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

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

[Out]

(a^2*(A*b - a*B)*(a + b*x^2)^(5/2))/(5*b^4) - (a*(2*A*b - 3*a*B)*(a + b*x^2)^(7/2))/(7*b^4) + ((A*b - 3*a*B)*(
a + b*x^2)^(9/2))/(9*b^4) + (B*(a + b*x^2)^(11/2))/(11*b^4)

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Rubi [A]  time = 0.0773654, antiderivative size = 103, 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} \[ \frac{a^2 \left (a+b x^2\right )^{5/2} (A b-a B)}{5 b^4}+\frac{\left (a+b x^2\right )^{9/2} (A b-3 a B)}{9 b^4}-\frac{a \left (a+b x^2\right )^{7/2} (2 A b-3 a B)}{7 b^4}+\frac{B \left (a+b x^2\right )^{11/2}}{11 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(A*b - a*B)*(a + b*x^2)^(5/2))/(5*b^4) - (a*(2*A*b - 3*a*B)*(a + b*x^2)^(7/2))/(7*b^4) + ((A*b - 3*a*B)*(
a + b*x^2)^(9/2))/(9*b^4) + (B*(a + b*x^2)^(11/2))/(11*b^4)

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

Mathematica [A]  time = 0.058488, size = 78, normalized size = 0.76 \[ \frac{\left (a+b x^2\right )^{5/2} \left (8 a^2 b \left (11 A+15 B x^2\right )-48 a^3 B-10 a b^2 x^2 \left (22 A+21 B x^2\right )+35 b^3 x^4 \left (11 A+9 B x^2\right )\right )}{3465 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(5/2)*(-48*a^3*B + 35*b^3*x^4*(11*A + 9*B*x^2) + 8*a^2*b*(11*A + 15*B*x^2) - 10*a*b^2*x^2*(22*A +
21*B*x^2)))/(3465*b^4)

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Maple [A]  time = 0.006, size = 77, normalized size = 0.8 \begin{align*}{\frac{315\,B{x}^{6}{b}^{3}+385\,A{b}^{3}{x}^{4}-210\,Ba{b}^{2}{x}^{4}-220\,Aa{b}^{2}{x}^{2}+120\,B{a}^{2}b{x}^{2}+88\,A{a}^{2}b-48\,B{a}^{3}}{3465\,{b}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(b*x^2+a)^(5/2)*(315*B*b^3*x^6+385*A*b^3*x^4-210*B*a*b^2*x^4-220*A*a*b^2*x^2+120*B*a^2*b*x^2+88*A*a^2*b
-48*B*a^3)/b^4

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61489, size = 284, normalized size = 2.76 \begin{align*} \frac{{\left (315 \, B b^{5} x^{10} + 35 \,{\left (12 \, B a b^{4} + 11 \, A b^{5}\right )} x^{8} + 5 \,{\left (3 \, B a^{2} b^{3} + 110 \, A a b^{4}\right )} x^{6} - 48 \, B a^{5} + 88 \, A a^{4} b - 3 \,{\left (6 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{4} + 4 \,{\left (6 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{3465 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(315*B*b^5*x^10 + 35*(12*B*a*b^4 + 11*A*b^5)*x^8 + 5*(3*B*a^2*b^3 + 110*A*a*b^4)*x^6 - 48*B*a^5 + 88*A*
a^4*b - 3*(6*B*a^3*b^2 - 11*A*a^2*b^3)*x^4 + 4*(6*B*a^4*b - 11*A*a^3*b^2)*x^2)*sqrt(b*x^2 + a)/b^4

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Sympy [A]  time = 3.99679, size = 260, normalized size = 2.52 \begin{align*} \begin{cases} \frac{8 A a^{4} \sqrt{a + b x^{2}}}{315 b^{3}} - \frac{4 A a^{3} x^{2} \sqrt{a + b x^{2}}}{315 b^{2}} + \frac{A a^{2} x^{4} \sqrt{a + b x^{2}}}{105 b} + \frac{10 A a x^{6} \sqrt{a + b x^{2}}}{63} + \frac{A b x^{8} \sqrt{a + b x^{2}}}{9} - \frac{16 B a^{5} \sqrt{a + b x^{2}}}{1155 b^{4}} + \frac{8 B a^{4} x^{2} \sqrt{a + b x^{2}}}{1155 b^{3}} - \frac{2 B a^{3} x^{4} \sqrt{a + b x^{2}}}{385 b^{2}} + \frac{B a^{2} x^{6} \sqrt{a + b x^{2}}}{231 b} + \frac{4 B a x^{8} \sqrt{a + b x^{2}}}{33} + \frac{B b x^{10} \sqrt{a + b x^{2}}}{11} & \text{for}\: b \neq 0 \\a^{\frac{3}{2}} \left (\frac{A x^{6}}{6} + \frac{B x^{8}}{8}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*A*a**4*sqrt(a + b*x**2)/(315*b**3) - 4*A*a**3*x**2*sqrt(a + b*x**2)/(315*b**2) + A*a**2*x**4*sqrt
(a + b*x**2)/(105*b) + 10*A*a*x**6*sqrt(a + b*x**2)/63 + A*b*x**8*sqrt(a + b*x**2)/9 - 16*B*a**5*sqrt(a + b*x*
*2)/(1155*b**4) + 8*B*a**4*x**2*sqrt(a + b*x**2)/(1155*b**3) - 2*B*a**3*x**4*sqrt(a + b*x**2)/(385*b**2) + B*a
**2*x**6*sqrt(a + b*x**2)/(231*b) + 4*B*a*x**8*sqrt(a + b*x**2)/33 + B*b*x**10*sqrt(a + b*x**2)/11, Ne(b, 0)),
(a**(3/2)*(A*x**6/6 + B*x**8/8), True))

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Giac [B]  time = 1.1141, size = 323, normalized size = 3.14 \begin{align*} \frac{\frac{33 \,{\left (15 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{2}\right )} A a}{b^{2}} + \frac{11 \,{\left (35 \,{\left (b x^{2} + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3}\right )} B a}{b^{3}} + \frac{11 \,{\left (35 \,{\left (b x^{2} + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3}\right )} A}{b^{2}} + \frac{{\left (315 \,{\left (b x^{2} + a\right )}^{\frac{11}{2}} - 1540 \,{\left (b x^{2} + a\right )}^{\frac{9}{2}} a + 2970 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} a^{2} - 2772 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a^{3} + 1155 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{4}\right )} B}{b^{3}}}{3465 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(33*(15*(b*x^2 + a)^(7/2) - 42*(b*x^2 + a)^(5/2)*a + 35*(b*x^2 + a)^(3/2)*a^2)*A*a/b^2 + 11*(35*(b*x^2
+ a)^(9/2) - 135*(b*x^2 + a)^(7/2)*a + 189*(b*x^2 + a)^(5/2)*a^2 - 105*(b*x^2 + a)^(3/2)*a^3)*B*a/b^3 + 11*(35
*(b*x^2 + a)^(9/2) - 135*(b*x^2 + a)^(7/2)*a + 189*(b*x^2 + a)^(5/2)*a^2 - 105*(b*x^2 + a)^(3/2)*a^3)*A/b^2 +
(315*(b*x^2 + a)^(11/2) - 1540*(b*x^2 + a)^(9/2)*a + 2970*(b*x^2 + a)^(7/2)*a^2 - 2772*(b*x^2 + a)^(5/2)*a^3 +
1155*(b*x^2 + a)^(3/2)*a^4)*B/b^3)/b