\(\int \frac {x (A+B x^2)}{(a+b x^2)^{5/2}} \, dx\) [590]

Optimal result

Integrand size = 20, antiderivative size = 44 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-A b+a B}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {B}{b^2 \sqrt {a+b x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {455, 45} \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {A b-a B}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {B}{b^2 \sqrt {a+b x^2}} \]

[In]

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

[Out]

-1/3*(A*b - a*B)/(b^2*(a + b*x^2)^(3/2)) - B/(b^2*Sqrt[a + b*x^2])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-A b-2 a B-3 b B x^2}{3 b^2 \left (a+b x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.68

method	result	size
gosper	\(-\frac {3 b B \,x^{2}+A b +2 B a}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{2}}\)	\(30\)
trager	\(-\frac {3 b B \,x^{2}+A b +2 B a}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{2}}\)	\(30\)
pseudoelliptic	\(-\frac {\left (3 x^{2} B +A \right ) b +2 B a}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{2}}\)	\(30\)
default	\(B \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )-\frac {A}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.18 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {{\left (3 \, B b x^{2} + 2 \, B a + A b\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (37) = 74\).

Time = 0.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 3.25 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\begin {cases} - \frac {A b}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} - \frac {2 B a}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} - \frac {3 B b x^{2}}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{2}}{2} + \frac {B x^{4}}{4}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.14 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {B x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {2 \, B a}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} - \frac {A}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {3 \, {\left (b x^{2} + a\right )} B - B a + A b}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.54 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {A\,b-B\,a+3\,B\,\left (b\,x^2+a\right )}{3\,b^2\,{\left (b\,x^2+a\right )}^{3/2}} \]

[In]

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

[Out]

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