3.2.87 \(\int \frac {x^8 (A+B x^3)}{\sqrt {a+b x^3}} \, dx\)

Optimal. Leaf size=103 \[ \frac {2 a^2 \sqrt {a+b x^3} (A b-a B)}{3 b^4}+\frac {2 \left (a+b x^3\right )^{5/2} (A b-3 a B)}{15 b^4}-\frac {2 a \left (a+b x^3\right )^{3/2} (2 A b-3 a B)}{9 b^4}+\frac {2 B \left (a+b x^3\right )^{7/2}}{21 b^4} \]

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Rubi [A]  time = 0.07, 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 \sqrt {a+b x^3} (A b-a B)}{3 b^4}+\frac {2 \left (a+b x^3\right )^{5/2} (A b-3 a B)}{15 b^4}-\frac {2 a \left (a+b x^3\right )^{3/2} (2 A b-3 a B)}{9 b^4}+\frac {2 B \left (a+b x^3\right )^{7/2}}{21 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(A*b - a*B)*Sqrt[a + b*x^3])/(3*b^4) - (2*a*(2*A*b - 3*a*B)*(a + b*x^3)^(3/2))/(9*b^4) + (2*(A*b - 3*a*
B)*(a + b*x^3)^(5/2))/(15*b^4) + (2*B*(a + b*x^3)^(7/2))/(21*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 )}{\sqrt {a+b x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2 (A+B x)}{\sqrt {a+b x}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (-\frac {a^2 (-A b+a B)}{b^3 \sqrt {a+b x}}+\frac {a (-2 A b+3 a B) \sqrt {a+b x}}{b^3}+\frac {(A b-3 a B) (a+b x)^{3/2}}{b^3}+\frac {B (a+b x)^{5/2}}{b^3}\right ) \, dx,x,x^3\right )\\ &=\frac {2 a^2 (A b-a B) \sqrt {a+b x^3}}{3 b^4}-\frac {2 a (2 A b-3 a B) \left (a+b x^3\right )^{3/2}}{9 b^4}+\frac {2 (A b-3 a B) \left (a+b x^3\right )^{5/2}}{15 b^4}+\frac {2 B \left (a+b x^3\right )^{7/2}}{21 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 78, normalized size = 0.76 \begin {gather*} \frac {2 \sqrt {a+b x^3} \left (-48 a^3 B+8 a^2 b \left (7 A+3 B x^3\right )-2 a b^2 x^3 \left (14 A+9 B x^3\right )+3 b^3 x^6 \left (7 A+5 B x^3\right )\right )}{315 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x^3]*(-48*a^3*B + 8*a^2*b*(7*A + 3*B*x^3) + 3*b^3*x^6*(7*A + 5*B*x^3) - 2*a*b^2*x^3*(14*A + 9*B*
x^3)))/(315*b^4)

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IntegrateAlgebraic [A]  time = 0.05, size = 80, normalized size = 0.78 \begin {gather*} -\frac {2 \sqrt {a+b x^3} \left (48 a^3 B-56 a^2 A b-24 a^2 b B x^3+28 a A b^2 x^3+18 a b^2 B x^6-21 A b^3 x^6-15 b^3 B x^9\right )}{315 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a + b*x^3]*(-56*a^2*A*b + 48*a^3*B + 28*a*A*b^2*x^3 - 24*a^2*b*B*x^3 - 21*A*b^3*x^6 + 18*a*b^2*B*x^6
- 15*b^3*B*x^9))/(315*b^4)

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fricas [A]  time = 0.92, size = 76, normalized size = 0.74 \begin {gather*} \frac {2 \, {\left (15 \, B b^{3} x^{9} - 3 \, {\left (6 \, B a b^{2} - 7 \, A b^{3}\right )} x^{6} - 48 \, B a^{3} + 56 \, A a^{2} b + 4 \, {\left (6 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{315 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

2/315*(15*B*b^3*x^9 - 3*(6*B*a*b^2 - 7*A*b^3)*x^6 - 48*B*a^3 + 56*A*a^2*b + 4*(6*B*a^2*b - 7*A*a*b^2)*x^3)*sqr
t(b*x^3 + a)/b^4

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giac [A]  time = 0.16, size = 101, normalized size = 0.98 \begin {gather*} -\frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {b x^{3} + a}}{3 \, b^{4}} + \frac {2 \, {\left (15 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} B - 63 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B a + 105 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B a^{2} + 21 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} A b - 70 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A a b\right )}}{315 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

-2/3*(B*a^3 - A*a^2*b)*sqrt(b*x^3 + a)/b^4 + 2/315*(15*(b*x^3 + a)^(7/2)*B - 63*(b*x^3 + a)^(5/2)*B*a + 105*(b
*x^3 + a)^(3/2)*B*a^2 + 21*(b*x^3 + a)^(5/2)*A*b - 70*(b*x^3 + a)^(3/2)*A*a*b)/b^4

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maple [A]  time = 0.04, size = 77, normalized size = 0.75 \begin {gather*} \frac {2 \sqrt {b \,x^{3}+a}\, \left (15 B \,x^{9} b^{3}+21 A \,b^{3} x^{6}-18 B a \,b^{2} x^{6}-28 A a \,b^{2} x^{3}+24 B \,a^{2} b \,x^{3}+56 A \,a^{2} b -48 B \,a^{3}\right )}{315 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)^(1/2),x)

[Out]

2/315*(b*x^3+a)^(1/2)*(15*B*b^3*x^9+21*A*b^3*x^6-18*B*a*b^2*x^6-28*A*a*b^2*x^3+24*B*a^2*b*x^3+56*A*a^2*b-48*B*
a^3)/b^4

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maxima [A]  time = 0.59, size = 118, normalized size = 1.15 \begin {gather*} \frac {2}{105} \, B {\left (\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}}}{b^{4}} - \frac {21 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a}{b^{4}} + \frac {35 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}}{b^{4}} - \frac {35 \, \sqrt {b x^{3} + a} a^{3}}{b^{4}}\right )} + \frac {2}{45} \, A {\left (\frac {3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}}}{b^{3}} - \frac {10 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a}{b^{3}} + \frac {15 \, \sqrt {b x^{3} + a} a^{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)^(1/2),x, algorithm="maxima")

[Out]

2/105*B*(5*(b*x^3 + a)^(7/2)/b^4 - 21*(b*x^3 + a)^(5/2)*a/b^4 + 35*(b*x^3 + a)^(3/2)*a^2/b^4 - 35*sqrt(b*x^3 +
a)*a^3/b^4) + 2/45*A*(3*(b*x^3 + a)^(5/2)/b^3 - 10*(b*x^3 + a)^(3/2)*a/b^3 + 15*sqrt(b*x^3 + a)*a^2/b^3)

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mupad [B]  time = 2.68, size = 104, normalized size = 1.01 \begin {gather*} \frac {8\,a^2\,\sqrt {b\,x^3+a}\,\left (2\,A-\frac {12\,B\,a}{7\,b}\right )}{45\,b^3}+\frac {x^6\,\sqrt {b\,x^3+a}\,\left (2\,A-\frac {12\,B\,a}{7\,b}\right )}{15\,b}+\frac {2\,B\,x^9\,\sqrt {b\,x^3+a}}{21\,b}-\frac {4\,a\,x^3\,\sqrt {b\,x^3+a}\,\left (2\,A-\frac {12\,B\,a}{7\,b}\right )}{45\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8*(A + B*x^3))/(a + b*x^3)^(1/2),x)

[Out]

(8*a^2*(a + b*x^3)^(1/2)*(2*A - (12*B*a)/(7*b)))/(45*b^3) + (x^6*(a + b*x^3)^(1/2)*(2*A - (12*B*a)/(7*b)))/(15
*b) + (2*B*x^9*(a + b*x^3)^(1/2))/(21*b) - (4*a*x^3*(a + b*x^3)^(1/2)*(2*A - (12*B*a)/(7*b)))/(45*b^2)

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sympy [A]  time = 3.44, size = 175, normalized size = 1.70 \begin {gather*} \begin {cases} \frac {16 A a^{2} \sqrt {a + b x^{3}}}{45 b^{3}} - \frac {8 A a x^{3} \sqrt {a + b x^{3}}}{45 b^{2}} + \frac {2 A x^{6} \sqrt {a + b x^{3}}}{15 b} - \frac {32 B a^{3} \sqrt {a + b x^{3}}}{105 b^{4}} + \frac {16 B a^{2} x^{3} \sqrt {a + b x^{3}}}{105 b^{3}} - \frac {4 B a x^{6} \sqrt {a + b x^{3}}}{35 b^{2}} + \frac {2 B x^{9} \sqrt {a + b x^{3}}}{21 b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{9}}{9} + \frac {B x^{12}}{12}}{\sqrt {a}} & \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)**(1/2),x)

[Out]

Piecewise((16*A*a**2*sqrt(a + b*x**3)/(45*b**3) - 8*A*a*x**3*sqrt(a + b*x**3)/(45*b**2) + 2*A*x**6*sqrt(a + b*
x**3)/(15*b) - 32*B*a**3*sqrt(a + b*x**3)/(105*b**4) + 16*B*a**2*x**3*sqrt(a + b*x**3)/(105*b**3) - 4*B*a*x**6
*sqrt(a + b*x**3)/(35*b**2) + 2*B*x**9*sqrt(a + b*x**3)/(21*b), Ne(b, 0)), ((A*x**9/9 + B*x**12/12)/sqrt(a), T
rue))

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