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

Optimal. Leaf size=73 \[ \frac{2 \left (a+b x^3\right )^{7/2} (A b-2 a B)}{21 b^3}-\frac{2 a \left (a+b x^3\right )^{5/2} (A b-a B)}{15 b^3}+\frac{2 B \left (a+b x^3\right )^{9/2}}{27 b^3} \]

[Out]

(-2*a*(A*b - a*B)*(a + b*x^3)^(5/2))/(15*b^3) + (2*(A*b - 2*a*B)*(a + b*x^3)^(7/2))/(21*b^3) + (2*B*(a + b*x^3
)^(9/2))/(27*b^3)

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Rubi [A]  time = 0.0575149, antiderivative size = 73, normalized size of antiderivative = 1., 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} \[ \frac{2 \left (a+b x^3\right )^{7/2} (A b-2 a B)}{21 b^3}-\frac{2 a \left (a+b x^3\right )^{5/2} (A b-a B)}{15 b^3}+\frac{2 B \left (a+b x^3\right )^{9/2}}{27 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(A*b - a*B)*(a + b*x^3)^(5/2))/(15*b^3) + (2*(A*b - 2*a*B)*(a + b*x^3)^(7/2))/(21*b^3) + (2*B*(a + b*x^3
)^(9/2))/(27*b^3)

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]]

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]))))

Rubi steps

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

Mathematica [A]  time = 0.0377418, size = 57, normalized size = 0.78 \[ \frac{2 \left (a+b x^3\right )^{5/2} \left (8 a^2 B-2 a b \left (9 A+10 B x^3\right )+5 b^2 x^3 \left (9 A+7 B x^3\right )\right )}{945 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x^3)^(5/2)*(8*a^2*B + 5*b^2*x^3*(9*A + 7*B*x^3) - 2*a*b*(9*A + 10*B*x^3)))/(945*b^3)

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Maple [A]  time = 0.006, size = 53, normalized size = 0.7 \begin{align*} -{\frac{-70\,{b}^{2}B{x}^{6}-90\,A{x}^{3}{b}^{2}+40\,B{x}^{3}ab+36\,abA-16\,{a}^{2}B}{945\,{b}^{3}} \left ( b{x}^{3}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/945*(b*x^3+a)^(5/2)*(-35*B*b^2*x^6-45*A*b^2*x^3+20*B*a*b*x^3+18*A*a*b-8*B*a^2)/b^3

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Maxima [A]  time = 0.927493, size = 113, normalized size = 1.55 \begin{align*} \frac{2}{105} \,{\left (\frac{5 \,{\left (b x^{3} + a\right )}^{\frac{7}{2}}}{b^{2}} - \frac{7 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} a}{b^{2}}\right )} A + \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 )} B \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(5*(b*x^3 + a)^(7/2)/b^2 - 7*(b*x^3 + a)^(5/2)*a/b^2)*A + 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)*B

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Fricas [A]  time = 1.71325, size = 219, normalized size = 3. \begin{align*} \frac{2 \,{\left (35 \, B b^{4} x^{12} + 5 \,{\left (10 \, B a b^{3} + 9 \, A b^{4}\right )} x^{9} + 3 \,{\left (B a^{2} b^{2} + 24 \, A a b^{3}\right )} x^{6} + 8 \, B a^{4} - 18 \, A a^{3} b -{\left (4 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt{b x^{3} + a}}{945 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/945*(35*B*b^4*x^12 + 5*(10*B*a*b^3 + 9*A*b^4)*x^9 + 3*(B*a^2*b^2 + 24*A*a*b^3)*x^6 + 8*B*a^4 - 18*A*a^3*b -
(4*B*a^3*b - 9*A*a^2*b^2)*x^3)*sqrt(b*x^3 + a)/b^3

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Sympy [A]  time = 5.31557, size = 216, normalized size = 2.96 \begin{align*} \begin{cases} - \frac{4 A a^{3} \sqrt{a + b x^{3}}}{105 b^{2}} + \frac{2 A a^{2} x^{3} \sqrt{a + b x^{3}}}{105 b} + \frac{16 A a x^{6} \sqrt{a + b x^{3}}}{105} + \frac{2 A b x^{9} \sqrt{a + b x^{3}}}{21} + \frac{16 B a^{4} \sqrt{a + b x^{3}}}{945 b^{3}} - \frac{8 B a^{3} x^{3} \sqrt{a + b x^{3}}}{945 b^{2}} + \frac{2 B a^{2} x^{6} \sqrt{a + b x^{3}}}{315 b} + \frac{20 B a x^{9} \sqrt{a + b x^{3}}}{189} + \frac{2 B b x^{12} \sqrt{a + b x^{3}}}{27} & \text{for}\: b \neq 0 \\a^{\frac{3}{2}} \left (\frac{A x^{6}}{6} + \frac{B x^{9}}{9}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-4*A*a**3*sqrt(a + b*x**3)/(105*b**2) + 2*A*a**2*x**3*sqrt(a + b*x**3)/(105*b) + 16*A*a*x**6*sqrt(a
+ b*x**3)/105 + 2*A*b*x**9*sqrt(a + b*x**3)/21 + 16*B*a**4*sqrt(a + b*x**3)/(945*b**3) - 8*B*a**3*x**3*sqrt(a
+ b*x**3)/(945*b**2) + 2*B*a**2*x**6*sqrt(a + b*x**3)/(315*b) + 20*B*a*x**9*sqrt(a + b*x**3)/189 + 2*B*b*x**1
2*sqrt(a + b*x**3)/27, Ne(b, 0)), (a**(3/2)*(A*x**6/6 + B*x**9/9), True))

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Giac [B]  time = 1.20729, size = 247, normalized size = 3.38 \begin{align*} \frac{2 \,{\left (\frac{21 \,{\left (3 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a\right )} A a}{b} + \frac{3 \,{\left (15 \,{\left (b x^{3} + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a^{2}\right )} B a}{b^{2}} + \frac{3 \,{\left (15 \,{\left (b x^{3} + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a^{2}\right )} A}{b} + \frac{{\left (35 \,{\left (b x^{3} + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x^{3} + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a^{3}\right )} B}{b^{2}}\right )}}{945 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/945*(21*(3*(b*x^3 + a)^(5/2) - 5*(b*x^3 + a)^(3/2)*a)*A*a/b + 3*(15*(b*x^3 + a)^(7/2) - 42*(b*x^3 + a)^(5/2)
*a + 35*(b*x^3 + a)^(3/2)*a^2)*B*a/b^2 + 3*(15*(b*x^3 + a)^(7/2) - 42*(b*x^3 + a)^(5/2)*a + 35*(b*x^3 + a)^(3/
2)*a^2)*A/b + (35*(b*x^3 + a)^(9/2) - 135*(b*x^3 + a)^(7/2)*a + 189*(b*x^3 + a)^(5/2)*a^2 - 105*(b*x^3 + a)^(3
/2)*a^3)*B/b^2)/b