\(\int \frac {x (A+B x^2)}{(a+b x^2)^3} \, dx\) [93]

Optimal result

Integrand size = 18, antiderivative size = 32 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {\left (A+B x^2\right )^2}{4 (A b-a B) \left (a+b x^2\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {455, 37} \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {\left (A+B x^2\right )^2}{4 \left (a+b x^2\right )^2 (A b-a B)} \]

[In]

Int[(x*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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)^3} \, dx,x,x^2\right ) \\ & = -\frac {\left (A+B x^2\right )^2}{4 (A b-a B) \left (a+b x^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {A b+B \left (a+2 b x^2\right )}{4 b^2 \left (a+b x^2\right )^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91

method	result	size
gosper	\(-\frac {2 b B \,x^{2}+A b +B a}{4 \left (b \,x^{2}+a \right )^{2} b^{2}}\)	\(29\)
parallelrisch	\(-\frac {2 b B \,x^{2}+A b +B a}{4 \left (b \,x^{2}+a \right )^{2} b^{2}}\)	\(29\)
norman	\(\frac {-\frac {B \,x^{2}}{2 b}-\frac {A b +B a}{4 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}\)	\(33\)
risch	\(\frac {-\frac {B \,x^{2}}{2 b}-\frac {A b +B a}{4 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}\)	\(33\)
default	\(-\frac {A b -B a}{4 b^{2} \left (b \,x^{2}+a \right )^{2}}-\frac {B}{2 b^{2} \left (b \,x^{2}+a \right )}\)	\(39\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {2 \, B b x^{2} + B a + A b}{4 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {- A b - B a - 2 B b x^{2}}{4 a^{2} b^{2} + 8 a b^{3} x^{2} + 4 b^{4} x^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {2 \, B b x^{2} + B a + A b}{4 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {2 \, B b x^{2} + B a + A b}{4 \, {\left (b x^{2} + a\right )}^{2} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {\frac {A\,b+B\,a}{4\,b^2}+\frac {B\,x^2}{2\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4} \]

[In]

int((x*(A + B*x^2))/(a + b*x^2)^3,x)

[Out]

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