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

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

[Out]

-1/3*a*(A*b-B*a)*(b*x^2+a)^(3/2)/b^3+1/5*(A*b-2*B*a)*(b*x^2+a)^(5/2)/b^3+1/7*B*(b*x^2+a)^(7/2)/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} \[ \frac {\left (a+b x^2\right )^{5/2} (A b-2 a B)}{5 b^3}-\frac {a \left (a+b x^2\right )^{3/2} (A b-a B)}{3 b^3}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.78 \[ \frac {\left (a+b x^2\right )^{3/2} \left (8 a^2 B-2 a b \left (7 A+6 B x^2\right )+3 b^2 x^2 \left (7 A+5 B x^2\right )\right )}{105 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(3/2)*(8*a^2*B + 3*b^2*x^2*(7*A + 5*B*x^2) - 2*a*b*(7*A + 6*B*x^2)))/(105*b^3)

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fricas [A]  time = 0.62, size = 75, normalized size = 1.03 \[ \frac {{\left (15 \, B b^{3} x^{6} + 3 \, {\left (B a b^{2} + 7 \, A b^{3}\right )} x^{4} + 8 \, B a^{3} - 14 \, A a^{2} b - {\left (4 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(15*B*b^3*x^6 + 3*(B*a*b^2 + 7*A*b^3)*x^4 + 8*B*a^3 - 14*A*a^2*b - (4*B*a^2*b - 7*A*a*b^2)*x^2)*sqrt(b*x
^2 + a)/b^3

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giac [A]  time = 0.30, size = 73, normalized size = 1.00 \[ \frac {15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B - 42 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a + 35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} + 21 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b - 35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a b}{105 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(15*(b*x^2 + a)^(7/2)*B - 42*(b*x^2 + a)^(5/2)*B*a + 35*(b*x^2 + a)^(3/2)*B*a^2 + 21*(b*x^2 + a)^(5/2)*A
*b - 35*(b*x^2 + a)^(3/2)*A*a*b)/b^3

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maple [A]  time = 0.01, size = 53, normalized size = 0.73 \[ -\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (-15 B \,b^{2} x^{4}-21 A \,b^{2} x^{2}+12 B a b \,x^{2}+14 a b A -8 a^{2} B \right )}{105 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(b*x^2+a)^(3/2)*(-15*B*b^2*x^4-21*A*b^2*x^2+12*B*a*b*x^2+14*A*a*b-8*B*a^2)/b^3

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maxima [A]  time = 1.09, size = 90, normalized size = 1.23 \[ \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{4}}{7 \, b} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x^{2}}{35 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x^{2}}{5 \, b} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2}}{105 \, b^{3}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a}{15 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*(b*x^2 + a)^(3/2)*B*x^4/b - 4/35*(b*x^2 + a)^(3/2)*B*a*x^2/b^2 + 1/5*(b*x^2 + a)^(3/2)*A*x^2/b + 8/105*(b*
x^2 + a)^(3/2)*B*a^2/b^3 - 2/15*(b*x^2 + a)^(3/2)*A*a/b^2

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mupad [B]  time = 0.56, size = 76, normalized size = 1.04 \[ \sqrt {b\,x^2+a}\,\left (\frac {B\,x^6}{7}+\frac {8\,B\,a^3-14\,A\,a^2\,b}{105\,b^3}+\frac {x^4\,\left (21\,A\,b^3+3\,B\,a\,b^2\right )}{105\,b^3}+\frac {a\,x^2\,\left (7\,A\,b-4\,B\,a\right )}{105\,b^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a + b*x^2)^(1/2)*((B*x^6)/7 + (8*B*a^3 - 14*A*a^2*b)/(105*b^3) + (x^4*(21*A*b^3 + 3*B*a*b^2))/(105*b^3) + (a*
x^2*(7*A*b - 4*B*a))/(105*b^2))

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sympy [A]  time = 0.95, size = 162, normalized size = 2.22 \[ \begin {cases} - \frac {2 A a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {A a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {A x^{4} \sqrt {a + b x^{2}}}{5} + \frac {8 B a^{3} \sqrt {a + b x^{2}}}{105 b^{3}} - \frac {4 B a^{2} x^{2} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {B a x^{4} \sqrt {a + b x^{2}}}{35 b} + \frac {B x^{6} \sqrt {a + b x^{2}}}{7} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{4}}{4} + \frac {B x^{6}}{6}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x**2+A)*(b*x**2+a)**(1/2),x)

[Out]

Piecewise((-2*A*a**2*sqrt(a + b*x**2)/(15*b**2) + A*a*x**2*sqrt(a + b*x**2)/(15*b) + A*x**4*sqrt(a + b*x**2)/5
+ 8*B*a**3*sqrt(a + b*x**2)/(105*b**3) - 4*B*a**2*x**2*sqrt(a + b*x**2)/(105*b**2) + B*a*x**4*sqrt(a + b*x**2
)/(35*b) + B*x**6*sqrt(a + b*x**2)/7, Ne(b, 0)), (sqrt(a)*(A*x**4/4 + B*x**6/6), True))

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