3.180 \(\int x^5 \sqrt{a+b x^3} (A+B x^3) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0598291, 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} \[ \frac{2 \left (a+b x^3\right )^{5/2} (A b-2 a B)}{15 b^3}-\frac{2 a \left (a+b x^3\right )^{3/2} (A b-a B)}{9 b^3}+\frac{2 B \left (a+b x^3\right )^{7/2}}{21 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0387175, size = 57, normalized size = 0.78 \[ \frac{2 \left (a+b x^3\right )^{3/2} \left (8 a^2 B-2 a b \left (7 A+6 B x^3\right )+3 b^2 x^3 \left (7 A+5 B x^3\right )\right )}{315 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x^3)^(3/2)*(8*a^2*B + 3*b^2*x^3*(7*A + 5*B*x^3) - 2*a*b*(7*A + 6*B*x^3)))/(315*b^3)

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Maple [A]  time = 0.005, size = 53, normalized size = 0.7 \begin{align*} -{\frac{-30\,{b}^{2}B{x}^{6}-42\,A{x}^{3}{b}^{2}+24\,B{x}^{3}ab+28\,abA-16\,{a}^{2}B}{315\,{b}^{3}} \left ( b{x}^{3}+a \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^3+A)*(b*x^3+a)^(1/2),x)

[Out]

-2/315*(b*x^3+a)^(3/2)*(-15*B*b^2*x^6-21*A*b^2*x^3+12*B*a*b*x^3+14*A*a*b-8*B*a^2)/b^3

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Maxima [A]  time = 0.937005, size = 113, normalized size = 1.55 \begin{align*} \frac{2}{315} \, B{\left (\frac{15 \,{\left (b x^{3} + a\right )}^{\frac{7}{2}}}{b^{3}} - \frac{42 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} a}{b^{3}} + \frac{35 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a^{2}}{b^{3}}\right )} + \frac{2}{45} \, A{\left (\frac{3 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}}}{b^{2}} - \frac{5 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a}{b^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)*(b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

2/315*B*(15*(b*x^3 + a)^(7/2)/b^3 - 42*(b*x^3 + a)^(5/2)*a/b^3 + 35*(b*x^3 + a)^(3/2)*a^2/b^3) + 2/45*A*(3*(b*
x^3 + a)^(5/2)/b^2 - 5*(b*x^3 + a)^(3/2)*a/b^2)

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Fricas [A]  time = 1.69199, size = 166, normalized size = 2.27 \begin{align*} \frac{2 \,{\left (15 \, B b^{3} x^{9} + 3 \,{\left (B a b^{2} + 7 \, A b^{3}\right )} x^{6} + 8 \, B a^{3} - 14 \, A a^{2} b -{\left (4 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \sqrt{b x^{3} + a}}{315 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)*(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

2/315*(15*B*b^3*x^9 + 3*(B*a*b^2 + 7*A*b^3)*x^6 + 8*B*a^3 - 14*A*a^2*b - (4*B*a^2*b - 7*A*a*b^2)*x^3)*sqrt(b*x
^3 + a)/b^3

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Sympy [A]  time = 1.61222, size = 168, normalized size = 2.3 \begin{align*} \begin{cases} - \frac{4 A a^{2} \sqrt{a + b x^{3}}}{45 b^{2}} + \frac{2 A a x^{3} \sqrt{a + b x^{3}}}{45 b} + \frac{2 A x^{6} \sqrt{a + b x^{3}}}{15} + \frac{16 B a^{3} \sqrt{a + b x^{3}}}{315 b^{3}} - \frac{8 B a^{2} x^{3} \sqrt{a + b x^{3}}}{315 b^{2}} + \frac{2 B a x^{6} \sqrt{a + b x^{3}}}{105 b} + \frac{2 B x^{9} \sqrt{a + b x^{3}}}{21} & \text{for}\: b \neq 0 \\\sqrt{a} \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)*(b*x**3+a)**(1/2),x)

[Out]

Piecewise((-4*A*a**2*sqrt(a + b*x**3)/(45*b**2) + 2*A*a*x**3*sqrt(a + b*x**3)/(45*b) + 2*A*x**6*sqrt(a + b*x**
3)/15 + 16*B*a**3*sqrt(a + b*x**3)/(315*b**3) - 8*B*a**2*x**3*sqrt(a + b*x**3)/(315*b**2) + 2*B*a*x**6*sqrt(a
+ b*x**3)/(105*b) + 2*B*x**9*sqrt(a + b*x**3)/21, Ne(b, 0)), (sqrt(a)*(A*x**6/6 + B*x**9/9), True))

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Giac [A]  time = 1.10447, size = 107, normalized size = 1.47 \begin{align*} \frac{2 \,{\left (\frac{7 \,{\left (3 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a\right )} A}{b} + \frac{{\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}{b^{2}}\right )}}{315 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)*(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

2/315*(7*(3*(b*x^3 + a)^(5/2) - 5*(b*x^3 + a)^(3/2)*a)*A/b + (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/b^2)/b