3.5.86 \(\int x^5 \sqrt {a+b x^2} (A+B x^2) \, dx\)

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 103, 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 {a^2 \left (a+b x^2\right )^{3/2} (A b-a B)}{3 b^4}+\frac {\left (a+b x^2\right )^{7/2} (A b-3 a B)}{7 b^4}-\frac {a \left (a+b x^2\right )^{5/2} (2 A b-3 a B)}{5 b^4}+\frac {B \left (a+b x^2\right )^{9/2}}{9 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 75, normalized size = 0.73 \begin {gather*} \frac {\left (a+b x^2\right )^{3/2} \left (-16 a^3 B+24 a^2 b \left (A+B x^2\right )-6 a b^2 x^2 \left (6 A+5 B x^2\right )+5 b^3 x^4 \left (9 A+7 B x^2\right )\right )}{315 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(3/2)*(-16*a^3*B + 24*a^2*b*(A + B*x^2) - 6*a*b^2*x^2*(6*A + 5*B*x^2) + 5*b^3*x^4*(9*A + 7*B*x^2)
))/(315*b^4)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.05, size = 80, normalized size = 0.78 \begin {gather*} \frac {\left (a+b x^2\right )^{3/2} \left (-16 a^3 B+24 a^2 A b+24 a^2 b B x^2-36 a A b^2 x^2-30 a b^2 B x^4+45 A b^3 x^4+35 b^3 B x^6\right )}{315 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5*Sqrt[a + b*x^2]*(A + B*x^2),x]

[Out]

((a + b*x^2)^(3/2)*(24*a^2*A*b - 16*a^3*B - 36*a*A*b^2*x^2 + 24*a^2*b*B*x^2 + 45*A*b^3*x^4 - 30*a*b^2*B*x^4 +
35*b^3*B*x^6))/(315*b^4)

________________________________________________________________________________________

fricas [A]  time = 1.37, size = 99, normalized size = 0.96 \begin {gather*} \frac {{\left (35 \, B b^{4} x^{8} + 5 \, {\left (B a b^{3} + 9 \, A b^{4}\right )} x^{6} - 16 \, B a^{4} + 24 \, A a^{3} b - 3 \, {\left (2 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + 4 \, {\left (2 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{315 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)*(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

1/315*(35*B*b^4*x^8 + 5*(B*a*b^3 + 9*A*b^4)*x^6 - 16*B*a^4 + 24*A*a^3*b - 3*(2*B*a^2*b^2 - 3*A*a*b^3)*x^4 + 4*
(2*B*a^3*b - 3*A*a^2*b^2)*x^2)*sqrt(b*x^2 + a)/b^4

________________________________________________________________________________________

giac [A]  time = 0.31, size = 104, normalized size = 1.01 \begin {gather*} \frac {35 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B - 135 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a + 189 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} - 105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} + 45 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b - 126 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a b + 105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} b}{315 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)*(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/315*(35*(b*x^2 + a)^(9/2)*B - 135*(b*x^2 + a)^(7/2)*B*a + 189*(b*x^2 + a)^(5/2)*B*a^2 - 105*(b*x^2 + a)^(3/2
)*B*a^3 + 45*(b*x^2 + a)^(7/2)*A*b - 126*(b*x^2 + a)^(5/2)*A*a*b + 105*(b*x^2 + a)^(3/2)*A*a^2*b)/b^4

________________________________________________________________________________________

maple [A]  time = 0.01, size = 77, normalized size = 0.75 \begin {gather*} \frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (35 B \,x^{6} b^{3}+45 A \,b^{3} x^{4}-30 B a \,b^{2} x^{4}-36 A a \,b^{2} x^{2}+24 B \,a^{2} b \,x^{2}+24 A \,a^{2} b -16 B \,a^{3}\right )}{315 b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^2+A)*(b*x^2+a)^(1/2),x)

[Out]

1/315*(b*x^2+a)^(3/2)*(35*B*b^3*x^6+45*A*b^3*x^4-30*B*a*b^2*x^4-36*A*a*b^2*x^2+24*B*a^2*b*x^2+24*A*a^2*b-16*B*
a^3)/b^4

________________________________________________________________________________________

maxima [A]  time = 0.97, size = 132, normalized size = 1.28 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{6}}{9 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x^{4}}{21 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x^{4}}{7 \, b} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x^{2}}{105 \, b^{3}} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x^{2}}{35 \, b^{2}} - \frac {16 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3}}{315 \, b^{4}} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2}}{105 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)*(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

1/9*(b*x^2 + a)^(3/2)*B*x^6/b - 2/21*(b*x^2 + a)^(3/2)*B*a*x^4/b^2 + 1/7*(b*x^2 + a)^(3/2)*A*x^4/b + 8/105*(b*
x^2 + a)^(3/2)*B*a^2*x^2/b^3 - 4/35*(b*x^2 + a)^(3/2)*A*a*x^2/b^2 - 16/315*(b*x^2 + a)^(3/2)*B*a^3/b^4 + 8/105
*(b*x^2 + a)^(3/2)*A*a^2/b^3

________________________________________________________________________________________

mupad [B]  time = 0.64, size = 96, normalized size = 0.93 \begin {gather*} \sqrt {b\,x^2+a}\,\left (\frac {B\,x^8}{9}-\frac {16\,B\,a^4-24\,A\,a^3\,b}{315\,b^4}+\frac {x^6\,\left (45\,A\,b^4+5\,B\,a\,b^3\right )}{315\,b^4}-\frac {4\,a^2\,x^2\,\left (3\,A\,b-2\,B\,a\right )}{315\,b^3}+\frac {a\,x^4\,\left (3\,A\,b-2\,B\,a\right )}{105\,b^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(A + B*x^2)*(a + b*x^2)^(1/2),x)

[Out]

(a + b*x^2)^(1/2)*((B*x^8)/9 - (16*B*a^4 - 24*A*a^3*b)/(315*b^4) + (x^6*(45*A*b^4 + 5*B*a*b^3))/(315*b^4) - (4
*a^2*x^2*(3*A*b - 2*B*a))/(315*b^3) + (a*x^4*(3*A*b - 2*B*a))/(105*b^2))

________________________________________________________________________________________

sympy [A]  time = 2.22, size = 212, normalized size = 2.06 \begin {gather*} \begin {cases} \frac {8 A a^{3} \sqrt {a + b x^{2}}}{105 b^{3}} - \frac {4 A a^{2} x^{2} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {A a x^{4} \sqrt {a + b x^{2}}}{35 b} + \frac {A x^{6} \sqrt {a + b x^{2}}}{7} - \frac {16 B a^{4} \sqrt {a + b x^{2}}}{315 b^{4}} + \frac {8 B a^{3} x^{2} \sqrt {a + b x^{2}}}{315 b^{3}} - \frac {2 B a^{2} x^{4} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {B a x^{6} \sqrt {a + b x^{2}}}{63 b} + \frac {B x^{8} \sqrt {a + b x^{2}}}{9} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{6}}{6} + \frac {B x^{8}}{8}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**2+A)*(b*x**2+a)**(1/2),x)

[Out]

Piecewise((8*A*a**3*sqrt(a + b*x**2)/(105*b**3) - 4*A*a**2*x**2*sqrt(a + b*x**2)/(105*b**2) + A*a*x**4*sqrt(a
+ b*x**2)/(35*b) + A*x**6*sqrt(a + b*x**2)/7 - 16*B*a**4*sqrt(a + b*x**2)/(315*b**4) + 8*B*a**3*x**2*sqrt(a +
b*x**2)/(315*b**3) - 2*B*a**2*x**4*sqrt(a + b*x**2)/(105*b**2) + B*a*x**6*sqrt(a + b*x**2)/(63*b) + B*x**8*sqr
t(a + b*x**2)/9, Ne(b, 0)), (sqrt(a)*(A*x**6/6 + B*x**8/8), True))

________________________________________________________________________________________