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

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

[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)

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Rubi [A]  time = 0.0873286, antiderivative size = 103, 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} \[ \frac{a^2 \left (a+b x^2\right )^{3/2} (A b-a B)}{3 b^4}+\frac{\left (a+b x^2\right )^{7/2} (A b-3 a B)}{7 b^4}-\frac{a \left (a+b x^2\right )^{5/2} (2 A b-3 a B)}{5 b^4}+\frac{B \left (a+b x^2\right )^{9/2}}{9 b^4} \]

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 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^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 = 0.055869, size = 75, normalized size = 0.73 \[ \frac{\left (a+b x^2\right )^{3/2} \left (24 a^2 b \left (A+B x^2\right )-16 a^3 B-6 a b^2 x^2 \left (6 A+5 B x^2\right )+5 b^3 x^4 \left (9 A+7 B x^2\right )\right )}{315 b^4} \]

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)

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Maple [A]  time = 0.006, size = 77, normalized size = 0.8 \begin{align*}{\frac{35\,B{x}^{6}{b}^{3}+45\,A{b}^{3}{x}^{4}-30\,Ba{b}^{2}{x}^{4}-36\,Aa{b}^{2}{x}^{2}+24\,B{a}^{2}b{x}^{2}+24\,A{a}^{2}b-16\,B{a}^{3}}{315\,{b}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}} \end{align*}

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.28993, size = 219, normalized size = 2.13 \begin{align*} \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{align*}

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

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Sympy [A]  time = 1.44617, size = 212, normalized size = 2.06 \begin{align*} \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{align*}

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))

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Giac [A]  time = 1.11333, size = 144, normalized size = 1.4 \begin{align*} \frac{\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^{2}} + \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^{3}}}{315 \, b} \end{align*}

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*(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^2 + (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^3)/b