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

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

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Rubi [A]  time = 0.06, antiderivative size = 73, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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]))))

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]]

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*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.78 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.04, size = 56, normalized size = 0.77 \begin {gather*} \frac {\left (a+b x^2\right )^{5/2} \left (8 a^2 B-18 a A b-20 a b B x^2+45 A b^2 x^2+35 b^2 B x^4\right )}{315 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(5/2)*(-18*a*A*b + 8*a^2*B + 45*A*b^2*x^2 - 20*a*b*B*x^2 + 35*b^2*B*x^4))/(315*b^3)

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fricas [A]  time = 0.84, size = 99, normalized size = 1.36 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.28, size = 73, normalized size = 1.00 \begin {gather*} \frac {35 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B - 90 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a + 63 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} + 45 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b - 63 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a b}{315 \, b^{3}} \end {gather*}

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*(35*(b*x^2 + a)^(9/2)*B - 90*(b*x^2 + a)^(7/2)*B*a + 63*(b*x^2 + a)^(5/2)*B*a^2 + 45*(b*x^2 + a)^(7/2)*A
*b - 63*(b*x^2 + a)^(5/2)*A*a*b)/b^3

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maple [A]  time = 0.01, size = 53, normalized size = 0.73 \begin {gather*} -\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (-35 B \,b^{2} x^{4}-45 A \,b^{2} x^{2}+20 B a b \,x^{2}+18 a b A -8 a^{2} B \right )}{315 b^{3}} \end {gather*}

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 [A]  time = 1.00, size = 90, normalized size = 1.23 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x^{4}}{9 \, b} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a x^{2}}{63 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A x^{2}}{7 \, b} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2}}{315 \, b^{3}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a}{35 \, b^{2}} \end {gather*}

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]

1/9*(b*x^2 + a)^(5/2)*B*x^4/b - 4/63*(b*x^2 + a)^(5/2)*B*a*x^2/b^2 + 1/7*(b*x^2 + a)^(5/2)*A*x^2/b + 8/315*(b*
x^2 + a)^(5/2)*B*a^2/b^3 - 2/35*(b*x^2 + a)^(5/2)*A*a/b^2

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mupad [B]  time = 0.63, size = 96, normalized size = 1.32 \begin {gather*} \sqrt {b\,x^2+a}\,\left (\frac {8\,B\,a^4-18\,A\,a^3\,b}{315\,b^3}+\frac {x^6\,\left (45\,A\,b^4+50\,B\,a\,b^3\right )}{315\,b^3}+\frac {B\,b\,x^8}{9}+\frac {a^2\,x^2\,\left (9\,A\,b-4\,B\,a\right )}{315\,b^2}+\frac {a\,x^4\,\left (24\,A\,b+B\,a\right )}{105\,b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a + b*x^2)^(1/2)*((8*B*a^4 - 18*A*a^3*b)/(315*b^3) + (x^6*(45*A*b^4 + 50*B*a*b^3))/(315*b^3) + (B*b*x^8)/9 +
(a^2*x^2*(9*A*b - 4*B*a))/(315*b^2) + (a*x^4*(24*A*b + B*a))/(105*b))

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sympy [A]  time = 2.72, size = 209, normalized size = 2.86 \begin {gather*} \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 {gather*}

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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