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

Optimal. Leaf size=103 \[ \frac {2 a^2 \left (a+b x^3\right )^{5/2} (A b-a B)}{15 b^4}+\frac {2 \left (a+b x^3\right )^{9/2} (A b-3 a B)}{27 b^4}-\frac {2 a \left (a+b x^3\right )^{7/2} (2 A b-3 a B)}{21 b^4}+\frac {2 B \left (a+b x^3\right )^{11/2}}{33 b^4} \]

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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 \left (a+b x^3\right )^{5/2} (A b-a B)}{15 b^4}+\frac {2 \left (a+b x^3\right )^{9/2} (A b-3 a B)}{27 b^4}-\frac {2 a \left (a+b x^3\right )^{7/2} (2 A b-3 a B)}{21 b^4}+\frac {2 B \left (a+b x^3\right )^{11/2}}{33 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(A*b - a*B)*(a + b*x^3)^(5/2))/(15*b^4) - (2*a*(2*A*b - 3*a*B)*(a + b*x^3)^(7/2))/(21*b^4) + (2*(A*b -
3*a*B)*(a + b*x^3)^(9/2))/(27*b^4) + (2*B*(a + b*x^3)^(11/2))/(33*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 x^8 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int x^2 (a+b x)^{3/2} (A+B x) \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (-\frac {a^2 (-A b+a B) (a+b x)^{3/2}}{b^3}+\frac {a (-2 A b+3 a B) (a+b x)^{5/2}}{b^3}+\frac {(A b-3 a B) (a+b x)^{7/2}}{b^3}+\frac {B (a+b x)^{9/2}}{b^3}\right ) \, dx,x,x^3\right )\\ &=\frac {2 a^2 (A b-a B) \left (a+b x^3\right )^{5/2}}{15 b^4}-\frac {2 a (2 A b-3 a B) \left (a+b x^3\right )^{7/2}}{21 b^4}+\frac {2 (A b-3 a B) \left (a+b x^3\right )^{9/2}}{27 b^4}+\frac {2 B \left (a+b x^3\right )^{11/2}}{33 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 78, normalized size = 0.76 \begin {gather*} \frac {2 \left (a+b x^3\right )^{5/2} \left (-48 a^3 B+8 a^2 b \left (11 A+15 B x^3\right )-10 a b^2 x^3 \left (22 A+21 B x^3\right )+35 b^3 x^6 \left (11 A+9 B x^3\right )\right )}{10395 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x^3)^(5/2)*(-48*a^3*B + 35*b^3*x^6*(11*A + 9*B*x^3) + 8*a^2*b*(11*A + 15*B*x^3) - 10*a*b^2*x^3*(22*A
+ 21*B*x^3)))/(10395*b^4)

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IntegrateAlgebraic [A]  time = 0.06, size = 80, normalized size = 0.78 \begin {gather*} -\frac {2 \left (a+b x^3\right )^{5/2} \left (48 a^3 B-88 a^2 A b-120 a^2 b B x^3+220 a A b^2 x^3+210 a b^2 B x^6-385 A b^3 x^6-315 b^3 B x^9\right )}{10395 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*x^3)^(5/2)*(-88*a^2*A*b + 48*a^3*B + 220*a*A*b^2*x^3 - 120*a^2*b*B*x^3 - 385*A*b^3*x^6 + 210*a*b^2*
B*x^6 - 315*b^3*B*x^9))/(10395*b^4)

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fricas [A]  time = 1.11, size = 124, normalized size = 1.20 \begin {gather*} \frac {2 \, {\left (315 \, B b^{5} x^{15} + 35 \, {\left (12 \, B a b^{4} + 11 \, A b^{5}\right )} x^{12} + 5 \, {\left (3 \, B a^{2} b^{3} + 110 \, A a b^{4}\right )} x^{9} - 3 \, {\left (6 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{6} - 48 \, B a^{5} + 88 \, A a^{4} b + 4 \, {\left (6 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{10395 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/10395*(315*B*b^5*x^15 + 35*(12*B*a*b^4 + 11*A*b^5)*x^12 + 5*(3*B*a^2*b^3 + 110*A*a*b^4)*x^9 - 3*(6*B*a^3*b^2
- 11*A*a^2*b^3)*x^6 - 48*B*a^5 + 88*A*a^4*b + 4*(6*B*a^4*b - 11*A*a^3*b^2)*x^3)*sqrt(b*x^3 + a)/b^4

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giac [A]  time = 0.16, size = 104, normalized size = 1.01 \begin {gather*} \frac {2 \, {\left (315 \, {\left (b x^{3} + a\right )}^{\frac {11}{2}} B - 1155 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}} B a + 1485 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} B a^{2} - 693 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B a^{3} + 385 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}} A b - 990 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} A a b + 693 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} A a^{2} b\right )}}{10395 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/10395*(315*(b*x^3 + a)^(11/2)*B - 1155*(b*x^3 + a)^(9/2)*B*a + 1485*(b*x^3 + a)^(7/2)*B*a^2 - 693*(b*x^3 + a
)^(5/2)*B*a^3 + 385*(b*x^3 + a)^(9/2)*A*b - 990*(b*x^3 + a)^(7/2)*A*a*b + 693*(b*x^3 + a)^(5/2)*A*a^2*b)/b^4

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maple [A]  time = 0.05, size = 77, normalized size = 0.75 \begin {gather*} \frac {2 \left (b \,x^{3}+a \right )^{\frac {5}{2}} \left (315 B \,x^{9} b^{3}+385 A \,b^{3} x^{6}-210 B a \,b^{2} x^{6}-220 A a \,b^{2} x^{3}+120 B \,a^{2} b \,x^{3}+88 A \,a^{2} b -48 B \,a^{3}\right )}{10395 b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/10395*(b*x^3+a)^(5/2)*(315*B*b^3*x^9+385*A*b^3*x^6-210*B*a*b^2*x^6-220*A*a*b^2*x^3+120*B*a^2*b*x^3+88*A*a^2*
b-48*B*a^3)/b^4

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maxima [A]  time = 0.61, size = 118, normalized size = 1.15 \begin {gather*} \frac {2}{945} \, {\left (\frac {35 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}}}{b^{3}} - \frac {90 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} a}{b^{3}} + \frac {63 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{2}}{b^{3}}\right )} A + \frac {2}{3465} \, {\left (\frac {105 \, {\left (b x^{3} + a\right )}^{\frac {11}{2}}}{b^{4}} - \frac {385 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}} a}{b^{4}} + \frac {495 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} a^{2}}{b^{4}} - \frac {231 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{3}}{b^{4}}\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/945*(35*(b*x^3 + a)^(9/2)/b^3 - 90*(b*x^3 + a)^(7/2)*a/b^3 + 63*(b*x^3 + a)^(5/2)*a^2/b^3)*A + 2/3465*(105*(
b*x^3 + a)^(11/2)/b^4 - 385*(b*x^3 + a)^(9/2)*a/b^4 + 495*(b*x^3 + a)^(7/2)*a^2/b^4 - 231*(b*x^3 + a)^(5/2)*a^
3/b^4)*B

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mupad [B]  time = 2.65, size = 206, normalized size = 2.00 \begin {gather*} \frac {20\,A\,a\,x^9\,\sqrt {b\,x^3+a}}{189}+\frac {2\,A\,b\,x^{12}\,\sqrt {b\,x^3+a}}{27}+\frac {8\,B\,a\,x^{12}\,\sqrt {b\,x^3+a}}{99}+\frac {2\,B\,b\,x^{15}\,\sqrt {b\,x^3+a}}{33}+\frac {16\,A\,a^4\,\sqrt {b\,x^3+a}}{945\,b^3}-\frac {32\,B\,a^5\,\sqrt {b\,x^3+a}}{3465\,b^4}-\frac {8\,A\,a^3\,x^3\,\sqrt {b\,x^3+a}}{945\,b^2}+\frac {2\,A\,a^2\,x^6\,\sqrt {b\,x^3+a}}{315\,b}+\frac {16\,B\,a^4\,x^3\,\sqrt {b\,x^3+a}}{3465\,b^3}-\frac {4\,B\,a^3\,x^6\,\sqrt {b\,x^3+a}}{1155\,b^2}+\frac {2\,B\,a^2\,x^9\,\sqrt {b\,x^3+a}}{693\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(20*A*a*x^9*(a + b*x^3)^(1/2))/189 + (2*A*b*x^12*(a + b*x^3)^(1/2))/27 + (8*B*a*x^12*(a + b*x^3)^(1/2))/99 + (
2*B*b*x^15*(a + b*x^3)^(1/2))/33 + (16*A*a^4*(a + b*x^3)^(1/2))/(945*b^3) - (32*B*a^5*(a + b*x^3)^(1/2))/(3465
*b^4) - (8*A*a^3*x^3*(a + b*x^3)^(1/2))/(945*b^2) + (2*A*a^2*x^6*(a + b*x^3)^(1/2))/(315*b) + (16*B*a^4*x^3*(a
+ b*x^3)^(1/2))/(3465*b^3) - (4*B*a^3*x^6*(a + b*x^3)^(1/2))/(1155*b^2) + (2*B*a^2*x^9*(a + b*x^3)^(1/2))/(69
3*b)

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sympy [A]  time = 8.44, size = 267, normalized size = 2.59 \begin {gather*} \begin {cases} \frac {16 A a^{4} \sqrt {a + b x^{3}}}{945 b^{3}} - \frac {8 A a^{3} x^{3} \sqrt {a + b x^{3}}}{945 b^{2}} + \frac {2 A a^{2} x^{6} \sqrt {a + b x^{3}}}{315 b} + \frac {20 A a x^{9} \sqrt {a + b x^{3}}}{189} + \frac {2 A b x^{12} \sqrt {a + b x^{3}}}{27} - \frac {32 B a^{5} \sqrt {a + b x^{3}}}{3465 b^{4}} + \frac {16 B a^{4} x^{3} \sqrt {a + b x^{3}}}{3465 b^{3}} - \frac {4 B a^{3} x^{6} \sqrt {a + b x^{3}}}{1155 b^{2}} + \frac {2 B a^{2} x^{9} \sqrt {a + b x^{3}}}{693 b} + \frac {8 B a x^{12} \sqrt {a + b x^{3}}}{99} + \frac {2 B b x^{15} \sqrt {a + b x^{3}}}{33} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{9}}{9} + \frac {B x^{12}}{12}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((16*A*a**4*sqrt(a + b*x**3)/(945*b**3) - 8*A*a**3*x**3*sqrt(a + b*x**3)/(945*b**2) + 2*A*a**2*x**6*s
qrt(a + b*x**3)/(315*b) + 20*A*a*x**9*sqrt(a + b*x**3)/189 + 2*A*b*x**12*sqrt(a + b*x**3)/27 - 32*B*a**5*sqrt(
a + b*x**3)/(3465*b**4) + 16*B*a**4*x**3*sqrt(a + b*x**3)/(3465*b**3) - 4*B*a**3*x**6*sqrt(a + b*x**3)/(1155*b
**2) + 2*B*a**2*x**9*sqrt(a + b*x**3)/(693*b) + 8*B*a*x**12*sqrt(a + b*x**3)/99 + 2*B*b*x**15*sqrt(a + b*x**3)
/33, Ne(b, 0)), (a**(3/2)*(A*x**9/9 + B*x**12/12), True))

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