\(\int \frac {x^5 (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\) [570]

Optimal result

Integrand size = 22, antiderivative size = 99 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {a^2 (A b-a B)}{b^4 \sqrt {a+b x^2}}-\frac {a (2 A b-3 a B) \sqrt {a+b x^2}}{b^4}+\frac {(A b-3 a B) \left (a+b x^2\right )^{3/2}}{3 b^4}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b^4} \]

[Out]

1/3*(A*b-3*B*a)*(b*x^2+a)^(3/2)/b^4+1/5*B*(b*x^2+a)^(5/2)/b^4-a^2*(A*b-B*a)/b^4/(b*x^2+a)^(1/2)-a*(2*A*b-3*B*a
)*(b*x^2+a)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 78} \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {a^2 (A b-a B)}{b^4 \sqrt {a+b x^2}}+\frac {\left (a+b x^2\right )^{3/2} (A b-3 a B)}{3 b^4}-\frac {a \sqrt {a+b x^2} (2 A b-3 a B)}{b^4}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b^4} \]

[In]

Int[(x^5*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

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

Rule 78

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 457

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {48 a^3 B-8 a^2 b \left (5 A-3 B x^2\right )+b^3 x^4 \left (5 A+3 B x^2\right )-2 a b^2 x^2 \left (10 A+3 B x^2\right )}{15 b^4 \sqrt {a+b x^2}} \]

[In]

Integrate[(x^5*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

(48*a^3*B - 8*a^2*b*(5*A - 3*B*x^2) + b^3*x^4*(5*A + 3*B*x^2) - 2*a*b^2*x^2*(10*A + 3*B*x^2))/(15*b^4*Sqrt[a +
b*x^2])

Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(-\frac {8 \left (-\frac {x^{4} \left (\frac {3 x^{2} B}{5}+A \right ) b^{3}}{8}+\frac {x^{2} \left (\frac {3 x^{2} B}{10}+A \right ) a \,b^{2}}{2}+a^{2} \left (-\frac {3 x^{2} B}{5}+A \right ) b -\frac {6 a^{3} B}{5}\right )}{3 \sqrt {b \,x^{2}+a}\, b^{4}}\)	\(68\)
gosper	\(-\frac {-3 b^{3} B \,x^{6}-5 A \,b^{3} x^{4}+6 B a \,b^{2} x^{4}+20 a A \,b^{2} x^{2}-24 B \,a^{2} b \,x^{2}+40 a^{2} b A -48 a^{3} B}{15 \sqrt {b \,x^{2}+a}\, b^{4}}\)	\(77\)
trager	\(-\frac {-3 b^{3} B \,x^{6}-5 A \,b^{3} x^{4}+6 B a \,b^{2} x^{4}+20 a A \,b^{2} x^{2}-24 B \,a^{2} b \,x^{2}+40 a^{2} b A -48 a^{3} B}{15 \sqrt {b \,x^{2}+a}\, b^{4}}\)	\(77\)
risch	\(-\frac {\left (-3 b^{2} B \,x^{4}-5 A \,b^{2} x^{2}+9 B a b \,x^{2}+25 a b A -33 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{15 b^{4}}-\frac {a^{2} \left (A b -B a \right )}{b^{4} \sqrt {b \,x^{2}+a}}\)	\(79\)
default	\(B \left (\frac {x^{6}}{5 b \sqrt {b \,x^{2}+a}}-\frac {6 a \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{\sqrt {b \,x^{2}+a}\, b}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )}{5 b}\right )+A \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{\sqrt {b \,x^{2}+a}\, b}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )\)	\(142\)

[In]

int(x^5*(B*x^2+A)/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-8/3/(b*x^2+a)^(1/2)*(-1/8*x^4*(3/5*x^2*B+A)*b^3+1/2*x^2*(3/10*x^2*B+A)*a*b^2+a^2*(-3/5*x^2*B+A)*b-6/5*a^3*B)/
b^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left (3 \, B b^{3} x^{6} - {\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} x^{4} + 48 \, B a^{3} - 40 \, A a^{2} b + 4 \, {\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, {\left (b^{5} x^{2} + a b^{4}\right )}} \]

[In]

integrate(x^5*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

1/15*(3*B*b^3*x^6 - (6*B*a*b^2 - 5*A*b^3)*x^4 + 48*B*a^3 - 40*A*a^2*b + 4*(6*B*a^2*b - 5*A*a*b^2)*x^2)*sqrt(b*
x^2 + a)/(b^5*x^2 + a*b^4)

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.74 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\begin {cases} - \frac {8 A a^{2}}{3 b^{3} \sqrt {a + b x^{2}}} - \frac {4 A a x^{2}}{3 b^{2} \sqrt {a + b x^{2}}} + \frac {A x^{4}}{3 b \sqrt {a + b x^{2}}} + \frac {16 B a^{3}}{5 b^{4} \sqrt {a + b x^{2}}} + \frac {8 B a^{2} x^{2}}{5 b^{3} \sqrt {a + b x^{2}}} - \frac {2 B a x^{4}}{5 b^{2} \sqrt {a + b x^{2}}} + \frac {B x^{6}}{5 b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{6}}{6} + \frac {B x^{8}}{8}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5*(B*x**2+A)/(b*x**2+a)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.33 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B x^{6}}{5 \, \sqrt {b x^{2} + a} b} - \frac {2 \, B a x^{4}}{5 \, \sqrt {b x^{2} + a} b^{2}} + \frac {A x^{4}}{3 \, \sqrt {b x^{2} + a} b} + \frac {8 \, B a^{2} x^{2}}{5 \, \sqrt {b x^{2} + a} b^{3}} - \frac {4 \, A a x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {16 \, B a^{3}}{5 \, \sqrt {b x^{2} + a} b^{4}} - \frac {8 \, A a^{2}}{3 \, \sqrt {b x^{2} + a} b^{3}} \]

[In]

integrate(x^5*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.14 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B a^{3} - A a^{2} b}{\sqrt {b x^{2} + a} b^{4}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{16} - 15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b^{16} + 45 \, \sqrt {b x^{2} + a} B a^{2} b^{16} + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{17} - 30 \, \sqrt {b x^{2} + a} A a b^{17}}{15 \, b^{20}} \]

[In]

integrate(x^5*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

(B*a^3 - A*a^2*b)/(sqrt(b*x^2 + a)*b^4) + 1/15*(3*(b*x^2 + a)^(5/2)*B*b^16 - 15*(b*x^2 + a)^(3/2)*B*a*b^16 + 4
5*sqrt(b*x^2 + a)*B*a^2*b^16 + 5*(b*x^2 + a)^(3/2)*A*b^17 - 30*sqrt(b*x^2 + a)*A*a*b^17)/b^20

Mupad [B] (verification not implemented)

Time = 5.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.90 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\frac {B\,{\left (b\,x^2+a\right )}^3}{5}+B\,a^3+\frac {A\,b\,{\left (b\,x^2+a\right )}^2}{3}-B\,a\,{\left (b\,x^2+a\right )}^2+3\,B\,a^2\,\left (b\,x^2+a\right )-A\,a^2\,b-2\,A\,a\,b\,\left (b\,x^2+a\right )}{b^4\,\sqrt {b\,x^2+a}} \]

[In]

int((x^5*(A + B*x^2))/(a + b*x^2)^(3/2),x)

[Out]

((B*(a + b*x^2)^3)/5 + B*a^3 + (A*b*(a + b*x^2)^2)/3 - B*a*(a + b*x^2)^2 + 3*B*a^2*(a + b*x^2) - A*a^2*b - 2*A
*a*b*(a + b*x^2))/(b^4*(a + b*x^2)^(1/2))