(3*b^4) + (2*B*(a + b*x^3)^(3/2))/(9*b^4)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2 (A+B x)}{(a+b x)^{5/2}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (-\frac {a^2 (-A b+a B)}{b^3 (a+b x)^{5/2}}+\frac {a (-2 A b+3 a B)}{b^3 (a+b x)^{3/2}}+\frac {A b-3 a B}{b^3 \sqrt {a+b x}}+\frac {B \sqrt {a+b x}}{b^3}\right ) \, dx,x,x^3\right )\\ &=-\frac {2 a^2 (A b-a B)}{9 b^4 \left (a+b x^3\right )^{3/2}}+\frac {2 a (2 A b-3 a B)}{3 b^4 \sqrt {a+b x^3}}+\frac {2 (A b-3 a B) \sqrt {a+b x^3}}{3 b^4}+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 73, normalized size = 0.71 \begin {gather*} \frac {2 \left (-16 a^3 B+8 a^2 b \left (A-3 B x^3\right )-6 a b^2 x^3 \left (B x^3-2 A\right )+b^3 x^6 \left (3 A+B x^3\right )\right )}{9 b^4 \left (a+b x^3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(A + B*x^3))/(a + b*x^3)^(5/2),x]

[Out]

(2*(-16*a^3*B + 8*a^2*b*(A - 3*B*x^3) - 6*a*b^2*x^3*(-2*A + B*x^3) + b^3*x^6*(3*A + B*x^3)))/(9*b^4*(a + b*x^3
)^(3/2))

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IntegrateAlgebraic [A]  time = 0.06, size = 80, normalized size = 0.78 \begin {gather*} -\frac {2 \left (16 a^3 B-8 a^2 A b+24 a^2 b B x^3-12 a A b^2 x^3+6 a b^2 B x^6-3 A b^3 x^6-b^3 B x^9\right )}{9 b^4 \left (a+b x^3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^8*(A + B*x^3))/(a + b*x^3)^(5/2),x]

[Out]

(-2*(-8*a^2*A*b + 16*a^3*B - 12*a*A*b^2*x^3 + 24*a^2*b*B*x^3 - 3*A*b^3*x^6 + 6*a*b^2*B*x^6 - b^3*B*x^9))/(9*b^
4*(a + b*x^3)^(3/2))

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fricas [A]  time = 0.84, size = 98, normalized size = 0.95 \begin {gather*} \frac {2 \, {\left (B b^{3} x^{9} - 3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} x^{6} - 16 \, B a^{3} + 8 \, A a^{2} b - 12 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{9 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="fricas")

[Out]

2/9*(B*b^3*x^9 - 3*(2*B*a*b^2 - A*b^3)*x^6 - 16*B*a^3 + 8*A*a^2*b - 12*(2*B*a^2*b - A*a*b^2)*x^3)*sqrt(b*x^3 +
a)/(b^6*x^6 + 2*a*b^5*x^3 + a^2*b^4)

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giac [A]  time = 0.17, size = 104, normalized size = 1.01 \begin {gather*} -\frac {2 \, {\left (9 \, {\left (b x^{3} + a\right )} B a^{2} - B a^{3} - 6 \, {\left (b x^{3} + a\right )} A a b + A a^{2} b\right )}}{9 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{4}} + \frac {2 \, {\left ({\left (b x^{3} + a\right )}^{\frac {3}{2}} B b^{8} - 9 \, \sqrt {b x^{3} + a} B a b^{8} + 3 \, \sqrt {b x^{3} + a} A b^{9}\right )}}{9 \, b^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="giac")

[Out]

-2/9*(9*(b*x^3 + a)*B*a^2 - B*a^3 - 6*(b*x^3 + a)*A*a*b + A*a^2*b)/((b*x^3 + a)^(3/2)*b^4) + 2/9*((b*x^3 + a)^
(3/2)*B*b^8 - 9*sqrt(b*x^3 + a)*B*a*b^8 + 3*sqrt(b*x^3 + a)*A*b^9)/b^12

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maple [A]  time = 0.05, size = 76, normalized size = 0.74 \begin {gather*} \frac {\frac {2}{9} B \,x^{9} b^{3}+\frac {2}{3} A \,b^{3} x^{6}-\frac {4}{3} B a \,b^{2} x^{6}+\frac {8}{3} A a \,b^{2} x^{3}-\frac {16}{3} B \,a^{2} b \,x^{3}+\frac {16}{9} A \,a^{2} b -\frac {32}{9} B \,a^{3}}{\left (b \,x^{3}+a \right )^{\frac {3}{2}} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(B*x^3+A)/(b*x^3+a)^(5/2),x)

[Out]

2/9/(b*x^3+a)^(3/2)*(B*b^3*x^9+3*A*b^3*x^6-6*B*a*b^2*x^6+12*A*a*b^2*x^3-24*B*a^2*b*x^3+8*A*a^2*b-16*B*a^3)/b^4

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maxima [A]  time = 0.59, size = 116, normalized size = 1.13 \begin {gather*} \frac {2}{9} \, B {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}}}{b^{4}} - \frac {9 \, \sqrt {b x^{3} + a} a}{b^{4}} - \frac {9 \, a^{2}}{\sqrt {b x^{3} + a} b^{4}} + \frac {a^{3}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{4}}\right )} + \frac {2}{9} \, A {\left (\frac {3 \, \sqrt {b x^{3} + a}}{b^{3}} + \frac {6 \, a}{\sqrt {b x^{3} + a} b^{3}} - \frac {a^{2}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="maxima")

[Out]

2/9*B*((b*x^3 + a)^(3/2)/b^4 - 9*sqrt(b*x^3 + a)*a/b^4 - 9*a^2/(sqrt(b*x^3 + a)*b^4) + a^3/((b*x^3 + a)^(3/2)*
b^4)) + 2/9*A*(3*sqrt(b*x^3 + a)/b^3 + 6*a/(sqrt(b*x^3 + a)*b^3) - a^2/((b*x^3 + a)^(3/2)*b^3))

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mupad [B]  time = 2.80, size = 145, normalized size = 1.41 \begin {gather*} \frac {\sqrt {b\,x^3+a}\,\left (\frac {2\,\left (A\,b-2\,B\,a\right )}{b^3}-\frac {4\,B\,a}{3\,b^3}\right )}{3\,b}-\frac {\frac {2\,B\,a^2-2\,A\,a\,b}{3\,b^4}-\frac {a\,\left (\frac {2\,A\,b^2-2\,B\,a\,b}{3\,b^4}-\frac {2\,B\,a}{3\,b^3}\right )}{b}}{\sqrt {b\,x^3+a}}-\frac {a^2\,\left (\frac {2\,A}{9\,b}-\frac {2\,B\,a}{9\,b^2}\right )}{b^2\,{\left (b\,x^3+a\right )}^{3/2}}+\frac {2\,B\,x^3\,\sqrt {b\,x^3+a}}{9\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8*(A + B*x^3))/(a + b*x^3)^(5/2),x)

[Out]

((a + b*x^3)^(1/2)*((2*(A*b - 2*B*a))/b^3 - (4*B*a)/(3*b^3)))/(3*b) - ((2*B*a^2 - 2*A*a*b)/(3*b^4) - (a*((2*A*
b^2 - 2*B*a*b)/(3*b^4) - (2*B*a)/(3*b^3)))/b)/(a + b*x^3)^(1/2) - (a^2*((2*A)/(9*b) - (2*B*a)/(9*b^2)))/(b^2*(
a + b*x^3)^(3/2)) + (2*B*x^3*(a + b*x^3)^(1/2))/(9*b^3)

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sympy [A]  time = 5.18, size = 338, normalized size = 3.28 \begin {gather*} \begin {cases} \frac {16 A a^{2} b}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} + \frac {24 A a b^{2} x^{3}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} + \frac {6 A b^{3} x^{6}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} - \frac {32 B a^{3}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} - \frac {48 B a^{2} b x^{3}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} - \frac {12 B a b^{2} x^{6}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} + \frac {2 B b^{3} x^{9}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{9}}{9} + \frac {B x^{12}}{12}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(B*x**3+A)/(b*x**3+a)**(5/2),x)

[Out]

Piecewise((16*A*a**2*b/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3*sqrt(a + b*x**3)) + 24*A*a*b**2*x**3/(9*a*b**4
*sqrt(a + b*x**3) + 9*b**5*x**3*sqrt(a + b*x**3)) + 6*A*b**3*x**6/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3*sqr
t(a + b*x**3)) - 32*B*a**3/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3*sqrt(a + b*x**3)) - 48*B*a**2*b*x**3/(9*a*
b**4*sqrt(a + b*x**3) + 9*b**5*x**3*sqrt(a + b*x**3)) - 12*B*a*b**2*x**6/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x
**3*sqrt(a + b*x**3)) + 2*B*b**3*x**9/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3*sqrt(a + b*x**3)), Ne(b, 0)), (
(A*x**9/9 + B*x**12/12)/a**(5/2), True))

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