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

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

[Out]

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

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Rubi [A]  time = 0.0592361, 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} \[ \frac{\left (a+b x^2\right )^{7/2} (A b-2 a B)}{7 b^3}-\frac{a \left (a+b x^2\right )^{5/2} (A b-a B)}{5 b^3}+\frac{B \left (a+b x^2\right )^{9/2}}{9 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0393702, size = 57, normalized size = 0.78 \[ \frac{\left (a+b x^2\right )^{5/2} \left (8 a^2 B-2 a b \left (9 A+10 B x^2\right )+5 b^2 x^2 \left (9 A+7 B x^2\right )\right )}{315 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(5/2)*(8*a^2*B + 5*b^2*x^2*(9*A + 7*B*x^2) - 2*a*b*(9*A + 10*B*x^2)))/(315*b^3)

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Maple [A]  time = 0.005, size = 53, normalized size = 0.7 \begin{align*} -{\frac{-35\,{b}^{2}B{x}^{4}-45\,A{b}^{2}{x}^{2}+20\,Bab{x}^{2}+18\,abA-8\,{a}^{2}B}{315\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/315*(b*x^2+a)^(5/2)*(-35*B*b^2*x^4-45*A*b^2*x^2+20*B*a*b*x^2+18*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)^(3/2)*(B*x^2+A),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62572, size = 217, normalized size = 2.97 \begin{align*} \frac{{\left (35 \, B b^{4} x^{8} + 5 \,{\left (10 \, B a b^{3} + 9 \, A b^{4}\right )} x^{6} + 8 \, B a^{4} - 18 \, A a^{3} b + 3 \,{\left (B a^{2} b^{2} + 24 \, A a b^{3}\right )} x^{4} -{\left (4 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{315 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(35*B*b^4*x^8 + 5*(10*B*a*b^3 + 9*A*b^4)*x^6 + 8*B*a^4 - 18*A*a^3*b + 3*(B*a^2*b^2 + 24*A*a*b^3)*x^4 - (
4*B*a^3*b - 9*A*a^2*b^2)*x^2)*sqrt(b*x^2 + a)/b^3

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Sympy [A]  time = 2.32155, size = 209, normalized size = 2.86 \begin{align*} \begin{cases} - \frac{2 A a^{3} \sqrt{a + b x^{2}}}{35 b^{2}} + \frac{A a^{2} x^{2} \sqrt{a + b x^{2}}}{35 b} + \frac{8 A a x^{4} \sqrt{a + b x^{2}}}{35} + \frac{A b x^{6} \sqrt{a + b x^{2}}}{7} + \frac{8 B a^{4} \sqrt{a + b x^{2}}}{315 b^{3}} - \frac{4 B a^{3} x^{2} \sqrt{a + b x^{2}}}{315 b^{2}} + \frac{B a^{2} x^{4} \sqrt{a + b x^{2}}}{105 b} + \frac{10 B a x^{6} \sqrt{a + b x^{2}}}{63} + \frac{B b x^{8} \sqrt{a + b x^{2}}}{9} & \text{for}\: b \neq 0 \\a^{\frac{3}{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)**(3/2)*(B*x**2+A),x)

[Out]

Piecewise((-2*A*a**3*sqrt(a + b*x**2)/(35*b**2) + A*a**2*x**2*sqrt(a + b*x**2)/(35*b) + 8*A*a*x**4*sqrt(a + b*
x**2)/35 + A*b*x**6*sqrt(a + b*x**2)/7 + 8*B*a**4*sqrt(a + b*x**2)/(315*b**3) - 4*B*a**3*x**2*sqrt(a + b*x**2)
/(315*b**2) + B*a**2*x**4*sqrt(a + b*x**2)/(105*b) + 10*B*a*x**6*sqrt(a + b*x**2)/63 + B*b*x**8*sqrt(a + b*x**
2)/9, Ne(b, 0)), (a**(3/2)*(A*x**4/4 + B*x**6/6), True))

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Giac [B]  time = 1.12812, size = 247, normalized size = 3.38 \begin{align*} \frac{\frac{21 \,{\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}{b} + \frac{3 \,{\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}{b^{2}} + \frac{3 \,{\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}{b} + \frac{{\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}{b^{2}}}{315 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(21*(3*(b*x^2 + a)^(5/2) - 5*(b*x^2 + a)^(3/2)*a)*A*a/b + 3*(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/b^2 + 3*(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/b + (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/b^2)/b