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

Optimal. Leaf size=100 \[ \frac{a^2 \sqrt{a+b x^2} (A b-a B)}{b^4}+\frac{\left (a+b x^2\right )^{5/2} (A b-3 a B)}{5 b^4}-\frac{a \left (a+b x^2\right )^{3/2} (2 A b-3 a B)}{3 b^4}+\frac{B \left (a+b x^2\right )^{7/2}}{7 b^4} \]

[Out]

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

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Rubi [A]  time = 0.0760649, antiderivative size = 100, 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} \[ \frac{a^2 \sqrt{a+b x^2} (A b-a B)}{b^4}+\frac{\left (a+b x^2\right )^{5/2} (A b-3 a B)}{5 b^4}-\frac{a \left (a+b x^2\right )^{3/2} (2 A b-3 a B)}{3 b^4}+\frac{B \left (a+b x^2\right )^{7/2}}{7 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0547371, size = 78, normalized size = 0.78 \[ \frac{\sqrt{a+b x^2} \left (8 a^2 b \left (7 A+3 B x^2\right )-48 a^3 B-2 a b^2 x^2 \left (14 A+9 B x^2\right )+3 b^3 x^4 \left (7 A+5 B x^2\right )\right )}{105 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-48*a^3*B + 8*a^2*b*(7*A + 3*B*x^2) + 3*b^3*x^4*(7*A + 5*B*x^2) - 2*a*b^2*x^2*(14*A + 9*B*x^
2)))/(105*b^4)

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Maple [A]  time = 0.005, size = 77, normalized size = 0.8 \begin{align*}{\frac{15\,{x}^{6}B{b}^{3}+21\,A{b}^{3}{x}^{4}-18\,Ba{b}^{2}{x}^{4}-28\,Aa{b}^{2}{x}^{2}+24\,B{a}^{2}b{x}^{2}+56\,A{a}^{2}b-48\,B{a}^{3}}{105\,{b}^{4}}\sqrt{b{x}^{2}+a}} \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]

1/105*(b*x^2+a)^(1/2)*(15*B*b^3*x^6+21*A*b^3*x^4-18*B*a*b^2*x^4-28*A*a*b^2*x^2+24*B*a^2*b*x^2+56*A*a^2*b-48*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 = 1.59341, size = 173, normalized size = 1.73 \begin{align*} \frac{{\left (15 \, B b^{3} x^{6} - 3 \,{\left (6 \, B a b^{2} - 7 \, A b^{3}\right )} x^{4} - 48 \, B a^{3} + 56 \, A a^{2} b + 4 \,{\left (6 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{105 \, 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]

1/105*(15*B*b^3*x^6 - 3*(6*B*a*b^2 - 7*A*b^3)*x^4 - 48*B*a^3 + 56*A*a^2*b + 4*(6*B*a^2*b - 7*A*a*b^2)*x^2)*sqr
t(b*x^2 + a)/b^4

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Sympy [A]  time = 1.37767, size = 172, normalized size = 1.72 \begin{align*} \begin{cases} \frac{8 A a^{2} \sqrt{a + b x^{2}}}{15 b^{3}} - \frac{4 A a x^{2} \sqrt{a + b x^{2}}}{15 b^{2}} + \frac{A x^{4} \sqrt{a + b x^{2}}}{5 b} - \frac{16 B a^{3} \sqrt{a + b x^{2}}}{35 b^{4}} + \frac{8 B a^{2} x^{2} \sqrt{a + b x^{2}}}{35 b^{3}} - \frac{6 B a x^{4} \sqrt{a + b x^{2}}}{35 b^{2}} + \frac{B x^{6} \sqrt{a + b x^{2}}}{7 b} & \text{for}\: b \neq 0 \\\frac{\frac{A x^{6}}{6} + \frac{B x^{8}}{8}}{\sqrt{a}} & \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**2*sqrt(a + b*x**2)/(15*b**3) - 4*A*a*x**2*sqrt(a + b*x**2)/(15*b**2) + A*x**4*sqrt(a + b*x**
2)/(5*b) - 16*B*a**3*sqrt(a + b*x**2)/(35*b**4) + 8*B*a**2*x**2*sqrt(a + b*x**2)/(35*b**3) - 6*B*a*x**4*sqrt(a
+ b*x**2)/(35*b**2) + B*x**6*sqrt(a + b*x**2)/(7*b), Ne(b, 0)), ((A*x**6/6 + B*x**8/8)/sqrt(a), True))

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Giac [A]  time = 1.10432, size = 140, normalized size = 1.4 \begin{align*} \frac{15 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} B - 63 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} B a + 105 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a^{2} - 105 \, \sqrt{b x^{2} + a} B a^{3} + 21 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} A b - 70 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A a b + 105 \, \sqrt{b x^{2} + a} A a^{2} b}{105 \, 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="giac")

[Out]

1/105*(15*(b*x^2 + a)^(7/2)*B - 63*(b*x^2 + a)^(5/2)*B*a + 105*(b*x^2 + a)^(3/2)*B*a^2 - 105*sqrt(b*x^2 + a)*B
*a^3 + 21*(b*x^2 + a)^(5/2)*A*b - 70*(b*x^2 + a)^(3/2)*A*a*b + 105*sqrt(b*x^2 + a)*A*a^2*b)/b^4