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

Optimal. Leaf size=97 \[ -\frac {a^2 (A b-a B)}{3 b^4 \left (a+b x^2\right )^{3/2}}+\frac {a (2 A b-3 a B)}{b^4 \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2} (A b-3 a B)}{b^4}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b^4} \]

[Out]

-1/3*a^2*(A*b-B*a)/b^4/(b*x^2+a)^(3/2)+1/3*B*(b*x^2+a)^(3/2)/b^4+a*(2*A*b-3*B*a)/b^4/(b*x^2+a)^(1/2)+(A*b-3*B*
a)*(b*x^2+a)^(1/2)/b^4

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Rubi [A]  time = 0.08, antiderivative size = 97, 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} \[ -\frac {a^2 (A b-a B)}{3 b^4 \left (a+b x^2\right )^{3/2}}+\frac {a (2 A b-3 a B)}{b^4 \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2} (A b-3 a B)}{b^4}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.05, size = 73, normalized size = 0.75 \[ \frac {-16 a^3 B+8 a^2 b \left (A-3 B x^2\right )-6 a b^2 x^2 \left (B x^2-2 A\right )+b^3 x^4 \left (3 A+B x^2\right )}{3 b^4 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*a^3*B + 8*a^2*b*(A - 3*B*x^2) - 6*a*b^2*x^2*(-2*A + B*x^2) + b^3*x^4*(3*A + B*x^2))/(3*b^4*(a + b*x^2)^(3
/2))

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fricas [A]  time = 0.68, size = 98, normalized size = 1.01 \[ \frac {{\left (B b^{3} x^{6} - 3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} x^{4} - 16 \, B a^{3} + 8 \, A a^{2} b - 12 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(B*b^3*x^6 - 3*(2*B*a*b^2 - A*b^3)*x^4 - 16*B*a^3 + 8*A*a^2*b - 12*(2*B*a^2*b - A*a*b^2)*x^2)*sqrt(b*x^2 +
a)/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4)

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giac [A]  time = 0.45, size = 104, normalized size = 1.07 \[ -\frac {9 \, {\left (b x^{2} + a\right )} B a^{2} - B a^{3} - 6 \, {\left (b x^{2} + a\right )} A a b + A a^{2} b}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{8} - 9 \, \sqrt {b x^{2} + a} B a b^{8} + 3 \, \sqrt {b x^{2} + a} A b^{9}}{3 \, b^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(9*(b*x^2 + a)*B*a^2 - B*a^3 - 6*(b*x^2 + a)*A*a*b + A*a^2*b)/((b*x^2 + a)^(3/2)*b^4) + 1/3*((b*x^2 + a)^
(3/2)*B*b^8 - 9*sqrt(b*x^2 + a)*B*a*b^8 + 3*sqrt(b*x^2 + a)*A*b^9)/b^12

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maple [A]  time = 0.01, size = 76, normalized size = 0.78 \[ \frac {B \,x^{6} b^{3}+3 A \,b^{3} x^{4}-6 B a \,b^{2} x^{4}+12 A a \,b^{2} x^{2}-24 B \,a^{2} b \,x^{2}+8 A \,a^{2} b -16 B \,a^{3}}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(B*b^3*x^6+3*A*b^3*x^4-6*B*a*b^2*x^4+12*A*a*b^2*x^2-24*B*a^2*b*x^2+8*A*a^2*b-16*B*a^3)/(b*x^2+a)^(3/2)/b^4

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maxima [A]  time = 1.16, size = 131, normalized size = 1.35 \[ \frac {B x^{6}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {2 \, B a x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {A x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {8 \, B a^{2} x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} + \frac {4 \, A a x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} - \frac {16 \, B a^{3}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}} + \frac {8 \, A a^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*x^6/((b*x^2 + a)^(3/2)*b) - 2*B*a*x^4/((b*x^2 + a)^(3/2)*b^2) + A*x^4/((b*x^2 + a)^(3/2)*b) - 8*B*a^2*x^
2/((b*x^2 + a)^(3/2)*b^3) + 4*A*a*x^2/((b*x^2 + a)^(3/2)*b^2) - 16/3*B*a^3/((b*x^2 + a)^(3/2)*b^4) + 8/3*A*a^2
/((b*x^2 + a)^(3/2)*b^3)

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mupad [B]  time = 0.86, size = 89, normalized size = 0.92 \[ \frac {B\,{\left (b\,x^2+a\right )}^3+B\,a^3+3\,A\,b\,{\left (b\,x^2+a\right )}^2-9\,B\,a\,{\left (b\,x^2+a\right )}^2-9\,B\,a^2\,\left (b\,x^2+a\right )-A\,a^2\,b+6\,A\,a\,b\,\left (b\,x^2+a\right )}{3\,b^4\,{\left (b\,x^2+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*(a + b*x^2)^3 + B*a^3 + 3*A*b*(a + b*x^2)^2 - 9*B*a*(a + b*x^2)^2 - 9*B*a^2*(a + b*x^2) - A*a^2*b + 6*A*a*b
*(a + b*x^2))/(3*b^4*(a + b*x^2)^(3/2))

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sympy [A]  time = 1.99, size = 337, normalized size = 3.47 \[ \begin {cases} \frac {8 A a^{2} b}{3 a b^{4} \sqrt {a + b x^{2}} + 3 b^{5} x^{2} \sqrt {a + b x^{2}}} + \frac {12 A a b^{2} x^{2}}{3 a b^{4} \sqrt {a + b x^{2}} + 3 b^{5} x^{2} \sqrt {a + b x^{2}}} + \frac {3 A b^{3} x^{4}}{3 a b^{4} \sqrt {a + b x^{2}} + 3 b^{5} x^{2} \sqrt {a + b x^{2}}} - \frac {16 B a^{3}}{3 a b^{4} \sqrt {a + b x^{2}} + 3 b^{5} x^{2} \sqrt {a + b x^{2}}} - \frac {24 B a^{2} b x^{2}}{3 a b^{4} \sqrt {a + b x^{2}} + 3 b^{5} x^{2} \sqrt {a + b x^{2}}} - \frac {6 B a b^{2} x^{4}}{3 a b^{4} \sqrt {a + b x^{2}} + 3 b^{5} x^{2} \sqrt {a + b x^{2}}} + \frac {B b^{3} x^{6}}{3 a b^{4} \sqrt {a + b x^{2}} + 3 b^{5} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{6}}{6} + \frac {B x^{8}}{8}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*A*a**2*b/(3*a*b**4*sqrt(a + b*x**2) + 3*b**5*x**2*sqrt(a + b*x**2)) + 12*A*a*b**2*x**2/(3*a*b**4*
sqrt(a + b*x**2) + 3*b**5*x**2*sqrt(a + b*x**2)) + 3*A*b**3*x**4/(3*a*b**4*sqrt(a + b*x**2) + 3*b**5*x**2*sqrt
(a + b*x**2)) - 16*B*a**3/(3*a*b**4*sqrt(a + b*x**2) + 3*b**5*x**2*sqrt(a + b*x**2)) - 24*B*a**2*b*x**2/(3*a*b
**4*sqrt(a + b*x**2) + 3*b**5*x**2*sqrt(a + b*x**2)) - 6*B*a*b**2*x**4/(3*a*b**4*sqrt(a + b*x**2) + 3*b**5*x**
2*sqrt(a + b*x**2)) + B*b**3*x**6/(3*a*b**4*sqrt(a + b*x**2) + 3*b**5*x**2*sqrt(a + b*x**2)), Ne(b, 0)), ((A*x
**6/6 + B*x**8/8)/a**(5/2), True))

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