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

Optimal. Leaf size=73 \[ \frac{\left (a+b x^2\right )^{9/2} (A b-2 a B)}{9 b^3}-\frac{a \left (a+b x^2\right )^{7/2} (A b-a B)}{7 b^3}+\frac{B \left (a+b x^2\right )^{11/2}}{11 b^3} \]

[Out]

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

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Rubi [A]  time = 0.0556794, 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} \[ \frac{\left (a+b x^2\right )^{9/2} (A b-2 a B)}{9 b^3}-\frac{a \left (a+b x^2\right )^{7/2} (A b-a B)}{7 b^3}+\frac{B \left (a+b x^2\right )^{11/2}}{11 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0424254, size = 57, normalized size = 0.78 \[ \frac{\left (a+b x^2\right )^{7/2} \left (8 a^2 B-2 a b \left (11 A+14 B x^2\right )+7 b^2 x^2 \left (11 A+9 B x^2\right )\right )}{693 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(7/2)*(8*a^2*B + 7*b^2*x^2*(11*A + 9*B*x^2) - 2*a*b*(11*A + 14*B*x^2)))/(693*b^3)

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Maple [A]  time = 0.005, size = 53, normalized size = 0.7 \begin{align*} -{\frac{-63\,{b}^{2}B{x}^{4}-77\,A{b}^{2}{x}^{2}+28\,Bab{x}^{2}+22\,abA-8\,{a}^{2}B}{693\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/693*(b*x^2+a)^(7/2)*(-63*B*b^2*x^4-77*A*b^2*x^2+28*B*a*b*x^2+22*A*a*b-8*B*a^2)/b^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61244, size = 273, normalized size = 3.74 \begin{align*} \frac{{\left (63 \, B b^{5} x^{10} + 7 \,{\left (23 \, B a b^{4} + 11 \, A b^{5}\right )} x^{8} +{\left (113 \, B a^{2} b^{3} + 209 \, A a b^{4}\right )} x^{6} + 8 \, B a^{5} - 22 \, A a^{4} b + 3 \,{\left (B a^{3} b^{2} + 55 \, A a^{2} b^{3}\right )} x^{4} -{\left (4 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{693 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/693*(63*B*b^5*x^10 + 7*(23*B*a*b^4 + 11*A*b^5)*x^8 + (113*B*a^2*b^3 + 209*A*a*b^4)*x^6 + 8*B*a^5 - 22*A*a^4*
b + 3*(B*a^3*b^2 + 55*A*a^2*b^3)*x^4 - (4*B*a^4*b - 11*A*a^3*b^2)*x^2)*sqrt(b*x^2 + a)/b^3

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Sympy [A]  time = 5.9656, size = 260, normalized size = 3.56 \begin{align*} \begin{cases} - \frac{2 A a^{4} \sqrt{a + b x^{2}}}{63 b^{2}} + \frac{A a^{3} x^{2} \sqrt{a + b x^{2}}}{63 b} + \frac{5 A a^{2} x^{4} \sqrt{a + b x^{2}}}{21} + \frac{19 A a b x^{6} \sqrt{a + b x^{2}}}{63} + \frac{A b^{2} x^{8} \sqrt{a + b x^{2}}}{9} + \frac{8 B a^{5} \sqrt{a + b x^{2}}}{693 b^{3}} - \frac{4 B a^{4} x^{2} \sqrt{a + b x^{2}}}{693 b^{2}} + \frac{B a^{3} x^{4} \sqrt{a + b x^{2}}}{231 b} + \frac{113 B a^{2} x^{6} \sqrt{a + b x^{2}}}{693} + \frac{23 B a b x^{8} \sqrt{a + b x^{2}}}{99} + \frac{B b^{2} x^{10} \sqrt{a + b x^{2}}}{11} & \text{for}\: b \neq 0 \\a^{\frac{5}{2}} \left (\frac{A x^{4}}{4} + \frac{B x^{6}}{6}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*A*a**4*sqrt(a + b*x**2)/(63*b**2) + A*a**3*x**2*sqrt(a + b*x**2)/(63*b) + 5*A*a**2*x**4*sqrt(a +
b*x**2)/21 + 19*A*a*b*x**6*sqrt(a + b*x**2)/63 + A*b**2*x**8*sqrt(a + b*x**2)/9 + 8*B*a**5*sqrt(a + b*x**2)/(
693*b**3) - 4*B*a**4*x**2*sqrt(a + b*x**2)/(693*b**2) + B*a**3*x**4*sqrt(a + b*x**2)/(231*b) + 113*B*a**2*x**6
*sqrt(a + b*x**2)/693 + 23*B*a*b*x**8*sqrt(a + b*x**2)/99 + B*b**2*x**10*sqrt(a + b*x**2)/11, Ne(b, 0)), (a**(
5/2)*(A*x**4/4 + B*x**6/6), True))

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Giac [B]  time = 1.13096, size = 431, normalized size = 5.9 \begin{align*} \frac{\frac{231 \,{\left (3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a\right )} A a^{2}}{b} + \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 )} B a^{2}}{b^{2}} + \frac{66 \,{\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} + \frac{22 \,{\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^{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 )} A}{b} + \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^{2}}}{3465 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(231*(3*(b*x^2 + a)^(5/2) - 5*(b*x^2 + a)^(3/2)*a)*A*a^2/b + 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)*B*a^2/b^2 + 66*(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 + 22*(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^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)*A/b + (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^2)/b